Sec 2 G1 Maths Tutor Bukit Timah

Did you know? When a formula says distance = speed × time, it also quietly tells you that time = distance ÷ speed—provided the speed is nonzero. Changing the subject of a formula is not magic. It is the disciplined process of isolating one quantity without changing the relationship between all the quantities. For families exploring a Secondary 2 G1 Mathematics tutor in Bukit Timah, it offers an accessible bridge from solving one-variable equations to later mathematical modelling.

A simple Secondary 2 G1 Maths foundation is 3x+5=20: subtract five, divide by three and check x=5. In a general formula y=3x+5, the same operations produce x=(y−5)/3. Now x is the subject, meaning its value is expressed in terms of the other symbols. This differs from evaluating a known formula with given numbers: we rearrange the relationship symbolically before inserting particular values.

The MOE 2020 Secondary Two G1 content explicitly includes linear equations in one variable, algebraic-expression simplification, rates and prism/cylinder mensuration. The full 2027 G1 SEC Mathematics K110 syllabus also names changing the subject of a simple formula. This is therefore a later-stage G1 bridge for learners ready for it, not a claim that every 2026 Secondary 2 school paper already assesses general formula rearrangement. School scope and prerequisites take precedence.

Quick answer: make the chosen letter the subject

Write the complete formula and underline the requested subject.
Undo addition or subtraction that is outside the variable’s current product or fraction.
Undo multiplication or division by the full nonzero factor attached to the variable.
Keep brackets around grouped numerators, negative values and fractions.
Check by substituting the rearranged expression into the original formula and verifying the identity or given numerical output.

One-minute diagnostic: two different kinds of answer

Ask the pupil to solve 2x+3=11, then to make x the subject of y=2x+3. The first gives a numerical answer x=4; the second gives the general expression x=(y−3)/2. Next ask what happens if the coefficient of x is zero: division by zero is not permitted, and a unique inverse may not exist. A pupil who can do the first task but not the second may need practice treating other letters as numbers, rather than another worksheet of fixed integer answers.

Subject means the quantity isolated on one side

In V=Ah, volume V is already the subject. If we need height h, divide by nonzero cross-sectional area A: h=V/A. The original and rearranged formulae encode one relationship, not two different laws. Check the units as well: volume cm³ divided by area cm² returns height cm. A formula rearrangement can be algebraically sound but physically inapplicable if the given quantities are missing, inconsistent or outside a meaningful domain.

247 original changing-the-subject clinics

The examples below include familiar G1 foundations—linear rules, rectangle area, average speed and unknown means—as well as later-stage symbolic manipulations and formula restrictions. The numbers are original examples, not past examination items. Choose the group relevant to the student’s current syllabus and follow the first-error check rather than assigning the entire collection as one homework task.

Isolate a variable in a linear formula

When making isolate a variable in a linear formula work, the most useful first question is ‘Which operation is applied to the variable last?’ Undo that operation on both complete sides before moving inward. A correct subject form should allow the original to be reconstructed. Watch for zero denominators, physical units and a distinction between a symbolic formula and a numerical answer.

1. Linear formula 1: make x the subject

Question. A general formula reads y=2x−13, with a nonzero coefficient 2. Rearrange the formula to make x the subject. Then check it when y=-31.

Why the inverse works. To isolate x, first remove the signed constant -13 from both entire sides, then divide by the nonzero coefficient 2. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=2x+(-13), subtract -13: y−(-13)=2x. Divide by 2: x=(y−(-13))/2. With y=-31, the numerator becomes -18, giving x=-9. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 2 before removing -13, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-9 into the original formula: 2×(-9)+(-13)=-31. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-29. Since the coefficient is 2, the new x should be -8. Explain the one-unit change using the original linear relationship.

2. Linear formula 2: make x the subject

Question. A general formula reads y=3x−6, with a nonzero coefficient 3. Rearrange the formula to make x the subject. Then check it when y=-18.

Why the inverse works. To isolate x, first remove the signed constant -6 from both entire sides, then divide by the nonzero coefficient 3. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=3x+(-6), subtract -6: y−(-6)=3x. Divide by 3: x=(y−(-6))/3. With y=-18, the numerator becomes -12, giving x=-4. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 3 before removing -6, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-4 into the original formula: 3×(-4)+(-6)=-18. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-15. Since the coefficient is 3, the new x should be -3. Explain the one-unit change using the original linear relationship.

3. Linear formula 3: make x the subject

Question. A general formula reads y=4x+1, with a nonzero coefficient 4. Rearrange the formula to make x the subject. Then check it when y=5.

Why the inverse works. To isolate x, first remove the signed constant 1 from both entire sides, then divide by the nonzero coefficient 4. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=4x+(1), subtract 1: y−(1)=4x. Divide by 4: x=(y−(1))/4. With y=5, the numerator becomes 4, giving x=1. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 4 before removing 1, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=1 into the original formula: 4×(1)+(1)=5. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=9. Since the coefficient is 4, the new x should be 2. Explain the one-unit change using the original linear relationship.

4. Linear formula 4: make x the subject

Question. A general formula reads y=5x+8, with a nonzero coefficient 5. Rearrange the formula to make x the subject. Then check it when y=38.

Why the inverse works. To isolate x, first remove the signed constant 8 from both entire sides, then divide by the nonzero coefficient 5. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=5x+(8), subtract 8: y−(8)=5x. Divide by 5: x=(y−(8))/5. With y=38, the numerator becomes 30, giving x=6. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 5 before removing 8, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=6 into the original formula: 5×(6)+(8)=38. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=43. Since the coefficient is 5, the new x should be 7. Explain the one-unit change using the original linear relationship.

5. Linear formula 5: make x the subject

Question. A general formula reads y=-2x−12, with a nonzero coefficient -2. Rearrange the formula to make x the subject. Then check it when y=4.

Why the inverse works. To isolate x, first remove the signed constant -12 from both entire sides, then divide by the nonzero coefficient -2. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-2x+(-12), subtract -12: y−(-12)=-2x. Divide by -2: x=(y−(-12))/-2. With y=4, the numerator becomes 16, giving x=-8. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -2 before removing -12, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-8 into the original formula: -2×(-8)+(-12)=4. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=2. Since the coefficient is -2, the new x should be -7. Explain the one-unit change using the original linear relationship.

6. Linear formula 6: make x the subject

Question. A general formula reads y=-3x−5, with a nonzero coefficient -3. Rearrange the formula to make x the subject. Then check it when y=4.

Why the inverse works. To isolate x, first remove the signed constant -5 from both entire sides, then divide by the nonzero coefficient -3. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-3x+(-5), subtract -5: y−(-5)=-3x. Divide by -3: x=(y−(-5))/-3. With y=4, the numerator becomes 9, giving x=-3. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -3 before removing -5, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-3 into the original formula: -3×(-3)+(-5)=4. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=1. Since the coefficient is -3, the new x should be -2. Explain the one-unit change using the original linear relationship.

7. Linear formula 7: make x the subject

Question. A general formula reads y=-4x+2, with a nonzero coefficient -4. Rearrange the formula to make x the subject. Then check it when y=-6.

Why the inverse works. To isolate x, first remove the signed constant 2 from both entire sides, then divide by the nonzero coefficient -4. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-4x+(2), subtract 2: y−(2)=-4x. Divide by -4: x=(y−(2))/-4. With y=-6, the numerator becomes -8, giving x=2. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -4 before removing 2, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=2 into the original formula: -4×(2)+(2)=-6. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-10. Since the coefficient is -4, the new x should be 3. Explain the one-unit change using the original linear relationship.

8. Linear formula 8: make x the subject

Question. A general formula reads y=-5x+9, with a nonzero coefficient -5. Rearrange the formula to make x the subject. Then check it when y=-26.

Why the inverse works. To isolate x, first remove the signed constant 9 from both entire sides, then divide by the nonzero coefficient -5. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-5x+(9), subtract 9: y−(9)=-5x. Divide by -5: x=(y−(9))/-5. With y=-26, the numerator becomes -35, giving x=7. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -5 before removing 9, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=7 into the original formula: -5×(7)+(9)=-26. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-31. Since the coefficient is -5, the new x should be 8. Explain the one-unit change using the original linear relationship.

9. Linear formula 9: make x the subject

Question. A general formula reads y=6x−11, with a nonzero coefficient 6. Rearrange the formula to make x the subject. Then check it when y=-53.

Why the inverse works. To isolate x, first remove the signed constant -11 from both entire sides, then divide by the nonzero coefficient 6. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=6x+(-11), subtract -11: y−(-11)=6x. Divide by 6: x=(y−(-11))/6. With y=-53, the numerator becomes -42, giving x=-7. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 6 before removing -11, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-7 into the original formula: 6×(-7)+(-11)=-53. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-47. Since the coefficient is 6, the new x should be -6. Explain the one-unit change using the original linear relationship.

10. Linear formula 10: make x the subject

Question. A general formula reads y=-6x−4, with a nonzero coefficient -6. Rearrange the formula to make x the subject. Then check it when y=8.

Why the inverse works. To isolate x, first remove the signed constant -4 from both entire sides, then divide by the nonzero coefficient -6. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-6x+(-4), subtract -4: y−(-4)=-6x. Divide by -6: x=(y−(-4))/-6. With y=8, the numerator becomes 12, giving x=-2. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -6 before removing -4, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-2 into the original formula: -6×(-2)+(-4)=8. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=2. Since the coefficient is -6, the new x should be -1. Explain the one-unit change using the original linear relationship.

11. Linear formula 11: make x the subject

Question. A general formula reads y=2x+3, with a nonzero coefficient 2. Rearrange the formula to make x the subject. Then check it when y=9.

Why the inverse works. To isolate x, first remove the signed constant 3 from both entire sides, then divide by the nonzero coefficient 2. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=2x+(3), subtract 3: y−(3)=2x. Divide by 2: x=(y−(3))/2. With y=9, the numerator becomes 6, giving x=3. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 2 before removing 3, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=3 into the original formula: 2×(3)+(3)=9. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=11. Since the coefficient is 2, the new x should be 4. Explain the one-unit change using the original linear relationship.

12. Linear formula 12: make x the subject

Question. A general formula reads y=3x+10, with a nonzero coefficient 3. Rearrange the formula to make x the subject. Then check it when y=34.

Why the inverse works. To isolate x, first remove the signed constant 10 from both entire sides, then divide by the nonzero coefficient 3. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=3x+(10), subtract 10: y−(10)=3x. Divide by 3: x=(y−(10))/3. With y=34, the numerator becomes 24, giving x=8. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 3 before removing 10, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=8 into the original formula: 3×(8)+(10)=34. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=37. Since the coefficient is 3, the new x should be 9. Explain the one-unit change using the original linear relationship.

13. Linear formula 13: make x the subject

Question. A general formula reads y=4x−10, with a nonzero coefficient 4. Rearrange the formula to make x the subject. Then check it when y=-34.

Why the inverse works. To isolate x, first remove the signed constant -10 from both entire sides, then divide by the nonzero coefficient 4. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=4x+(-10), subtract -10: y−(-10)=4x. Divide by 4: x=(y−(-10))/4. With y=-34, the numerator becomes -24, giving x=-6. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 4 before removing -10, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-6 into the original formula: 4×(-6)+(-10)=-34. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-30. Since the coefficient is 4, the new x should be -5. Explain the one-unit change using the original linear relationship.

14. Linear formula 14: make x the subject

Question. A general formula reads y=5x−3, with a nonzero coefficient 5. Rearrange the formula to make x the subject. Then check it when y=-8.

Why the inverse works. To isolate x, first remove the signed constant -3 from both entire sides, then divide by the nonzero coefficient 5. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=5x+(-3), subtract -3: y−(-3)=5x. Divide by 5: x=(y−(-3))/5. With y=-8, the numerator becomes -5, giving x=-1. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 5 before removing -3, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-1 into the original formula: 5×(-1)+(-3)=-8. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-3. Since the coefficient is 5, the new x should be 0. Explain the one-unit change using the original linear relationship.

15. Linear formula 15: make x the subject

Question. A general formula reads y=-2x+4, with a nonzero coefficient -2. Rearrange the formula to make x the subject. Then check it when y=-4.

Why the inverse works. To isolate x, first remove the signed constant 4 from both entire sides, then divide by the nonzero coefficient -2. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-2x+(4), subtract 4: y−(4)=-2x. Divide by -2: x=(y−(4))/-2. With y=-4, the numerator becomes -8, giving x=4. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -2 before removing 4, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=4 into the original formula: -2×(4)+(4)=-4. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-6. Since the coefficient is -2, the new x should be 5. Explain the one-unit change using the original linear relationship.

16. Linear formula 16: make x the subject

Question. A general formula reads y=-3x+11, with a nonzero coefficient -3. Rearrange the formula to make x the subject. Then check it when y=-16.

Why the inverse works. To isolate x, first remove the signed constant 11 from both entire sides, then divide by the nonzero coefficient -3. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-3x+(11), subtract 11: y−(11)=-3x. Divide by -3: x=(y−(11))/-3. With y=-16, the numerator becomes -27, giving x=9. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -3 before removing 11, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=9 into the original formula: -3×(9)+(11)=-16. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-19. Since the coefficient is -3, the new x should be 10. Explain the one-unit change using the original linear relationship.

17. Linear formula 17: make x the subject

Question. A general formula reads y=-4x−9, with a nonzero coefficient -4. Rearrange the formula to make x the subject. Then check it when y=11.

