Sec 2 G1 Maths Tutor Bukit Timah | Variables Both Sides

Did you know? The same unknown can appear on both sides of a Mathematics equation without turning it into a two-variable problem. For example, 3x+5=2x+9 still asks for just one number, and the answer is x=4. This matters because many pupils know how to solve 3x+5=17 but hesitate as soon as another x appears after the equals sign. The correction begins with a simple rule: whatever you do to one side, do to the other.

Quick answer: solving x on both sides

In 3x+5=2x+9, subtract 2x from both sides to leave x+5=9, then subtract five to obtain x=4. Check: 3(4)+5=17 and 2(4)+9=17. An equation can also have no solution, as in 2x+5=2x+7, or every real number as a solution, as in 2x+5=2x+5. Never divide by a variable coefficient that has become zero.

This guide supports pupils and parents searching for a Secondary 2 G1 Mathematics tutor in Bukit Timah. The MOE Secondary Two G1 Mathematics outline includes solving linear equations in one variable and formulating them from word problems. The focus here is the special technique of keeping equality valid when x appears on both sides, building on ordinary signed numbers and single-bracket expansion. The pupil’s school’s current G1 syllabus and teacher instructions determine the immediate question difficulty.

Jump to the part that fixes the mistake

First-lesson diagnostic
Keep both sides balanced
Handle brackets and negative coefficients
One, none or infinitely many solutions
Original equations and real-world clinics
Five common errors
A three-pupil tutorial model
Parent practice routine
Frequently asked questions
Official sources

The one-minute diagnostic

Ask the pupil to solve 4x+3=2x+11 and explain each step. The correct answer is x=4: subtract 2x to get 2x+3=11, then 2x=8. Now change the equation to 4x+3=4x+11. After subtracting 4x, it says 3=11, which cannot hold, so there is no solution. This contrasting pair quickly shows whether the child knows the difference between balancing an equation and blindly dividing coefficients.

The equals sign means the two totals match

Collect the variable terms without changing only one side

For 5x+2=3x+10, subtract 3x from both sides, leaving 2x+2=10. Then remove the constant and divide: 2x=8, x=4. In written working it is common to say ‘move 3x across and change the sign’, but the underlying reason is that subtracting an equal quantity from equal totals preserves equality. Teaching that reason makes the method survive changes of sign.

Collect constants separately

The goal is a statement such as 2x=8, not a jumble of x and ordinary units. If the equation is 5x−3=2x+9, subtract 2x first to get 3x−3=9; add three to both sides; get x=4. Trying to add −3 to x-terms because everything appears on one line combines unlike quantities and loses the structure.

A negative coefficient is still part of x

In −3x+7=x−9, the difference between coefficients is −4. Rearrange to −4x=−16, hence x=4. The two negative signs dividing give a positive solution. The pupil should predict the likely sign and substitute into the original. Making up a rule such as ‘every negative sign disappears when it crosses equals’ causes repeated errors.

Expand the full bracket before collecting x

Consider 2(x+4)=x+13. Expand to 2x+8=x+13, then subtract x and subtract eight to get x=5. Each term inside a bracket must be multiplied. When a minus is outside a bracket, as in 4x−(x+2)=10, distribute −1 to both x and 2; the correct line is 3x−2=10, leading to x=4. Use parentheses during substitution to avoid accidentally altering the original grouping.

Why an equation can have one, no or many solutions

If subtracting variable terms leaves 0x=2, no real x makes the statement true. If it leaves 0x=0, every real x works. If a nonzero coefficient remains, dividing by that coefficient yields one real solution for an ordinary linear equation. These outcomes are not errors in the question. They are different mathematical descriptions of the relationship between the two sides.

A tutor can show three very similar examples: 2x+5=3x+3 has unique solution x=2; 2x+5=2x+7 has no solution; 2x+5=2x+5 has all real solutions. This is a stronger lesson than memorising ‘collect x on the left’, because the learner must think about the equality after the x terms cancel.

91 original linear-equation clinics

These cases include direct x-on-both-sides equations, brackets on both sides, imaginary equal-cost plans and short diagnostic checks. All numbers are original teaching examples, not copied school tests. Attempt a few at a time, hide the worked lines and check the found value in both original expressions. The first invalid transformation is the most useful tutoring clue.