Why the inverse works. To isolate x, first remove the signed constant -9 from both entire sides, then divide by the nonzero coefficient -4. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-4x+(-9), subtract -9: y−(-9)=-4x. Divide by -4: x=(y−(-9))/-4. With y=11, the numerator becomes 20, giving x=-5. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -4 before removing -9, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-5 into the original formula: -4×(-5)+(-9)=11. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=7. Since the coefficient is -4, the new x should be -4. Explain the one-unit change using the original linear relationship.

18. Linear formula 18: make x the subject

Question. A general formula reads y=-5x−2, with a nonzero coefficient -5. Rearrange the formula to make x the subject. Then check it when y=-2.

Why the inverse works. To isolate x, first remove the signed constant -2 from both entire sides, then divide by the nonzero coefficient -5. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-5x+(-2), subtract -2: y−(-2)=-5x. Divide by -5: x=(y−(-2))/-5. With y=-2, the numerator becomes 0, giving x=0. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -5 before removing -2, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=0 into the original formula: -5×(0)+(-2)=-2. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-7. Since the coefficient is -5, the new x should be 1. Explain the one-unit change using the original linear relationship.

19. Linear formula 19: make x the subject

Question. A general formula reads y=6x+5, with a nonzero coefficient 6. Rearrange the formula to make x the subject. Then check it when y=35.

Why the inverse works. To isolate x, first remove the signed constant 5 from both entire sides, then divide by the nonzero coefficient 6. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=6x+(5), subtract 5: y−(5)=6x. Divide by 6: x=(y−(5))/6. With y=35, the numerator becomes 30, giving x=5. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 6 before removing 5, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=5 into the original formula: 6×(5)+(5)=35. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=41. Since the coefficient is 6, the new x should be 6. Explain the one-unit change using the original linear relationship.

20. Linear formula 20: make x the subject

Question. A general formula reads y=-6x+12, with a nonzero coefficient -6. Rearrange the formula to make x the subject. Then check it when y=66.

Why the inverse works. To isolate x, first remove the signed constant 12 from both entire sides, then divide by the nonzero coefficient -6. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-6x+(12), subtract 12: y−(12)=-6x. Divide by -6: x=(y−(12))/-6. With y=66, the numerator becomes 54, giving x=-9. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -6 before removing 12, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-9 into the original formula: -6×(-9)+(12)=66. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=60. Since the coefficient is -6, the new x should be -8. Explain the one-unit change using the original linear relationship.

21. Linear formula 21: make x the subject

Question. A general formula reads y=2x−8, with a nonzero coefficient 2. Rearrange the formula to make x the subject. Then check it when y=-16.

Why the inverse works. To isolate x, first remove the signed constant -8 from both entire sides, then divide by the nonzero coefficient 2. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=2x+(-8), subtract -8: y−(-8)=2x. Divide by 2: x=(y−(-8))/2. With y=-16, the numerator becomes -8, giving x=-4. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 2 before removing -8, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-4 into the original formula: 2×(-4)+(-8)=-16. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-14. Since the coefficient is 2, the new x should be -3. Explain the one-unit change using the original linear relationship.

22. Linear formula 22: make x the subject

Question. A general formula reads y=3x−1, with a nonzero coefficient 3. Rearrange the formula to make x the subject. Then check it when y=2.

Why the inverse works. To isolate x, first remove the signed constant -1 from both entire sides, then divide by the nonzero coefficient 3. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=3x+(-1), subtract -1: y−(-1)=3x. Divide by 3: x=(y−(-1))/3. With y=2, the numerator becomes 3, giving x=1. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 3 before removing -1, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=1 into the original formula: 3×(1)+(-1)=2. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=5. Since the coefficient is 3, the new x should be 2. Explain the one-unit change using the original linear relationship.

23. Linear formula 23: make x the subject

Question. A general formula reads y=4x+6, with a nonzero coefficient 4. Rearrange the formula to make x the subject. Then check it when y=30.

Why the inverse works. To isolate x, first remove the signed constant 6 from both entire sides, then divide by the nonzero coefficient 4. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=4x+(6), subtract 6: y−(6)=4x. Divide by 4: x=(y−(6))/4. With y=30, the numerator becomes 24, giving x=6. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 4 before removing 6, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=6 into the original formula: 4×(6)+(6)=30. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=34. Since the coefficient is 4, the new x should be 7. Explain the one-unit change using the original linear relationship.

24. Linear formula 24: make x the subject

Question. A general formula reads y=5x+13, with a nonzero coefficient 5. Rearrange the formula to make x the subject. Then check it when y=-27.

Why the inverse works. To isolate x, first remove the signed constant 13 from both entire sides, then divide by the nonzero coefficient 5. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=5x+(13), subtract 13: y−(13)=5x. Divide by 5: x=(y−(13))/5. With y=-27, the numerator becomes -40, giving x=-8. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 5 before removing 13, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-8 into the original formula: 5×(-8)+(13)=-27. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-22. Since the coefficient is 5, the new x should be -7. Explain the one-unit change using the original linear relationship.

25. Linear formula 25: make x the subject

Question. A general formula reads y=-2x−7, with a nonzero coefficient -2. Rearrange the formula to make x the subject. Then check it when y=-1.

Why the inverse works. To isolate x, first remove the signed constant -7 from both entire sides, then divide by the nonzero coefficient -2. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-2x+(-7), subtract -7: y−(-7)=-2x. Divide by -2: x=(y−(-7))/-2. With y=-1, the numerator becomes 6, giving x=-3. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -2 before removing -7, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-3 into the original formula: -2×(-3)+(-7)=-1. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-3. Since the coefficient is -2, the new x should be -2. Explain the one-unit change using the original linear relationship.

26. Linear formula 26: make x the subject

Question. A general formula reads y=-3x+0, with a nonzero coefficient -3. Rearrange the formula to make x the subject. Then check it when y=-6.

Why the inverse works. To isolate x, first remove the signed constant 0 from both entire sides, then divide by the nonzero coefficient -3. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-3x+(0), subtract 0: y−(0)=-3x. Divide by -3: x=(y−(0))/-3. With y=-6, the numerator becomes -6, giving x=2. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -3 before removing 0, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=2 into the original formula: -3×(2)+(0)=-6. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-9. Since the coefficient is -3, the new x should be 3. Explain the one-unit change using the original linear relationship.

27. Linear formula 27: make x the subject

Question. A general formula reads y=-4x+7, with a nonzero coefficient -4. Rearrange the formula to make x the subject. Then check it when y=-21.

Why the inverse works. To isolate x, first remove the signed constant 7 from both entire sides, then divide by the nonzero coefficient -4. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-4x+(7), subtract 7: y−(7)=-4x. Divide by -4: x=(y−(7))/-4. With y=-21, the numerator becomes -28, giving x=7. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -4 before removing 7, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=7 into the original formula: -4×(7)+(7)=-21. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-25. Since the coefficient is -4, the new x should be 8. Explain the one-unit change using the original linear relationship.

28. Linear formula 28: make x the subject

Question. A general formula reads y=-5x−13, with a nonzero coefficient -5. Rearrange the formula to make x the subject. Then check it when y=22.

Why the inverse works. To isolate x, first remove the signed constant -13 from both entire sides, then divide by the nonzero coefficient -5. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-5x+(-13), subtract -13: y−(-13)=-5x. Divide by -5: x=(y−(-13))/-5. With y=22, the numerator becomes 35, giving x=-7. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -5 before removing -13, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-7 into the original formula: -5×(-7)+(-13)=22. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=17. Since the coefficient is -5, the new x should be -6. Explain the one-unit change using the original linear relationship.

29. Linear formula 29: make x the subject

Question. A general formula reads y=6x−6, with a nonzero coefficient 6. Rearrange the formula to make x the subject. Then check it when y=-18.

Why the inverse works. To isolate x, first remove the signed constant -6 from both entire sides, then divide by the nonzero coefficient 6. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=6x+(-6), subtract -6: y−(-6)=6x. Divide by 6: x=(y−(-6))/6. With y=-18, the numerator becomes -12, giving x=-2. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by 6 before removing -6, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=-2 into the original formula: 6×(-2)+(-6)=-18. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-12. Since the coefficient is 6, the new x should be -1. Explain the one-unit change using the original linear relationship.

30. Linear formula 30: make x the subject

Question. A general formula reads y=-6x+1, with a nonzero coefficient -6. Rearrange the formula to make x the subject. Then check it when y=-17.

Why the inverse works. To isolate x, first remove the signed constant 1 from both entire sides, then divide by the nonzero coefficient -6. The sign of a matters when simplifying, but an equality is preserved when both sides are divided by the same nonzero number.

Work it through. Starting y=-6x+(1), subtract 1: y−(1)=-6x. Divide by -6: x=(y−(1))/-6. With y=-17, the numerator becomes -18, giving x=3. This symbolic rearrangement holds for every real y because a≠0.

First potential mistake. A pupil may divide y by -6 before removing 1, leaving the constant improperly scaled. When a is negative, an incorrect sign on the divisor may also reverse the answer. Do not change only one side of the equality.

Verify in the original formula. Substitute x=3 into the original formula: -6×(3)+(1)=-17. The reconstructed output matches the given y, independently checking the rearrangement.

Changed formula. Change the output to y=-23. Since the coefficient is -6, the new x should be 4. Explain the one-unit change using the original linear relationship.

Rearrange area and perimeter formulae

When making rearrange area and perimeter formulae work, the most useful first question is ‘Which operation is applied to the variable last?’ Undo that operation on both complete sides before moving inward. A correct subject form should allow the original to be reconstructed. Watch for zero denominators, physical units and a distinction between a symbolic formula and a numerical answer.

31. Rectangle area 1: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=12 cm² and l=4 cm.

Why the inverse works. Area multiplies two perpendicular lengths. To undo multiplication by nonzero length l, divide both sides by l. The resulting unit is cm²/cm=cm, a length. The positivity of l is a physical condition and also ensures division is defined.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=12/4=3 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=12 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 4 cm×3 cm=12 cm², exactly the given area. Check the separate perimeter as 2(4+3)=14 cm; it is not the same output quantity.

Changed formula. If the area doubled to 24 cm² while l stayed 4 cm, the required width would double to 6 cm. Explain why the same algebra rearrangement predicts that proportional change.

32. Rectangle area 2: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=56 cm² and l=7 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=56/7=8 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=56 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 7 cm×8 cm=56 cm², exactly the given area. Check the separate perimeter as 2(7+8)=30 cm; it is not the same output quantity.

Changed formula. If the area doubled to 112 cm² while l stayed 7 cm, the required width would double to 16 cm. Explain why the same algebra rearrangement predicts that proportional change.

33. Rectangle area 3: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=130 cm² and l=10 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=130/10=13 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=130 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 10 cm×13 cm=130 cm², exactly the given area. Check the separate perimeter as 2(10+13)=46 cm; it is not the same output quantity.

Changed formula. If the area doubled to 260 cm² while l stayed 10 cm, the required width would double to 26 cm. Explain why the same algebra rearrangement predicts that proportional change.

34. Rectangle area 4: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=234 cm² and l=13 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=234/13=18 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=234 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 13 cm×18 cm=234 cm², exactly the given area. Check the separate perimeter as 2(13+18)=62 cm; it is not the same output quantity.

Changed formula. If the area doubled to 468 cm² while l stayed 13 cm, the required width would double to 36 cm. Explain why the same algebra rearrangement predicts that proportional change.

35. Rectangle area 5: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=96 cm² and l=16 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=96/16=6 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=96 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 16 cm×6 cm=96 cm², exactly the given area. Check the separate perimeter as 2(16+6)=44 cm; it is not the same output quantity.

Changed formula. If the area doubled to 192 cm² while l stayed 16 cm, the required width would double to 12 cm. Explain why the same algebra rearrangement predicts that proportional change.

36. Rectangle area 6: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=209 cm² and l=19 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=209/19=11 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=209 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 19 cm×11 cm=209 cm², exactly the given area. Check the separate perimeter as 2(19+11)=60 cm; it is not the same output quantity.

Changed formula. If the area doubled to 418 cm² while l stayed 19 cm, the required width would double to 22 cm. Explain why the same algebra rearrangement predicts that proportional change.

37. Rectangle area 7: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=352 cm² and l=22 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=352/22=16 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=352 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 22 cm×16 cm=352 cm², exactly the given area. Check the separate perimeter as 2(22+16)=76 cm; it is not the same output quantity.

Changed formula. If the area doubled to 704 cm² while l stayed 22 cm, the required width would double to 32 cm. Explain why the same algebra rearrangement predicts that proportional change.

38. Rectangle area 8: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=100 cm² and l=25 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=100/25=4 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=100 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 25 cm×4 cm=100 cm², exactly the given area. Check the separate perimeter as 2(25+4)=58 cm; it is not the same output quantity.

Changed formula. If the area doubled to 200 cm² while l stayed 25 cm, the required width would double to 8 cm. Explain why the same algebra rearrangement predicts that proportional change.

39. Rectangle area 9: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=45 cm² and l=5 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=45/5=9 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=45 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 5 cm×9 cm=45 cm², exactly the given area. Check the separate perimeter as 2(5+9)=28 cm; it is not the same output quantity.

Changed formula. If the area doubled to 90 cm² while l stayed 5 cm, the required width would double to 18 cm. Explain why the same algebra rearrangement predicts that proportional change.

40. Rectangle area 10: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=112 cm² and l=8 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=112/8=14 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=112 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 8 cm×14 cm=112 cm², exactly the given area. Check the separate perimeter as 2(8+14)=44 cm; it is not the same output quantity.

Changed formula. If the area doubled to 224 cm² while l stayed 8 cm, the required width would double to 28 cm. Explain why the same algebra rearrangement predicts that proportional change.

41. Rectangle area 11: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=209 cm² and l=11 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=209/11=19 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

Verify in the original formula. Forward reconstruction: 11 cm×19 cm=209 cm², exactly the given area. Check the separate perimeter as 2(11+19)=60 cm; it is not the same output quantity.

Changed formula. If the area doubled to 418 cm² while l stayed 11 cm, the required width would double to 38 cm. Explain why the same algebra rearrangement predicts that proportional change.

42. Rectangle area 12: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=98 cm² and l=14 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=98/14=7 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=98 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 14 cm×7 cm=98 cm², exactly the given area. Check the separate perimeter as 2(14+7)=42 cm; it is not the same output quantity.

Changed formula. If the area doubled to 196 cm² while l stayed 14 cm, the required width would double to 14 cm. Explain why the same algebra rearrangement predicts that proportional change.

43. Rectangle area 13: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=204 cm² and l=17 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=204/17=12 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=204 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 17 cm×12 cm=204 cm², exactly the given area. Check the separate perimeter as 2(17+12)=58 cm; it is not the same output quantity.

Changed formula. If the area doubled to 408 cm² while l stayed 17 cm, the required width would double to 24 cm. Explain why the same algebra rearrangement predicts that proportional change.

44. Rectangle area 14: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=340 cm² and l=20 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=340/20=17 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=340 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 20 cm×17 cm=340 cm², exactly the given area. Check the separate perimeter as 2(20+17)=74 cm; it is not the same output quantity.

Changed formula. If the area doubled to 680 cm² while l stayed 20 cm, the required width would double to 34 cm. Explain why the same algebra rearrangement predicts that proportional change.

45. Rectangle area 15: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=115 cm² and l=23 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=115/23=5 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=115 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 23 cm×5 cm=115 cm², exactly the given area. Check the separate perimeter as 2(23+5)=56 cm; it is not the same output quantity.

Changed formula. If the area doubled to 230 cm² while l stayed 23 cm, the required width would double to 10 cm. Explain why the same algebra rearrangement predicts that proportional change.

46. Rectangle area 16: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=260 cm² and l=26 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=260/26=10 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=260 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 26 cm×10 cm=260 cm², exactly the given area. Check the separate perimeter as 2(26+10)=72 cm; it is not the same output quantity.

Changed formula. If the area doubled to 520 cm² while l stayed 26 cm, the required width would double to 20 cm. Explain why the same algebra rearrangement predicts that proportional change.

47. Rectangle area 17: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=90 cm² and l=6 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=90/6=15 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=90 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 6 cm×15 cm=90 cm², exactly the given area. Check the separate perimeter as 2(6+15)=42 cm; it is not the same output quantity.

Changed formula. If the area doubled to 180 cm² while l stayed 6 cm, the required width would double to 30 cm. Explain why the same algebra rearrangement predicts that proportional change.

48. Rectangle area 18: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=27 cm² and l=9 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=27/9=3 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=27 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 9 cm×3 cm=27 cm², exactly the given area. Check the separate perimeter as 2(9+3)=24 cm; it is not the same output quantity.

Changed formula. If the area doubled to 54 cm² while l stayed 9 cm, the required width would double to 6 cm. Explain why the same algebra rearrangement predicts that proportional change.

49. Rectangle area 19: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=96 cm² and l=12 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=96/12=8 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

Verify in the original formula. Forward reconstruction: 12 cm×8 cm=96 cm², exactly the given area. Check the separate perimeter as 2(12+8)=40 cm; it is not the same output quantity.

Changed formula. If the area doubled to 192 cm² while l stayed 12 cm, the required width would double to 16 cm. Explain why the same algebra rearrangement predicts that proportional change.

50. Rectangle area 20: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=195 cm² and l=15 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=195/15=13 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=195 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 15 cm×13 cm=195 cm², exactly the given area. Check the separate perimeter as 2(15+13)=56 cm; it is not the same output quantity.

Changed formula. If the area doubled to 390 cm² while l stayed 15 cm, the required width would double to 26 cm. Explain why the same algebra rearrangement predicts that proportional change.

51. Rectangle area 21: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=324 cm² and l=18 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=324/18=18 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=324 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 18 cm×18 cm=324 cm², exactly the given area. Check the separate perimeter as 2(18+18)=72 cm; it is not the same output quantity.

Changed formula. If the area doubled to 648 cm² while l stayed 18 cm, the required width would double to 36 cm. Explain why the same algebra rearrangement predicts that proportional change.

52. Rectangle area 22: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=126 cm² and l=21 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=126/21=6 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=126 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 21 cm×6 cm=126 cm², exactly the given area. Check the separate perimeter as 2(21+6)=54 cm; it is not the same output quantity.

Changed formula. If the area doubled to 252 cm² while l stayed 21 cm, the required width would double to 12 cm. Explain why the same algebra rearrangement predicts that proportional change.

53. Rectangle area 23: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=264 cm² and l=24 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=264/24=11 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=264 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 24 cm×11 cm=264 cm², exactly the given area. Check the separate perimeter as 2(24+11)=70 cm; it is not the same output quantity.

Changed formula. If the area doubled to 528 cm² while l stayed 24 cm, the required width would double to 22 cm. Explain why the same algebra rearrangement predicts that proportional change.

54. Rectangle area 24: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=64 cm² and l=4 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=64/4=16 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=64 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 4 cm×16 cm=64 cm², exactly the given area. Check the separate perimeter as 2(4+16)=40 cm; it is not the same output quantity.

Changed formula. If the area doubled to 128 cm² while l stayed 4 cm, the required width would double to 32 cm. Explain why the same algebra rearrangement predicts that proportional change.

55. Rectangle area 25: make width the subject

Question. The area formula is A=lw. Rearrange it to make width w the subject and find w when A=28 cm² and l=7 cm.

Work it through. Divide A=lw by l to obtain w=A/l. With the supplied area and length, w=28/7=4 cm. This is more than a memorised reverse formula: it is equality balance applied to measurement.

First potential mistake. Subtracting l from A would confuse a product with an addition. Writing w=28 cm² would also use the wrong unit. Ask whether the child knows which part of the original formula is multiplied.

Verify in the original formula. Forward reconstruction: 7 cm×4 cm=28 cm², exactly the given area. Check the separate perimeter as 2(7+4)=22 cm; it is not the same output quantity.

Changed formula. If the area doubled to 56 cm² while l stayed 7 cm, the required width would double to 8 cm. Explain why the same algebra rearrangement predicts that proportional change.

56. Perimeter formula 1: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=16 cm and l=5 cm.

Why the inverse works. The outer factor two multiplies the entire length-plus-width sum. Undo it first by dividing the complete perimeter by two. Next subtract the known length l. This is a different inverse sequence from the area formula because perimeter uses addition.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=16, the half-perimeter is 8, leaving w=3 cm.

First potential mistake. Subtracting l before halving the perimeter incorrectly treats the outer two as if it multiplied only part of the bracket. Another error is using P=lw, which is an area relationship.

Verify in the original formula. Check 2(5+3)=16 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 5 cm. Explain the relation without re-solving every line.

57. Perimeter formula 2: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=40 cm and l=10 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=40, the half-perimeter is 20, leaving w=10 cm.

Verify in the original formula. Check 2(10+10)=40 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 12 cm. Explain the relation without re-solving every line.

58. Perimeter formula 3: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=64 cm and l=15 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=64, the half-perimeter is 32, leaving w=17 cm.

Verify in the original formula. Check 2(15+17)=64 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 19 cm. Explain the relation without re-solving every line.

59. Perimeter formula 4: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=52 cm and l=20 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=52, the half-perimeter is 26, leaving w=6 cm.

Verify in the original formula. Check 2(20+6)=52 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 8 cm. Explain the relation without re-solving every line.

60. Perimeter formula 5: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=76 cm and l=25 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=76, the half-perimeter is 38, leaving w=13 cm.

Verify in the original formula. Check 2(25+13)=76 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 15 cm. Explain the relation without re-solving every line.

61. Perimeter formula 6: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=56 cm and l=8 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=56, the half-perimeter is 28, leaving w=20 cm.

Verify in the original formula. Check 2(8+20)=56 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 22 cm. Explain the relation without re-solving every line.

62. Perimeter formula 7: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=44 cm and l=13 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=44, the half-perimeter is 22, leaving w=9 cm.

Verify in the original formula. Check 2(13+9)=44 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 11 cm. Explain the relation without re-solving every line.

63. Perimeter formula 8: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=68 cm and l=18 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=68, the half-perimeter is 34, leaving w=16 cm.

Verify in the original formula. Check 2(18+16)=68 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 18 cm. Explain the relation without re-solving every line.

64. Perimeter formula 9: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=56 cm and l=23 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=56, the half-perimeter is 28, leaving w=5 cm.

Verify in the original formula. Check 2(23+5)=56 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 7 cm. Explain the relation without re-solving every line.

65. Perimeter formula 10: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=36 cm and l=6 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=36, the half-perimeter is 18, leaving w=12 cm.

Verify in the original formula. Check 2(6+12)=36 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 14 cm. Explain the relation without re-solving every line.

66. Perimeter formula 11: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=60 cm and l=11 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=60, the half-perimeter is 30, leaving w=19 cm.

Verify in the original formula. Check 2(11+19)=60 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 21 cm. Explain the relation without re-solving every line.

67. Perimeter formula 12: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=48 cm and l=16 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=48, the half-perimeter is 24, leaving w=8 cm.

Verify in the original formula. Check 2(16+8)=48 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 10 cm. Explain the relation without re-solving every line.

68. Perimeter formula 13: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=72 cm and l=21 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=72, the half-perimeter is 36, leaving w=15 cm.

Verify in the original formula. Check 2(21+15)=72 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 17 cm. Explain the relation without re-solving every line.

69. Perimeter formula 14: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=60 cm and l=26 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=60, the half-perimeter is 30, leaving w=4 cm.

Verify in the original formula. Check 2(26+4)=60 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 6 cm. Explain the relation without re-solving every line.

70. Perimeter formula 15: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=40 cm and l=9 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=40, the half-perimeter is 20, leaving w=11 cm.

Verify in the original formula. Check 2(9+11)=40 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 13 cm. Explain the relation without re-solving every line.

71. Perimeter formula 16: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=64 cm and l=14 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=64, the half-perimeter is 32, leaving w=18 cm.

Verify in the original formula. Check 2(14+18)=64 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 20 cm. Explain the relation without re-solving every line.

72. Perimeter formula 17: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=52 cm and l=19 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=52, the half-perimeter is 26, leaving w=7 cm.

Verify in the original formula. Check 2(19+7)=52 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 9 cm. Explain the relation without re-solving every line.

73. Perimeter formula 18: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=76 cm and l=24 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=76, the half-perimeter is 38, leaving w=14 cm.

Verify in the original formula. Check 2(24+14)=76 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 16 cm. Explain the relation without re-solving every line.

74. Perimeter formula 19: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=20 cm and l=7 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=20, the half-perimeter is 10, leaving w=3 cm.

Verify in the original formula. Check 2(7+3)=20 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 5 cm. Explain the relation without re-solving every line.

75. Perimeter formula 20: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=44 cm and l=12 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=44, the half-perimeter is 22, leaving w=10 cm.

Verify in the original formula. Check 2(12+10)=44 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 12 cm. Explain the relation without re-solving every line.

76. Perimeter formula 21: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=68 cm and l=17 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=68, the half-perimeter is 34, leaving w=17 cm.

Verify in the original formula. Check 2(17+17)=68 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 19 cm. Explain the relation without re-solving every line.

77. Perimeter formula 22: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=56 cm and l=22 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=56, the half-perimeter is 28, leaving w=6 cm.

Verify in the original formula. Check 2(22+6)=56 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 8 cm. Explain the relation without re-solving every line.

78. Perimeter formula 23: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=36 cm and l=5 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=36, the half-perimeter is 18, leaving w=13 cm.

Verify in the original formula. Check 2(5+13)=36 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 15 cm. Explain the relation without re-solving every line.

79. Perimeter formula 24: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=60 cm and l=10 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=60, the half-perimeter is 30, leaving w=20 cm.

Verify in the original formula. Check 2(10+20)=60 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 22 cm. Explain the relation without re-solving every line.

80. Perimeter formula 25: make width the subject

Question. A rectangle’s perimeter formula is P=2(l+w). Rearrange it to make w the subject, then find w when P=48 cm and l=15 cm.

Work it through. From P=2(l+w), divide by two to get P/2=l+w. Subtract l: w=P/2−l. With P=48, the half-perimeter is 24, leaving w=9 cm.

Verify in the original formula. Check 2(15+9)=48 cm. The original formula is reconstructed with positive side lengths, and the result is in centimetres, not square centimetres.

Changed formula. Increase P by four centimetres while l remains fixed. The half-perimeter rises by two, so w increases to 11 cm. Explain the relation without re-solving every line.

Turn a fixed-fee total into a count

When making turn a fixed-fee total into a count work, the most useful first question is ‘Which operation is applied to the variable last?’ Undo that operation on both complete sides before moving inward. A correct subject form should allow the original to be reconstructed. Watch for zero denominators, physical units and a distinction between a symbolic formula and a numerical answer.

81. A fixed-fee formula 1: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$3, r=$2/use and C=$7.

Why the inverse works. The unknown n appears in the repeated-charge product rn. Remove the fixed fee once, then divide by the nonzero per-use rate. Units also guide the inverse: dollars divided by dollars per use returns a number of uses.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(7−3)/2=2 uses. This model is hypothetical and assumes every use has the same variable charge.

First potential mistake. Dividing the full cost C by the rate before subtracting the fixed fee makes the once-only charge behave as another use. Another mistake is multiplying F by n even though the story says the fee occurs once.

Verify in the original formula. Forward check: F+rn=3+2×2=$7, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $9 with other parameters unchanged, n would increase to 3. Use both the rearranged formula and the original charging story to explain.

82. A fixed-fee formula 2: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$8, r=$9/use and C=$53.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(53−8)/9=5 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=8+9×5=$53, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $62 with other parameters unchanged, n would increase to 6. Use both the rearranged formula and the original charging story to explain.

83. A fixed-fee formula 3: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$13, r=$7/use and C=$69.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(69−13)/7=8 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=13+7×8=$69, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $76 with other parameters unchanged, n would increase to 9. Use both the rearranged formula and the original charging story to explain.

84. A fixed-fee formula 4: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$18, r=$5/use and C=$73.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(73−18)/5=11 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=18+5×11=$73, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $78 with other parameters unchanged, n would increase to 12. Use both the rearranged formula and the original charging story to explain.

85. A fixed-fee formula 5: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$4, r=$3/use and C=$46.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(46−4)/3=14 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=4+3×14=$46, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $49 with other parameters unchanged, n would increase to 15. Use both the rearranged formula and the original charging story to explain.

86. A fixed-fee formula 6: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$9, r=$10/use and C=$179.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(179−9)/10=17 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=9+10×17=$179, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $189 with other parameters unchanged, n would increase to 18. Use both the rearranged formula and the original charging story to explain.

87. A fixed-fee formula 7: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$14, r=$8/use and C=$174.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(174−14)/8=20 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=14+8×20=$174, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $182 with other parameters unchanged, n would increase to 21. Use both the rearranged formula and the original charging story to explain.

88. A fixed-fee formula 8: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$19, r=$6/use and C=$157.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(157−19)/6=23 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=19+6×23=$157, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $163 with other parameters unchanged, n would increase to 24. Use both the rearranged formula and the original charging story to explain.

89. A fixed-fee formula 9: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$5, r=$4/use and C=$109.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(109−5)/4=26 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=5+4×26=$109, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $113 with other parameters unchanged, n would increase to 27. Use both the rearranged formula and the original charging story to explain.

90. A fixed-fee formula 10: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$10, r=$2/use and C=$16.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(16−10)/2=3 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=10+2×3=$16, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $18 with other parameters unchanged, n would increase to 4. Use both the rearranged formula and the original charging story to explain.

91. A fixed-fee formula 11: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$15, r=$9/use and C=$69.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(69−15)/9=6 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=15+9×6=$69, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $78 with other parameters unchanged, n would increase to 7. Use both the rearranged formula and the original charging story to explain.

92. A fixed-fee formula 12: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$20, r=$7/use and C=$83.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(83−20)/7=9 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=20+7×9=$83, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $90 with other parameters unchanged, n would increase to 10. Use both the rearranged formula and the original charging story to explain.

93. A fixed-fee formula 13: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$6, r=$5/use and C=$66.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(66−6)/5=12 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=6+5×12=$66, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $71 with other parameters unchanged, n would increase to 13. Use both the rearranged formula and the original charging story to explain.

94. A fixed-fee formula 14: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$11, r=$3/use and C=$56.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(56−11)/3=15 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=11+3×15=$56, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $59 with other parameters unchanged, n would increase to 16. Use both the rearranged formula and the original charging story to explain.

95. A fixed-fee formula 15: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$16, r=$10/use and C=$196.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(196−16)/10=18 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=16+10×18=$196, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $206 with other parameters unchanged, n would increase to 19. Use both the rearranged formula and the original charging story to explain.

96. A fixed-fee formula 16: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$21, r=$8/use and C=$189.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(189−21)/8=21 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=21+8×21=$189, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $197 with other parameters unchanged, n would increase to 22. Use both the rearranged formula and the original charging story to explain.

97. A fixed-fee formula 17: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$7, r=$6/use and C=$151.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(151−7)/6=24 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=7+6×24=$151, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $157 with other parameters unchanged, n would increase to 25. Use both the rearranged formula and the original charging story to explain.

98. A fixed-fee formula 18: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$12, r=$4/use and C=$120.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(120−12)/4=27 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=12+4×27=$120, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $124 with other parameters unchanged, n would increase to 28. Use both the rearranged formula and the original charging story to explain.

99. A fixed-fee formula 19: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$17, r=$2/use and C=$25.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(25−17)/2=4 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=17+2×4=$25, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $27 with other parameters unchanged, n would increase to 5. Use both the rearranged formula and the original charging story to explain.

100. A fixed-fee formula 20: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$3, r=$9/use and C=$66.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(66−3)/9=7 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=3+9×7=$66, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $75 with other parameters unchanged, n would increase to 8. Use both the rearranged formula and the original charging story to explain.

101. A fixed-fee formula 21: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$8, r=$7/use and C=$78.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(78−8)/7=10 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=8+7×10=$78, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $85 with other parameters unchanged, n would increase to 11. Use both the rearranged formula and the original charging story to explain.

102. A fixed-fee formula 22: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$13, r=$5/use and C=$78.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(78−13)/5=13 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=13+5×13=$78, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $83 with other parameters unchanged, n would increase to 14. Use both the rearranged formula and the original charging story to explain.

103. A fixed-fee formula 23: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$18, r=$3/use and C=$66.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(66−18)/3=16 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=18+3×16=$66, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $69 with other parameters unchanged, n would increase to 17. Use both the rearranged formula and the original charging story to explain.

104. A fixed-fee formula 24: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$4, r=$10/use and C=$194.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(194−4)/10=19 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=4+10×19=$194, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $204 with other parameters unchanged, n would increase to 20. Use both the rearranged formula and the original charging story to explain.

105. A fixed-fee formula 25: find number of uses

Question. A fictional service has total C=F+rn, where F is a once-only fee, r dollars per use and n the number of uses. Make n the subject and evaluate it for F=$9, r=$8/use and C=$185.

Work it through. Subtract F: C−F=rn. Divide by r: n=(C−F)/r. Numerically n=(185−9)/8=22 uses. This model is hypothetical and assumes every use has the same variable charge.

Verify in the original formula. Forward check: F+rn=9+8×22=$185, matching the original bill. The count is a positive whole number as required by the model.

Changed formula. If the final bill increased to $193 with other parameters unchanged, n would increase to 23. Use both the rearranged formula and the original charging story to explain.

Rearrange distance, rate and time

When making rearrange distance, rate and time work, the most useful first question is ‘Which operation is applied to the variable last?’ Undo that operation on both complete sides before moving inward. A correct subject form should allow the original to be reconstructed. Watch for zero denominators, physical units and a distinction between a symbolic formula and a numerical answer.

106. Average speed 1: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=10 km and average v=10 km/h.

Why the inverse works. A quotient v=d/t means distance divided by time. Multiply both sides by nonzero t to obtain vt=d, then divide by nonzero v. Both symbolic transformations preserve the equality under the given positive physical quantities.

Work it through. The rearranged expression is t=d/v. Substitute values: t=10/10=1 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

First potential mistake. Multiplying distance by speed would yield km²/h-like units, not time. Another possible error is claiming the average speed is guaranteed to hold at every instant; the formula concerns total distance and elapsed time.

Verify in the original formula. Check v×t=10 km/h×1 h=10 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 10 km/h but double the distance to 20 km. The elapsed time under this constant-average-rate model doubles to 2 hours. State the model assumption explicitly.

107. Average speed 2: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=102 km and average v=17 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=102/17=6 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=17 km/h×6 h=102 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 17 km/h but double the distance to 204 km. The elapsed time under this constant-average-rate model doubles to 12 hours. State the model assumption explicitly.

108. Average speed 3: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=264 km and average v=24 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=264/24=11 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=24 km/h×11 h=264 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 24 km/h but double the distance to 528 km. The elapsed time under this constant-average-rate model doubles to 22 hours. State the model assumption explicitly.

109. Average speed 4: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=124 km and average v=31 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=124/31=4 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=31 km/h×4 h=124 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 31 km/h but double the distance to 248 km. The elapsed time under this constant-average-rate model doubles to 8 hours. State the model assumption explicitly.

110. Average speed 5: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=342 km and average v=38 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=342/38=9 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=38 km/h×9 h=342 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 38 km/h but double the distance to 684 km. The elapsed time under this constant-average-rate model doubles to 18 hours. State the model assumption explicitly.

111. Average speed 6: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=90 km and average v=45 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=90/45=2 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=45 km/h×2 h=90 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 45 km/h but double the distance to 180 km. The elapsed time under this constant-average-rate model doubles to 4 hours. State the model assumption explicitly.

112. Average speed 7: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=364 km and average v=52 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=364/52=7 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=52 km/h×7 h=364 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 52 km/h but double the distance to 728 km. The elapsed time under this constant-average-rate model doubles to 14 hours. State the model assumption explicitly.

113. Average speed 8: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=708 km and average v=59 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=708/59=12 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=59 km/h×12 h=708 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 59 km/h but double the distance to 1416 km. The elapsed time under this constant-average-rate model doubles to 24 hours. State the model assumption explicitly.

114. Average speed 9: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=330 km and average v=66 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=330/66=5 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=66 km/h×5 h=330 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 66 km/h but double the distance to 660 km. The elapsed time under this constant-average-rate model doubles to 10 hours. State the model assumption explicitly.

115. Average speed 10: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=730 km and average v=73 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=730/73=10 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=73 km/h×10 h=730 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 73 km/h but double the distance to 1460 km. The elapsed time under this constant-average-rate model doubles to 20 hours. State the model assumption explicitly.

116. Average speed 11: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=240 km and average v=80 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=240/80=3 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=80 km/h×3 h=240 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 80 km/h but double the distance to 480 km. The elapsed time under this constant-average-rate model doubles to 6 hours. State the model assumption explicitly.

117. Average speed 12: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=696 km and average v=87 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=696/87=8 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=87 km/h×8 h=696 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 87 km/h but double the distance to 1392 km. The elapsed time under this constant-average-rate model doubles to 16 hours. State the model assumption explicitly.

118. Average speed 13: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=94 km and average v=94 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=94/94=1 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=94 km/h×1 h=94 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 94 km/h but double the distance to 188 km. The elapsed time under this constant-average-rate model doubles to 2 hours. State the model assumption explicitly.

119. Average speed 14: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=606 km and average v=101 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=606/101=6 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=101 km/h×6 h=606 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 101 km/h but double the distance to 1212 km. The elapsed time under this constant-average-rate model doubles to 12 hours. State the model assumption explicitly.

120. Average speed 15: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=165 km and average v=15 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=165/15=11 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=15 km/h×11 h=165 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 15 km/h but double the distance to 330 km. The elapsed time under this constant-average-rate model doubles to 22 hours. State the model assumption explicitly.

121. Average speed 16: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=88 km and average v=22 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=88/22=4 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=22 km/h×4 h=88 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 22 km/h but double the distance to 176 km. The elapsed time under this constant-average-rate model doubles to 8 hours. State the model assumption explicitly.

122. Average speed 17: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=261 km and average v=29 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=261/29=9 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=29 km/h×9 h=261 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 29 km/h but double the distance to 522 km. The elapsed time under this constant-average-rate model doubles to 18 hours. State the model assumption explicitly.

123. Average speed 18: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=72 km and average v=36 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=72/36=2 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=36 km/h×2 h=72 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 36 km/h but double the distance to 144 km. The elapsed time under this constant-average-rate model doubles to 4 hours. State the model assumption explicitly.

124. Average speed 19: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=301 km and average v=43 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=301/43=7 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=43 km/h×7 h=301 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 43 km/h but double the distance to 602 km. The elapsed time under this constant-average-rate model doubles to 14 hours. State the model assumption explicitly.

125. Average speed 20: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=600 km and average v=50 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=600/50=12 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=50 km/h×12 h=600 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 50 km/h but double the distance to 1200 km. The elapsed time under this constant-average-rate model doubles to 24 hours. State the model assumption explicitly.

126. Average speed 21: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=285 km and average v=57 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=285/57=5 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=57 km/h×5 h=285 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 57 km/h but double the distance to 570 km. The elapsed time under this constant-average-rate model doubles to 10 hours. State the model assumption explicitly.

127. Average speed 22: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=640 km and average v=64 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=640/64=10 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=64 km/h×10 h=640 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 64 km/h but double the distance to 1280 km. The elapsed time under this constant-average-rate model doubles to 20 hours. State the model assumption explicitly.

128. Average speed 23: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=213 km and average v=71 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=213/71=3 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=71 km/h×3 h=213 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 71 km/h but double the distance to 426 km. The elapsed time under this constant-average-rate model doubles to 6 hours. State the model assumption explicitly.

129. Average speed 24: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=624 km and average v=78 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=624/78=8 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=78 km/h×8 h=624 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 78 km/h but double the distance to 1248 km. The elapsed time under this constant-average-rate model doubles to 16 hours. State the model assumption explicitly.

130. Average speed 25: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=85 km and average v=85 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=85/85=1 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=85 km/h×1 h=85 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 85 km/h but double the distance to 170 km. The elapsed time under this constant-average-rate model doubles to 2 hours. State the model assumption explicitly.

131. Average speed 26: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=552 km and average v=92 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=552/92=6 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=92 km/h×6 h=552 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 92 km/h but double the distance to 1104 km. The elapsed time under this constant-average-rate model doubles to 12 hours. State the model assumption explicitly.

132. Average speed 27: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=1089 km and average v=99 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=1089/99=11 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=99 km/h×11 h=1089 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 99 km/h but double the distance to 2178 km. The elapsed time under this constant-average-rate model doubles to 22 hours. State the model assumption explicitly.

133. Average speed 28: make time the subject

Question. An ideal average-speed formula is v=d/t, where d is measured in kilometres and t in hours. Make t the subject and find the elapsed time if d=52 km and average v=13 km/h.

Work it through. The rearranged expression is t=d/v. Substitute values: t=52/13=4 hours. The unit km/(km/h) reduces to hours; this is a useful dimensional check on the algebra.

Verify in the original formula. Check v×t=13 km/h×4 h=52 km. This reconstructs the measured total and the stated average rate, independent of any details of movement within the interval.

Changed formula. Keep speed 13 km/h but double the distance to 104 km. The elapsed time under this constant-average-rate model doubles to 8 hours. State the model assumption explicitly.

Find a missing value from the mean

When making find a missing value from the mean work, the most useful first question is ‘Which operation is applied to the variable last?’ Undo that operation on both complete sides before moving inward. A correct subject form should allow the original to be reconstructed. Watch for zero denominators, physical units and a distinction between a symbolic formula and a numerical answer.

134. Mean of two values 1: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=2 and M=3.

Why the inverse works. The mean of two equally weighted readings is their sum divided by two. Undo division by two by multiplying both sides by two; then subtract the known reading a. Do not divide only one term of the numerator.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=3 and a=2; b=2(3)−2=4.

First potential mistake. A learner may write b=2(M−a), which doubles the subtraction of a as well as the mean. This produces a different formula and often an incorrect output. Grouping determines the order.

Verify in the original formula. Rebuild the mean: (2+4)/2=3. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=2 fixed. Then b rises by two to 6, because two readings must together increase by two to raise their mean by one.

135. Mean of two values 2: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=7 and M=9.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=9 and a=7; b=2(9)−7=11.

Verify in the original formula. Rebuild the mean: (7+11)/2=9. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=7 fixed. Then b rises by two to 13, because two readings must together increase by two to raise their mean by one.

136. Mean of two values 3: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=12 and M=15.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=15 and a=12; b=2(15)−12=18.

Verify in the original formula. Rebuild the mean: (12+18)/2=15. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=12 fixed. Then b rises by two to 20, because two readings must together increase by two to raise their mean by one.

137. Mean of two values 4: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=17 and M=21.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=21 and a=17; b=2(21)−17=25.

Verify in the original formula. Rebuild the mean: (17+25)/2=21. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=17 fixed. Then b rises by two to 27, because two readings must together increase by two to raise their mean by one.

138. Mean of two values 5: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=22 and M=27.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=27 and a=22; b=2(27)−22=32.

Verify in the original formula. Rebuild the mean: (22+32)/2=27. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=22 fixed. Then b rises by two to 34, because two readings must together increase by two to raise their mean by one.

139. Mean of two values 6: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=27 and M=33.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=33 and a=27; b=2(33)−27=39.

Verify in the original formula. Rebuild the mean: (27+39)/2=33. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=27 fixed. Then b rises by two to 41, because two readings must together increase by two to raise their mean by one.

140. Mean of two values 7: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=32 and M=20.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=20.5 and a=32; b=2(20.5)−32=9.

Verify in the original formula. Rebuild the mean: (32+9)/2=20.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=32 fixed. Then b rises by two to 11, because two readings must together increase by two to raise their mean by one.

141. Mean of two values 8: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=37 and M=26.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=26.5 and a=37; b=2(26.5)−37=16.

Verify in the original formula. Rebuild the mean: (37+16)/2=26.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=37 fixed. Then b rises by two to 18, because two readings must together increase by two to raise their mean by one.

142. Mean of two values 9: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=42 and M=32.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=32.5 and a=42; b=2(32.5)−42=23.

Verify in the original formula. Rebuild the mean: (42+23)/2=32.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=42 fixed. Then b rises by two to 25, because two readings must together increase by two to raise their mean by one.

143. Mean of two values 10: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=6 and M=18.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=18 and a=6; b=2(18)−6=30.

Verify in the original formula. Rebuild the mean: (6+30)/2=18. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=6 fixed. Then b rises by two to 32, because two readings must together increase by two to raise their mean by one.

144. Mean of two values 11: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=11 and M=24.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=24 and a=11; b=2(24)−11=37.

Verify in the original formula. Rebuild the mean: (11+37)/2=24. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=11 fixed. Then b rises by two to 39, because two readings must together increase by two to raise their mean by one.

145. Mean of two values 12: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=16 and M=11.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=11.5 and a=16; b=2(11.5)−16=7.

Verify in the original formula. Rebuild the mean: (16+7)/2=11.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=16 fixed. Then b rises by two to 9, because two readings must together increase by two to raise their mean by one.

146. Mean of two values 13: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=21 and M=17.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=17.5 and a=21; b=2(17.5)−21=14.

Verify in the original formula. Rebuild the mean: (21+14)/2=17.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=21 fixed. Then b rises by two to 16, because two readings must together increase by two to raise their mean by one.

147. Mean of two values 14: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=26 and M=23.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=23.5 and a=26; b=2(23.5)−26=21.

Verify in the original formula. Rebuild the mean: (26+21)/2=23.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=26 fixed. Then b rises by two to 23, because two readings must together increase by two to raise their mean by one.

148. Mean of two values 15: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=31 and M=29.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=29.5 and a=31; b=2(29.5)−31=28.

Verify in the original formula. Rebuild the mean: (31+28)/2=29.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=31 fixed. Then b rises by two to 30, because two readings must together increase by two to raise their mean by one.

149. Mean of two values 16: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=36 and M=35.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=35.5 and a=36; b=2(35.5)−36=35.

Verify in the original formula. Rebuild the mean: (36+35)/2=35.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=36 fixed. Then b rises by two to 37, because two readings must together increase by two to raise their mean by one.

150. Mean of two values 17: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=41 and M=23.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=23 and a=41; b=2(23)−41=5.

Verify in the original formula. Rebuild the mean: (41+5)/2=23. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=41 fixed. Then b rises by two to 7, because two readings must together increase by two to raise their mean by one.

151. Mean of two values 18: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=5 and M=8.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=8.5 and a=5; b=2(8.5)−5=12.

Verify in the original formula. Rebuild the mean: (5+12)/2=8.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=5 fixed. Then b rises by two to 14, because two readings must together increase by two to raise their mean by one.

152. Mean of two values 19: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=10 and M=14.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=14.5 and a=10; b=2(14.5)−10=19.

Verify in the original formula. Rebuild the mean: (10+19)/2=14.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=10 fixed. Then b rises by two to 21, because two readings must together increase by two to raise their mean by one.

153. Mean of two values 20: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=15 and M=20.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=20.5 and a=15; b=2(20.5)−15=26.

Verify in the original formula. Rebuild the mean: (15+26)/2=20.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=15 fixed. Then b rises by two to 28, because two readings must together increase by two to raise their mean by one.

154. Mean of two values 21: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=20 and M=26.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=26.5 and a=20; b=2(26.5)−20=33.

Verify in the original formula. Rebuild the mean: (20+33)/2=26.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=20 fixed. Then b rises by two to 35, because two readings must together increase by two to raise their mean by one.

155. Mean of two values 22: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=25 and M=32.5.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=32.5 and a=25; b=2(32.5)−25=40.

Verify in the original formula. Rebuild the mean: (25+40)/2=32.5. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=25 fixed. Then b rises by two to 42, because two readings must together increase by two to raise their mean by one.

156. Mean of two values 23: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=30 and M=20.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=20 and a=30; b=2(20)−30=10.

Verify in the original formula. Rebuild the mean: (30+10)/2=20. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=30 fixed. Then b rises by two to 12, because two readings must together increase by two to raise their mean by one.

157. Mean of two values 24: find one missing value

Question. Two abstract numerical readings have arithmetic mean M=(a+b)/2. Make b the subject, then calculate b when a=35 and M=26.

Work it through. From M=(a+b)/2, multiply by two: 2M=a+b. Subtract a: b=2M−a. Substitute M=26 and a=35; b=2(26)−35=17.

Verify in the original formula. Rebuild the mean: (35+17)/2=26. The recovered b fits the original average, not merely an intermediate rearrangement.

Changed formula. Increase mean M by one but keep a=35 fixed. Then b rises by two to 19, because two readings must together increase by two to raise their mean by one.

Make a cylinder height the subject

When making make a cylinder height the subject work, the most useful first question is ‘Which operation is applied to the variable last?’ Undo that operation on both complete sides before moving inward. A correct subject form should allow the original to be reconstructed. Watch for zero denominators, physical units and a distinction between a symbolic formula and a numerical answer.

158. Cylinder volume 1: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 2 cm and exact volume 12π cm³.

Why the inverse works. The factor multiplying h is πr², the cross-sectional area of the circular base. Divide the whole volume by that positive area to recover height. This is the same idea as dividing cuboid volume by a base area.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=12π cm³ and r=2 cm, h=(12π)/(π×4)=3 cm. The π factors cancel without rounding.

First potential mistake. The symbol r² means r×r, not 2r. A pupil who divides by 2πr will use cylinder circumference rather than base area and get the wrong height. Units help reveal the distinction.

Verify in the original formula. Check π×2²×3=12π cm³. The base area 4π cm² times height 3 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 24π cm³. The height would double to 6 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

159. Cylinder volume 2: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 5 cm and exact volume 200π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=200π cm³ and r=5 cm, h=(200π)/(π×25)=8 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×5²×8=200π cm³. The base area 25π cm² times height 8 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 400π cm³. The height would double to 16 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

160. Cylinder volume 3: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 8 cm and exact volume 832π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=832π cm³ and r=8 cm, h=(832π)/(π×64)=13 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×8²×13=832π cm³. The base area 64π cm² times height 13 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 1664π cm³. The height would double to 26 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

161. Cylinder volume 4: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 11 cm and exact volume 2178π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=2178π cm³ and r=11 cm, h=(2178π)/(π×121)=18 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×11²×18=2178π cm³. The base area 121π cm² times height 18 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 4356π cm³. The height would double to 36 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

162. Cylinder volume 5: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 14 cm and exact volume 4508π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=4508π cm³ and r=14 cm, h=(4508π)/(π×196)=23 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×14²×23=4508π cm³. The base area 196π cm² times height 23 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 9016π cm³. The height would double to 46 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

163. Cylinder volume 6: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 4 cm and exact volume 96π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=96π cm³ and r=4 cm, h=(96π)/(π×16)=6 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×4²×6=96π cm³. The base area 16π cm² times height 6 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 192π cm³. The height would double to 12 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

164. Cylinder volume 7: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 7 cm and exact volume 539π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=539π cm³ and r=7 cm, h=(539π)/(π×49)=11 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×7²×11=539π cm³. The base area 49π cm² times height 11 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 1078π cm³. The height would double to 22 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

165. Cylinder volume 8: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 10 cm and exact volume 1600π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=1600π cm³ and r=10 cm, h=(1600π)/(π×100)=16 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×10²×16=1600π cm³. The base area 100π cm² times height 16 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 3200π cm³. The height would double to 32 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

166. Cylinder volume 9: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 13 cm and exact volume 3549π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=3549π cm³ and r=13 cm, h=(3549π)/(π×169)=21 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×13²×21=3549π cm³. The base area 169π cm² times height 21 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 7098π cm³. The height would double to 42 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

167. Cylinder volume 10: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 3 cm and exact volume 36π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=36π cm³ and r=3 cm, h=(36π)/(π×9)=4 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×3²×4=36π cm³. The base area 9π cm² times height 4 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 72π cm³. The height would double to 8 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

168. Cylinder volume 11: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 6 cm and exact volume 324π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=324π cm³ and r=6 cm, h=(324π)/(π×36)=9 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×6²×9=324π cm³. The base area 36π cm² times height 9 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 648π cm³. The height would double to 18 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

169. Cylinder volume 12: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 9 cm and exact volume 1134π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=1134π cm³ and r=9 cm, h=(1134π)/(π×81)=14 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×9²×14=1134π cm³. The base area 81π cm² times height 14 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 2268π cm³. The height would double to 28 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

170. Cylinder volume 13: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 12 cm and exact volume 2736π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=2736π cm³ and r=12 cm, h=(2736π)/(π×144)=19 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×12²×19=2736π cm³. The base area 144π cm² times height 19 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 5472π cm³. The height would double to 38 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

171. Cylinder volume 14: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 2 cm and exact volume 96π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=96π cm³ and r=2 cm, h=(96π)/(π×4)=24 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×2²×24=96π cm³. The base area 4π cm² times height 24 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 192π cm³. The height would double to 48 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

172. Cylinder volume 15: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 5 cm and exact volume 175π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=175π cm³ and r=5 cm, h=(175π)/(π×25)=7 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×5²×7=175π cm³. The base area 25π cm² times height 7 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 350π cm³. The height would double to 14 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

173. Cylinder volume 16: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 8 cm and exact volume 768π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=768π cm³ and r=8 cm, h=(768π)/(π×64)=12 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×8²×12=768π cm³. The base area 64π cm² times height 12 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 1536π cm³. The height would double to 24 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

174. Cylinder volume 17: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 11 cm and exact volume 2057π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=2057π cm³ and r=11 cm, h=(2057π)/(π×121)=17 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×11²×17=2057π cm³. The base area 121π cm² times height 17 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 4114π cm³. The height would double to 34 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

175. Cylinder volume 18: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 14 cm and exact volume 4312π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=4312π cm³ and r=14 cm, h=(4312π)/(π×196)=22 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×14²×22=4312π cm³. The base area 196π cm² times height 22 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 8624π cm³. The height would double to 44 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

176. Cylinder volume 19: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 4 cm and exact volume 80π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=80π cm³ and r=4 cm, h=(80π)/(π×16)=5 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×4²×5=80π cm³. The base area 16π cm² times height 5 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 160π cm³. The height would double to 10 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

177. Cylinder volume 20: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 7 cm and exact volume 490π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=490π cm³ and r=7 cm, h=(490π)/(π×49)=10 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×7²×10=490π cm³. The base area 49π cm² times height 10 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 980π cm³. The height would double to 20 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

178. Cylinder volume 21: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 10 cm and exact volume 1500π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=1500π cm³ and r=10 cm, h=(1500π)/(π×100)=15 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×10²×15=1500π cm³. The base area 100π cm² times height 15 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 3000π cm³. The height would double to 30 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

179. Cylinder volume 22: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 13 cm and exact volume 3380π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=3380π cm³ and r=13 cm, h=(3380π)/(π×169)=20 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×13²×20=3380π cm³. The base area 169π cm² times height 20 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 6760π cm³. The height would double to 40 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

180. Cylinder volume 23: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 3 cm and exact volume 27π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=27π cm³ and r=3 cm, h=(27π)/(π×9)=3 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×3²×3=27π cm³. The base area 9π cm² times height 3 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 54π cm³. The height would double to 6 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

181. Cylinder volume 24: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 6 cm and exact volume 288π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=288π cm³ and r=6 cm, h=(288π)/(π×36)=8 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×6²×8=288π cm³. The base area 36π cm² times height 8 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 576π cm³. The height would double to 16 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

182. Cylinder volume 25: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 9 cm and exact volume 1053π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=1053π cm³ and r=9 cm, h=(1053π)/(π×81)=13 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×9²×13=1053π cm³. The base area 81π cm² times height 13 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 2106π cm³. The height would double to 26 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

183. Cylinder volume 26: make height the subject

Question. An ideal right circular cylinder has volume formula V=πr²h. Rearrange it for h, then find h for radius 12 cm and exact volume 2592π cm³.

Work it through. The subject form is h=V/(πr²), provided r>0. With V=2592π cm³ and r=12 cm, h=(2592π)/(π×144)=18 cm. The π factors cancel without rounding.

Verify in the original formula. Check π×12²×18=2592π cm³. The base area 144π cm² times height 18 cm produces cubic centimetres as required.

Changed formula. At unchanged radius, double the volume to 5184π cm³. The height would double to 36 cm. This is a later-stage formula-rearrangement extension tied to familiar cylinder mensuration.

Rearrange a grouped fraction formula

When making rearrange a grouped fraction formula work, the most useful first question is ‘Which operation is applied to the variable last?’ Undo that operation on both complete sides before moving inward. A correct subject form should allow the original to be reconstructed. Watch for zero denominators, physical units and a distinction between a symbolic formula and a numerical answer.

184. A whole numerator divided 1: isolate x

Question. A grouped formula is Q=(2x−10)/2. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-13.

Why the inverse works. The complete numerator is one quantity divided by 2. Multiply both sides by the nonzero denominator 2 first, then remove the signed constant -10, then divide by the nonzero coefficient 2.

Work it through. Starting Q=(2x+(-10))/2, multiply both sides by 2: 2Q=2x+(-10). Subtract -10: 2Q−(-10)=2x. Thus x=(2Q−(-10))/2. At Q=-13 the intended exact output is x=-8.

First potential mistake. A common mistake is to divide only the constant -10 by 2 and leave the 2x outside the fraction. Another is to divide by 2 before removing -10. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-8 back: (2×(-8)+(-10))/2=-26/2=-26/2. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 2/2. The rearranged x rises by one to -7, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

185. A whole numerator divided 2: isolate x

Question. A grouped formula is Q=(5x−3)/7. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=12/7.

Why the inverse works. The complete numerator is one quantity divided by 7. Multiply both sides by the nonzero denominator 7 first, then remove the signed constant -3, then divide by the nonzero coefficient 5.

Work it through. Starting Q=(5x+(-3))/7, multiply both sides by 7: 7Q=5x+(-3). Subtract -3: 7Q−(-3)=5x. Thus x=(7Q−(-3))/5. At Q=12/7 the intended exact output is x=3.

First potential mistake. A common mistake is to divide only the constant -3 by 7 and leave the 5x outside the fraction. Another is to divide by 5 before removing -3. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=3 back: (5×(3)+(-3))/7=12/7=12/7. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 5/7. The rearranged x rises by one to 4, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

186. A whole numerator divided 3: isolate x

Question. A grouped formula is Q=(8x+4)/4. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-5.

Why the inverse works. The complete numerator is one quantity divided by 4. Multiply both sides by the nonzero denominator 4 first, then remove the signed constant 4, then divide by the nonzero coefficient 8.

Work it through. Starting Q=(8x+(4))/4, multiply both sides by 4: 4Q=8x+(4). Subtract 4: 4Q−(4)=8x. Thus x=(4Q−(4))/8. At Q=-5 the intended exact output is x=-3.

First potential mistake. A common mistake is to divide only the constant 4 by 4 and leave the 8x outside the fraction. Another is to divide by 8 before removing 4. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-3 back: (8×(-3)+(4))/4=-20/4=-20/4. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 8/4. The rearranged x rises by one to -2, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

187. A whole numerator divided 4: isolate x

Question. A grouped formula is Q=(4x−10)/9. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=22/9.

Why the inverse works. The complete numerator is one quantity divided by 9. Multiply both sides by the nonzero denominator 9 first, then remove the signed constant -10, then divide by the nonzero coefficient 4.

Work it through. Starting Q=(4x+(-10))/9, multiply both sides by 9: 9Q=4x+(-10). Subtract -10: 9Q−(-10)=4x. Thus x=(9Q−(-10))/4. At Q=22/9 the intended exact output is x=8.

First potential mistake. A common mistake is to divide only the constant -10 by 9 and leave the 4x outside the fraction. Another is to divide by 4 before removing -10. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=8 back: (4×(8)+(-10))/9=22/9=22/9. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 4/9. The rearranged x rises by one to 9, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

188. A whole numerator divided 5: isolate x

Question. A grouped formula is Q=(7x−3)/6. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=11/6.

Why the inverse works. The complete numerator is one quantity divided by 6. Multiply both sides by the nonzero denominator 6 first, then remove the signed constant -3, then divide by the nonzero coefficient 7.

Work it through. Starting Q=(7x+(-3))/6, multiply both sides by 6: 6Q=7x+(-3). Subtract -3: 6Q−(-3)=7x. Thus x=(6Q−(-3))/7. At Q=11/6 the intended exact output is x=2.

First potential mistake. A common mistake is to divide only the constant -3 by 6 and leave the 7x outside the fraction. Another is to divide by 7 before removing -3. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=2 back: (7×(2)+(-3))/6=11/6=11/6. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 7/6. The rearranged x rises by one to 3, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

189. A whole numerator divided 6: isolate x

Question. A grouped formula is Q=(3x+4)/3. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-8/3.

Why the inverse works. The complete numerator is one quantity divided by 3. Multiply both sides by the nonzero denominator 3 first, then remove the signed constant 4, then divide by the nonzero coefficient 3.

Work it through. Starting Q=(3x+(4))/3, multiply both sides by 3: 3Q=3x+(4). Subtract 4: 3Q−(4)=3x. Thus x=(3Q−(4))/3. At Q=-8/3 the intended exact output is x=-4.

First potential mistake. A common mistake is to divide only the constant 4 by 3 and leave the 3x outside the fraction. Another is to divide by 3 before removing 4. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-4 back: (3×(-4)+(4))/3=-8/3=-8/3. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 3/3. The rearranged x rises by one to -3, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

190. A whole numerator divided 7: isolate x

Question. A grouped formula is Q=(6x−10)/8. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=4.

Why the inverse works. The complete numerator is one quantity divided by 8. Multiply both sides by the nonzero denominator 8 first, then remove the signed constant -10, then divide by the nonzero coefficient 6.

Work it through. Starting Q=(6x+(-10))/8, multiply both sides by 8: 8Q=6x+(-10). Subtract -10: 8Q−(-10)=6x. Thus x=(8Q−(-10))/6. At Q=4 the intended exact output is x=7.

First potential mistake. A common mistake is to divide only the constant -10 by 8 and leave the 6x outside the fraction. Another is to divide by 6 before removing -10. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=7 back: (6×(7)+(-10))/8=32/8=32/8. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 6/8. The rearranged x rises by one to 8, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

191. A whole numerator divided 8: isolate x

Question. A grouped formula is Q=(2x−3)/5. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-1/5.

Why the inverse works. The complete numerator is one quantity divided by 5. Multiply both sides by the nonzero denominator 5 first, then remove the signed constant -3, then divide by the nonzero coefficient 2.

Work it through. Starting Q=(2x+(-3))/5, multiply both sides by 5: 5Q=2x+(-3). Subtract -3: 5Q−(-3)=2x. Thus x=(5Q−(-3))/2. At Q=-1/5 the intended exact output is x=1.

First potential mistake. A common mistake is to divide only the constant -3 by 5 and leave the 2x outside the fraction. Another is to divide by 2 before removing -3. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=1 back: (2×(1)+(-3))/5=-1/5=-1/5. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 2/5. The rearranged x rises by one to 2, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

192. A whole numerator divided 9: isolate x

Question. A grouped formula is Q=(5x+4)/2. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-21/2.

Why the inverse works. The complete numerator is one quantity divided by 2. Multiply both sides by the nonzero denominator 2 first, then remove the signed constant 4, then divide by the nonzero coefficient 5.

Work it through. Starting Q=(5x+(4))/2, multiply both sides by 2: 2Q=5x+(4). Subtract 4: 2Q−(4)=5x. Thus x=(2Q−(4))/5. At Q=-21/2 the intended exact output is x=-5.

First potential mistake. A common mistake is to divide only the constant 4 by 2 and leave the 5x outside the fraction. Another is to divide by 5 before removing 4. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-5 back: (5×(-5)+(4))/2=-21/2=-21/2. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 5/2. The rearranged x rises by one to -4, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

193. A whole numerator divided 10: isolate x

Question. A grouped formula is Q=(8x−10)/7. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=38/7.

Why the inverse works. The complete numerator is one quantity divided by 7. Multiply both sides by the nonzero denominator 7 first, then remove the signed constant -10, then divide by the nonzero coefficient 8.

Work it through. Starting Q=(8x+(-10))/7, multiply both sides by 7: 7Q=8x+(-10). Subtract -10: 7Q−(-10)=8x. Thus x=(7Q−(-10))/8. At Q=38/7 the intended exact output is x=6.

First potential mistake. A common mistake is to divide only the constant -10 by 7 and leave the 8x outside the fraction. Another is to divide by 8 before removing -10. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=6 back: (8×(6)+(-10))/7=38/7=38/7. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 8/7. The rearranged x rises by one to 7, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

194. A whole numerator divided 11: isolate x

Question. A grouped formula is Q=(4x−3)/4. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-3/4.

Why the inverse works. The complete numerator is one quantity divided by 4. Multiply both sides by the nonzero denominator 4 first, then remove the signed constant -3, then divide by the nonzero coefficient 4.

Work it through. Starting Q=(4x+(-3))/4, multiply both sides by 4: 4Q=4x+(-3). Subtract -3: 4Q−(-3)=4x. Thus x=(4Q−(-3))/4. At Q=-3/4 the intended exact output is x=0.

First potential mistake. A common mistake is to divide only the constant -3 by 4 and leave the 4x outside the fraction. Another is to divide by 4 before removing -3. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=0 back: (4×(0)+(-3))/4=-3/4=-3/4. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 4/4. The rearranged x rises by one to 1, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

195. A whole numerator divided 12: isolate x

Question. A grouped formula is Q=(7x+4)/9. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-38/9.

Why the inverse works. The complete numerator is one quantity divided by 9. Multiply both sides by the nonzero denominator 9 first, then remove the signed constant 4, then divide by the nonzero coefficient 7.

Work it through. Starting Q=(7x+(4))/9, multiply both sides by 9: 9Q=7x+(4). Subtract 4: 9Q−(4)=7x. Thus x=(9Q−(4))/7. At Q=-38/9 the intended exact output is x=-6.

First potential mistake. A common mistake is to divide only the constant 4 by 9 and leave the 7x outside the fraction. Another is to divide by 7 before removing 4. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-6 back: (7×(-6)+(4))/9=-38/9=-38/9. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 7/9. The rearranged x rises by one to -5, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

196. A whole numerator divided 13: isolate x

Question. A grouped formula is Q=(3x−10)/6. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=5/6.

Why the inverse works. The complete numerator is one quantity divided by 6. Multiply both sides by the nonzero denominator 6 first, then remove the signed constant -10, then divide by the nonzero coefficient 3.

Work it through. Starting Q=(3x+(-10))/6, multiply both sides by 6: 6Q=3x+(-10). Subtract -10: 6Q−(-10)=3x. Thus x=(6Q−(-10))/3. At Q=5/6 the intended exact output is x=5.

First potential mistake. A common mistake is to divide only the constant -10 by 6 and leave the 3x outside the fraction. Another is to divide by 3 before removing -10. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=5 back: (3×(5)+(-10))/6=5/6=5/6. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 3/6. The rearranged x rises by one to 6, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

197. A whole numerator divided 14: isolate x

Question. A grouped formula is Q=(6x−3)/3. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-3.

Why the inverse works. The complete numerator is one quantity divided by 3. Multiply both sides by the nonzero denominator 3 first, then remove the signed constant -3, then divide by the nonzero coefficient 6.

Work it through. Starting Q=(6x+(-3))/3, multiply both sides by 3: 3Q=6x+(-3). Subtract -3: 3Q−(-3)=6x. Thus x=(3Q−(-3))/6. At Q=-3 the intended exact output is x=-1.

First potential mistake. A common mistake is to divide only the constant -3 by 3 and leave the 6x outside the fraction. Another is to divide by 6 before removing -3. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-1 back: (6×(-1)+(-3))/3=-9/3=-9/3. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 6/3. The rearranged x rises by one to 0, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

198. A whole numerator divided 15: isolate x

Question. A grouped formula is Q=(2x+4)/8. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-5/4.

Why the inverse works. The complete numerator is one quantity divided by 8. Multiply both sides by the nonzero denominator 8 first, then remove the signed constant 4, then divide by the nonzero coefficient 2.

Work it through. Starting Q=(2x+(4))/8, multiply both sides by 8: 8Q=2x+(4). Subtract 4: 8Q−(4)=2x. Thus x=(8Q−(4))/2. At Q=-5/4 the intended exact output is x=-7.

First potential mistake. A common mistake is to divide only the constant 4 by 8 and leave the 2x outside the fraction. Another is to divide by 2 before removing 4. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-7 back: (2×(-7)+(4))/8=-10/8=-10/8. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 2/8. The rearranged x rises by one to -6, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

199. A whole numerator divided 16: isolate x

Question. A grouped formula is Q=(5x−10)/5. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=2.

Why the inverse works. The complete numerator is one quantity divided by 5. Multiply both sides by the nonzero denominator 5 first, then remove the signed constant -10, then divide by the nonzero coefficient 5.

Work it through. Starting Q=(5x+(-10))/5, multiply both sides by 5: 5Q=5x+(-10). Subtract -10: 5Q−(-10)=5x. Thus x=(5Q−(-10))/5. At Q=2 the intended exact output is x=4.

First potential mistake. A common mistake is to divide only the constant -10 by 5 and leave the 5x outside the fraction. Another is to divide by 5 before removing -10. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=4 back: (5×(4)+(-10))/5=10/5=10/5. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 5/5. The rearranged x rises by one to 5, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

200. A whole numerator divided 17: isolate x

Question. A grouped formula is Q=(8x−3)/2. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-19/2.

Why the inverse works. The complete numerator is one quantity divided by 2. Multiply both sides by the nonzero denominator 2 first, then remove the signed constant -3, then divide by the nonzero coefficient 8.

Work it through. Starting Q=(8x+(-3))/2, multiply both sides by 2: 2Q=8x+(-3). Subtract -3: 2Q−(-3)=8x. Thus x=(2Q−(-3))/8. At Q=-19/2 the intended exact output is x=-2.

First potential mistake. A common mistake is to divide only the constant -3 by 2 and leave the 8x outside the fraction. Another is to divide by 8 before removing -3. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-2 back: (8×(-2)+(-3))/2=-19/2=-19/2. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 8/2. The rearranged x rises by one to -1, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

201. A whole numerator divided 18: isolate x

Question. A grouped formula is Q=(4x+4)/7. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-4.

Why the inverse works. The complete numerator is one quantity divided by 7. Multiply both sides by the nonzero denominator 7 first, then remove the signed constant 4, then divide by the nonzero coefficient 4.

Work it through. Starting Q=(4x+(4))/7, multiply both sides by 7: 7Q=4x+(4). Subtract 4: 7Q−(4)=4x. Thus x=(7Q−(4))/4. At Q=-4 the intended exact output is x=-8.

First potential mistake. A common mistake is to divide only the constant 4 by 7 and leave the 4x outside the fraction. Another is to divide by 4 before removing 4. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-8 back: (4×(-8)+(4))/7=-28/7=-28/7. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 4/7. The rearranged x rises by one to -7, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

202. A whole numerator divided 19: isolate x

Question. A grouped formula is Q=(7x−10)/4. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=11/4.

Why the inverse works. The complete numerator is one quantity divided by 4. Multiply both sides by the nonzero denominator 4 first, then remove the signed constant -10, then divide by the nonzero coefficient 7.

Work it through. Starting Q=(7x+(-10))/4, multiply both sides by 4: 4Q=7x+(-10). Subtract -10: 4Q−(-10)=7x. Thus x=(4Q−(-10))/7. At Q=11/4 the intended exact output is x=3.

First potential mistake. A common mistake is to divide only the constant -10 by 4 and leave the 7x outside the fraction. Another is to divide by 7 before removing -10. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=3 back: (7×(3)+(-10))/4=11/4=11/4. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 7/4. The rearranged x rises by one to 4, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

203. A whole numerator divided 20: isolate x

Question. A grouped formula is Q=(3x−3)/9. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-4/3.

Why the inverse works. The complete numerator is one quantity divided by 9. Multiply both sides by the nonzero denominator 9 first, then remove the signed constant -3, then divide by the nonzero coefficient 3.

Work it through. Starting Q=(3x+(-3))/9, multiply both sides by 9: 9Q=3x+(-3). Subtract -3: 9Q−(-3)=3x. Thus x=(9Q−(-3))/3. At Q=-4/3 the intended exact output is x=-3.

First potential mistake. A common mistake is to divide only the constant -3 by 9 and leave the 3x outside the fraction. Another is to divide by 3 before removing -3. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-3 back: (3×(-3)+(-3))/9=-12/9=-12/9. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 3/9. The rearranged x rises by one to -2, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

204. A whole numerator divided 21: isolate x

Question. A grouped formula is Q=(6x+4)/6. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=26/3.

Why the inverse works. The complete numerator is one quantity divided by 6. Multiply both sides by the nonzero denominator 6 first, then remove the signed constant 4, then divide by the nonzero coefficient 6.

Work it through. Starting Q=(6x+(4))/6, multiply both sides by 6: 6Q=6x+(4). Subtract 4: 6Q−(4)=6x. Thus x=(6Q−(4))/6. At Q=26/3 the intended exact output is x=8.

First potential mistake. A common mistake is to divide only the constant 4 by 6 and leave the 6x outside the fraction. Another is to divide by 6 before removing 4. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=8 back: (6×(8)+(4))/6=52/6=52/6. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 6/6. The rearranged x rises by one to 9, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

205. A whole numerator divided 22: isolate x

Question. A grouped formula is Q=(2x−10)/3. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-2.

Why the inverse works. The complete numerator is one quantity divided by 3. Multiply both sides by the nonzero denominator 3 first, then remove the signed constant -10, then divide by the nonzero coefficient 2.

Work it through. Starting Q=(2x+(-10))/3, multiply both sides by 3: 3Q=2x+(-10). Subtract -10: 3Q−(-10)=2x. Thus x=(3Q−(-10))/2. At Q=-2 the intended exact output is x=2.

First potential mistake. A common mistake is to divide only the constant -10 by 3 and leave the 2x outside the fraction. Another is to divide by 2 before removing -10. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=2 back: (2×(2)+(-10))/3=-6/3=-6/3. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 2/3. The rearranged x rises by one to 3, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

206. A whole numerator divided 23: isolate x

Question. A grouped formula is Q=(5x−3)/8. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-23/8.

Why the inverse works. The complete numerator is one quantity divided by 8. Multiply both sides by the nonzero denominator 8 first, then remove the signed constant -3, then divide by the nonzero coefficient 5.

Work it through. Starting Q=(5x+(-3))/8, multiply both sides by 8: 8Q=5x+(-3). Subtract -3: 8Q−(-3)=5x. Thus x=(8Q−(-3))/5. At Q=-23/8 the intended exact output is x=-4.

First potential mistake. A common mistake is to divide only the constant -3 by 8 and leave the 5x outside the fraction. Another is to divide by 5 before removing -3. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-4 back: (5×(-4)+(-3))/8=-23/8=-23/8. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 5/8. The rearranged x rises by one to -3, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

207. A whole numerator divided 24: isolate x

Question. A grouped formula is Q=(8x+4)/5. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=12.

Why the inverse works. The complete numerator is one quantity divided by 5. Multiply both sides by the nonzero denominator 5 first, then remove the signed constant 4, then divide by the nonzero coefficient 8.

Work it through. Starting Q=(8x+(4))/5, multiply both sides by 5: 5Q=8x+(4). Subtract 4: 5Q−(4)=8x. Thus x=(5Q−(4))/8. At Q=12 the intended exact output is x=7.

First potential mistake. A common mistake is to divide only the constant 4 by 5 and leave the 8x outside the fraction. Another is to divide by 8 before removing 4. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=7 back: (8×(7)+(4))/5=60/5=60/5. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 8/5. The rearranged x rises by one to 8, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

208. A whole numerator divided 25: isolate x

Question. A grouped formula is Q=(4x−10)/2. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-3.

Work it through. Starting Q=(4x+(-10))/2, multiply both sides by 2: 2Q=4x+(-10). Subtract -10: 2Q−(-10)=4x. Thus x=(2Q−(-10))/4. At Q=-3 the intended exact output is x=1.

First potential mistake. A common mistake is to divide only the constant -10 by 2 and leave the 4x outside the fraction. Another is to divide by 4 before removing -10. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=1 back: (4×(1)+(-10))/2=-6/2=-6/2. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 4/2. The rearranged x rises by one to 2, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

209. A whole numerator divided 26: isolate x

Question. A grouped formula is Q=(7x−3)/7. Rearrange it to make x the subject, preserving the fact that the entire numerator is divided. Verify the result when Q=-38/7.

Work it through. Starting Q=(7x+(-3))/7, multiply both sides by 7: 7Q=7x+(-3). Subtract -3: 7Q−(-3)=7x. Thus x=(7Q−(-3))/7. At Q=-38/7 the intended exact output is x=-5.

First potential mistake. A common mistake is to divide only the constant -3 by 7 and leave the 7x outside the fraction. Another is to divide by 7 before removing -3. A displayed rounded Q may cause a tiny rounding residual, so use its exact definition for verification.

Verify in the original formula. Substitute x=-5 back: (7×(-5)+(-3))/7=-38/7=-38/7. The exact fraction represents the given Q without premature rounding, so the verification remains mathematically precise.

Changed formula. Increase the exact Q by 7/7. The rearranged x rises by one to -4, because c×(a/c) divided by a equals one. Explain the change symbolically rather than from a rounded decimal.

Know when rearrangement is not unique

When making know when rearrangement is not unique work, the most useful first question is ‘Which operation is applied to the variable last?’ Undo that operation on both complete sides before moving inward. A correct subject form should allow the original to be reconstructed. Watch for zero denominators, physical units and a distinction between a symbolic formula and a numerical answer.

210. An expression is not an equation

Question. Can 3x+5 be ‘solved for x’ by itself?

Why the inverse works. To decide whether a formula can be rearranged safely, identify the requested variable, the operations containing it, the current domain and any divisor that could become zero. The original and new forms must describe the same permitted relationships.

Work it through. No. It is an expression with no equality or condition. A value of x may be chosen for evaluation, but the expression alone does not determine x.

First potential mistake. Adding an equals sign without a second quantity invents missing information.

Verify in the original formula. If y=3x+5 is supplied, x can be isolated symbolically.

Changed formula. Ask the learner to repeat the idea with a changed coefficient or a different requested subject. An independent check should reconstruct the original formula and name any restriction; speed is less important than the equality-preserving reason.

211. A coefficient must not be zero when dividing

Question. Can x be isolated uniquely from y=0x+5?

Work it through. No. The output is always five, so y=5 gives every x and y≠5 gives no x. Division by zero is undefined.

First potential mistake. Using x=(y−5)/0 is not legitimate.

Verify in the original formula. Check x=1 and x=100 both give y=5.

212. The output can be used to check the rearrangement

Question. What is the most useful check after making x the subject?

Work it through. Substitute the proposed x expression into the original and simplify to recover the output or equality.

First potential mistake. Checking only a later transformed line may conceal a prior error.

Verify in the original formula. For y=2x+3, x=(y−3)/2 reconstructs y exactly.

213. Multiplying both sides preserves equality

Question. Why is it valid to multiply y=d/t by t when t is nonzero?

Work it through. Both full sides receive the same multiplication, giving yt=d.

First potential mistake. Multiplying the numerator but not the entire left side changes the relation.

Verify in the original formula. Divide yt=d by t again to recover y=d/t.

214. Units can expose a wrong rearrangement

Question. For A=lw, would w=A×l have the correct physical units?

Work it through. No. It yields area×length with cubic dimensions; w=A/l gives a length.

First potential mistake. Multiplying instead of dividing is inconsistent with the original dimensions.

Verify in the original formula. cm²/cm=cm, as a width should be.

215. A negative algebraic input may be valid

Question. For y=2x+3, is x=−2 necessarily wrong?

Work it through. No. Algebraic inputs may be negative; y=−1 follows from substitution.

First potential mistake. Treating all variables as physical lengths without an assigned context is unjustified.

Verify in the original formula. 2(−2)+3=−1.

216. A physical length cannot be negative

Question. A rearrangement gives cylinder height h=−3 cm. Does that describe a valid physical cylinder height?

Work it through. No. It means the given physical model or inputs are inconsistent, even if a formal manipulation produced the number.

First potential mistake. Reporting a negative physical height as ordinary valid geometry confuses algebra and context.

Verify in the original formula. Height must be nonnegative and positive for a nondegenerate cylinder.

217. A fraction needs a nonzero denominator

Question. When is v=d/t defined?

Work it through. Only when t≠0 under ordinary real arithmetic.

First potential mistake. Cancelling t before noting its restriction may hide undefined division.

Verify in the original formula. State t>0 for elapsed time in a positive-duration model.

218. Changing subject does not change a formula’s meaning

Question. Are v=d/t and d=vt equivalent under t≠0?

Work it through. Yes. They encode the same relationship with different isolated variables.

First potential mistake. Treating the new subject as a new physical law is incorrect.

Verify in the original formula. Substitute example d=100,t=2,v=50.

219. Reverse the order of operations

Question. For y=3(x+2), find x.

Work it through. Divide by three, then subtract two: x=y/3−2.

First potential mistake. Subtracting two before dividing improperly changes the grouped term.

Verify in the original formula. 3((y/3−2)+2)=y.

220. A minus outside a bracket requires grouping

Question. Solve y=−(x+4) for x.

Work it through. Multiply by negative one: −y=x+4, so x=−y−4.

First potential mistake. Negating only x but not the constant changes the equation.

Verify in the original formula. At x=1, y=−5; inverse gives x=1.

221. Rearranging a ratio needs a stated relationship

Question. If A:B=2:3, can B be written in terms of A?

Work it through. Yes, B=(3/2)A for positive A and B with that ratio.

First potential mistake. Writing B=(2/3)A reverses the ratio order.

Verify in the original formula. If A=4, B=6.

222. A formula can involve a reciprocal

Question. For y=12/x and nonzero x,y, solve for x.

Work it through. xy=12, so x=12/y.

First potential mistake. Dividing 12 by x twice does not isolate the unknown.

Verify in the original formula. At y=3, x=4.

223. A given equation may have two valid inverse values

Question. For y=x², solve x in terms of y over real x.

Work it through. When y≥0, x=±√y; generally two possible x except at y=0.

First potential mistake. Using only positive principal square root and claiming uniqueness without a domain restriction.

Verify in the original formula. At y=9, both x=3 and −3 work.

224. A zero-rate question can be non-unique

Question. A constant total cost C=5 independent of n is given. Can n be found from C alone?

Work it through. No. Every n in the allowed domain gives same cost, so the model cannot identify one count.

First potential mistake. Dividing by a zero per-use rate is undefined.

Verify in the original formula. C=5+0n for any n.

225. An equation with two unknowns may only permit one rearrangement

Question. Can y=2x+3 make x the subject without knowing y numerically?

Work it through. Yes, x=(y−3)/2, a formula expression. It does not select one numeric x until y is given.

First potential mistake. Confusing symbolic subject change with solving for a particular numeric value.

Verify in the original formula. At y=11, x=4.

226. A height formula may require a square root in a different geometry

Question. If A=πr² and A>0, find the positive radius r.

Work it through. r=√(A/π). Taking the positive root is appropriate for physical radius.

First potential mistake. Claiming r=A/π forgets to reverse the square.

Verify in the original formula. If A=25π, r=5.

227. Brackets protect the complete numerator

Question. For y=(x+5)/3, find x.

Work it through. Multiply by three: 3y=x+5, so x=3y−5.

First potential mistake. Writing x=3(y−5) would multiply the constant too strongly.

Verify in the original formula. At y=4, x=7 and (7+5)/3=4.

228. A formula’s coefficient may itself be a variable

Question. For A=bh/2, rearrange for b assuming h≠0.

Work it through. b=2A/h.

First potential mistake. Treating b and h as fixed numeric coefficients rather than algebraic symbols can hide the general relation.

Verify in the original formula. ½(2A/h)h=A.

229. A zero area with positive base implies zero height

Question. For A=bh/2 with b>0, A=0. What is h?

Work it through. h=0, a degenerate triangle under ordinary geometry.

First potential mistake. Inferring negative height from zero area is unsupported.

Verify in the original formula. ½b×0=0.

230. A symbol for π does not mean a chosen decimal

Question. For C=2πr, find r in terms of C.

Work it through. r=C/(2π), exact when C is symbolic.

First potential mistake. Replacing π with 3.14 without instruction loses exactness.

Verify in the original formula. 2π(C/(2π))=C.

231. The inverse of addition differs from multiplication

Question. For C=F+rn, why not divide C by r first?

Work it through. F is added after rn, so remove F first and then divide by r.

First potential mistake. A wrong inverse order makes the fixed fee act like repeated usage.

Verify in the original formula. (C−F)/r reconstructs n.

232. Average and sum are not interchangeable

Question. For M=(a+b+c)/3, solve for c.

Work it through. c=3M−a−b.

First potential mistake. Subtracting a+b from M before multiplying by three changes the equation.

Verify in the original formula. Substituting back gives 3M/3=M.

233. Solve for a rate in a formula

Question. For d=vt with t>0, find v.

Work it through. v=d/t.

First potential mistake. Multiplying d×t rather than dividing produces the wrong unit.

Verify in the original formula. (d/t)t=d.

234. Reverse a temperature formula

Question. For T=12−2t, make t subject.

Work it through. t=(12−T)/2.

First potential mistake. Writing (T−12)/2 reverses the sign.

Verify in the original formula. At T=4, t=4.

235. A one-variable inequality is not a formula equality

Question. For 2x+3≤11, solve the condition.

Work it through. 2x≤8, so x≤4. This is a set/range, not one isolated numeric subject value.

First potential mistake. Writing x=4 alone omits other permitted values.

Verify in the original formula. x=3 works; x=5 fails.

236. Negative division reverses an inequality

Question. For −2x<6, what is the solution set?

Work it through. Divide by negative two and reverse < to >: x>−3.

First potential mistake. Keeping < after dividing by a negative changes the allowed set.

Verify in the original formula. x=0 works since 0<6, consistent with x>−3.

237. Rearranging an algebraic fraction must retain restrictions

Question. For y=1/x, can x=1/y be used at y=0?

Work it through. No. The original excludes x=0 and produces only nonzero y; inverse excludes y=0.

First potential mistake. Dropping the denominator restriction after rearrangement creates undefined values.

Verify in the original formula. At y=2, x=1/2.

238. A formula can include a constant term from measurement

Question. For P=2l+2w, is w=(P−2l)/2 equivalent to P/2−l?

Work it through. Yes, distribute division by two: (P−2l)/2=P/2−l.

First potential mistake. Assuming only one written appearance can be valid overlooks equivalent algebraic forms.

Verify in the original formula. Substitute P=30,l=9 to get w=6 both ways.

239. Units of mass density

Question. For m=ρV, make V subject if ρ>0.

Work it through. V=m/ρ. If m is g and ρ is g/cm³, the resulting volume unit is cm³.

First potential mistake. Writing V=mρ produces the wrong compound units.

Verify in the original formula. (g)/(g/cm³)=cm³.

240. Rearrange proportion safely

Question. If y=kx with k≠0, make x subject.

Work it through. x=y/k.

First potential mistake. Subtracting k from y confuses multiplication with addition.

Verify in the original formula. k(y/k)=y.

241. A correct subject change can be verified symbolically

Question. Show x=(y−b)/a is inverse of y=ax+b for a≠0.

Work it through. Substitute: a((y−b)/a)+b=y−b+b=y.

First potential mistake. Checking only a numerical case does not prove equivalence generally.

Verify in the original formula. The algebraic cancellation works for every allowed y.

242. A zero radius can be a special case

Question. For V=πr²h with r=0, can h be found uniquely from V=0?

Work it through. No. Every h gives V=0 under that degenerate formula, so dividing by πr²=0 is invalid.

First potential mistake. Announcing h=V/(πr²) without requiring r>0 hides undefined division.

Verify in the original formula. Try h=1, h=100: both V=0 at r=0.

243. An external measurement may differ from an internal formula value

Question. A box’s exterior length is known, but a formula uses internal volume. Can outside dimensions be substituted without wall information?

Work it through. Not safely. The formula needs the dimensions it defines; physical wall thickness may matter.

First potential mistake. A correct algebraic rearrangement cannot repair mismatched measurements.

Verify in the original formula. Check whether lengths are internal or external.

244. A ratio requires correct orientation

Question. For y=(a/b)x, make x the subject when a,b≠0.

Work it through. x=(b/a)y.

First potential mistake. Keeping a/b instead of reciprocal when solving reverses the direction.

Verify in the original formula. (a/b)(b/a)y=y.

245. Changing the subject should answer the requested letter

Question. A question says make h the subject of V=Ah. Is A=V/h the requested final result?

Work it through. No. It makes A the subject, whereas h=V/A is required for A≠0.

First potential mistake. A mathematically valid related formula can still answer a different question.

Verify in the original formula. V/A times A reconstructs V.

246. The word ‘formula’ doesn’t mean physical certainty

Question. A fictional line y=2t+3 fits a short set of readings. Does changing its subject prove the law lasts forever?

Work it through. No. Algebraic equivalence preserves the mathematical model, but the model’s real-world validity has separate assumptions and domain.

First potential mistake. Treating a derived expression as new empirical evidence overclaims.

Verify in the original formula. Distinguish mathematical relation from observation.

247. A delayed retest checks independent interpretation

Question. A pupil has rearranged P=2(l+w) once after tutoring. What next?

Work it through. Use a new rectangle with changed data, ask for the subject before showing steps, then verify with the original perimeter formula.

First potential mistake. Rewriting the same completed steps may only test copying.

Verify in the original formula. Change P and l; check new width independently.

Four rearrangement errors and their proper repair

Only one term is divided rather than a grouped sum

For y=(3x+5)/2, multiply the whole equality by two first: 2y=3x+5, so x=(2y−5)/3. Writing x=2y−5/3 divides only the five by three and changes the mathematical relationship. Ask the child to draw a long fraction bar covering the complete numerator, then rewrite with explicit brackets.

The operations are undone in the wrong order

When y=4x+7, the final step of the forward machine is +7; subtract seven from y before dividing by four. The correct subject form is x=(y−7)/4. Dividing y by four and then subtracting seven is not the inverse of this machine. Use arrows to illustrate reverse order, then translate the picture into balanced equation lines.

A zero divisor or physical restriction is ignored

From y=ax+b, x=(y−b)/a assumes a≠0. If a=0, the output is constant and may not determine a unique x. In a cylinder, r must be positive when dividing by πr² to isolate height. State these restrictions rather than treating an invalid division as a harmless shortcut.

The wrong letter is isolated

The instruction ‘make h the subject’ cannot be answered by giving A=V/h, even though that equation follows from V=Ah. Underline the requested variable and inspect the final form before submission. In applied questions, label what the subject measures, such as centimetres for height or hours for time.

What a three-pupil tutorial could teach

The eduKateSG Clementi small-group Mathematics programme describes 90-minute tutorials in groups of up to three near Sixth Avenue MRT. A suitable G1 session could place one pupil on ordinary linear-equation balance, another on making a rectangle’s width the subject, and a third on rearranging a grouped fraction with domain checks. The tutor can assess whether every child can rebuild the original, not merely copy a symbolic answer. This is a format illustration, not a confirmation of current class availability.

A possible lesson uses ten minutes of equation retrieval, twenty minutes comparing numeric and symbolic answers, twenty-five minutes of individual subject-changing exercises, fifteen minutes of area and rate applications, ten minutes of an unseen formula and ten minutes to correct the first invalid transformation. A Secondary 2 learner should not be rushed into later-stage manipulations when negative signs, fractions or one-variable equations are still unstable.

Five brief practice sessions for parents

Day one: solve 3x+5=20, then compare with making x the subject of y=3x+5.
Day two: make width the subject of area=length×width and check units.
Day three: rearrange a fixed-plus-repeated-cost formula from total cost to number of uses.
Day four: find time from average speed and distance, checking physical units.
Day five: compare one valid and one invalid rearrangement, explaining the first incorrect operation.

Frequently asked questions

What does ‘subject of a formula’ mean?

It is the variable isolated on one side of an equality, expressed in terms of the other quantities.

Is making x the subject the same as solving for one numeric x?

Not always. A subject form is often a general symbolic relation. A numerical value follows only after the other quantities are supplied.

Why must coefficients be nonzero for division?

Because division by zero is undefined; a zero coefficient may make the inverse non-unique or inconsistent.

Does G1 Mathematics include formula rearrangement?

The published 2027 K110 full graduating syllabus includes changing the subject of simple formulae. The MOE 2020 Secondary Two G1 content focuses on one-variable equations and expression simplification; current school teaching scope is the immediate guide.

How do I check a rearranged formula?

Substitute it into the original equality and simplify to recover the original relation, or insert suitable values in both forms and compare.

What if a physical result is negative?

Check whether a negative quantity is meaningful in context. A negative temperature can be meaningful, but a negative physical length or volume ordinarily cannot.

Official syllabus and reader notes

Reader note: All formulas, prices and measurements are original teaching illustrations. They are not official past-paper questions, physical specifications or assurances about current Secondary Two school topics. Division requires nonzero relevant factors, and physical interpretations require suitable domains. Current tuition fees, venue, teacher and availability must be confirmed directly.

Bring a school formula your child can substitute into but cannot rearrange to eduKate Singapore and ask which inverse operation a tutor would teach first.

Secondary 2 G1 Mathematics Tutor Bukit Timah | Changing the Subject of a Formula