1. Equation 1: collect variables and constants

Question. Solve -5x − 8 = −x + 12. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -1x from both sides to get -4x − 8=12. Then -4x=20; divide by -4 (which is nonzero) to obtain x=-5. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -1x from right to left without changing its sign, they might use -6x instead of (-5−-1)x. Equally, moving the constant -8 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-5, the left side becomes 17, and the right side becomes 17. Both are exactly 17. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 12 by 8. The resulting solution should be x=-4, because the difference between x-coefficients remains -4. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

2. Equation 2: collect variables and constants

Question. Solve 2x + 3 = 4x + 7. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 4x from both sides to get -2x + 3=7. Then -2x=4; divide by -2 (which is nonzero) to obtain x=-2. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 4x from right to left without changing its sign, they might use 6x instead of (2−4)x. Equally, moving the constant 3 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-2, the left side becomes -1, and the right side becomes -1. Both are exactly -1. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 7 by 5. The resulting solution should be x=-1, because the difference between x-coefficients remains -2. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

3. Equation 3: collect variables and constants

Question. Solve -2x − 9 = -11. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 0x from both sides to get -2x − 9=-11. Then -2x=-2; divide by -2 (which is nonzero) to obtain x=1. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 0x from right to left without changing its sign, they might use -2x instead of (-2−0)x. Equally, moving the constant -9 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=1, the left side becomes -11, and the right side becomes -11. Both are exactly -11. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -11 by -13. The resulting solution should be x=2, because the difference between x-coefficients remains -2. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

4. Equation 4: collect variables and constants

Question. Solve 5x + 2 = -4x + 38. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -4x from both sides to get 9x + 2=38. Then 9x=36; divide by 9 (which is nonzero) to obtain x=4. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -4x from right to left without changing its sign, they might use 1x instead of (5−-4)x. Equally, moving the constant 2 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=4, the left side becomes 22, and the right side becomes 22. Both are exactly 22. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 38 by 47. The resulting solution should be x=5, because the difference between x-coefficients remains 9. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

5. Equation 5: collect variables and constants

Question. Solve x − 10 = 3x − 24. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 3x from both sides to get -2x − 10=-24. Then -2x=-14; divide by -2 (which is nonzero) to obtain x=7. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 3x from right to left without changing its sign, they might use 4x instead of (1−3)x. Equally, moving the constant -10 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=7, the left side becomes -3, and the right side becomes -3. Both are exactly -3. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -24 by -26. The resulting solution should be x=8, because the difference between x-coefficients remains -2. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

6. Equation 6: collect variables and constants

Question. Solve -3x + 1 = −x + 11. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -1x from both sides to get -2x + 1=11. Then -2x=10; divide by -2 (which is nonzero) to obtain x=-5. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -1x from right to left without changing its sign, they might use -4x instead of (-3−-1)x. Equally, moving the constant 1 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-5, the left side becomes 16, and the right side becomes 16. Both are exactly 16. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 11 by 9. The resulting solution should be x=-4, because the difference between x-coefficients remains -2. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

7. Equation 7: collect variables and constants

Question. Solve 4x − 11 = 2x − 15. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 2x from both sides to get 2x − 11=-15. Then 2x=-4; divide by 2 (which is nonzero) to obtain x=-2. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 2x from right to left without changing its sign, they might use 6x instead of (4−2)x. Equally, moving the constant -11 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-2, the left side becomes -19, and the right side becomes -19. Both are exactly -19. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -15 by -13. The resulting solution should be x=-1, because the difference between x-coefficients remains 2. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

8. Equation 8: collect variables and constants

Question. Solve = -2x + 2. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -2x from both sides to get 2x=2. Then 2x=2; divide by 2 (which is nonzero) to obtain x=1. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -2x from right to left without changing its sign, they might use -2x instead of (0−-2)x. Equally, moving the constant 0 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=1, the left side becomes 0, and the right side becomes 0. Both are exactly 0. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 2 by 4. The resulting solution should be x=2, because the difference between x-coefficients remains 2. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

9. Equation 9: collect variables and constants

Question. Solve -4x + 11 = 3x − 17. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 3x from both sides to get -7x + 11=-17. Then -7x=-28; divide by -7 (which is nonzero) to obtain x=4. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 3x from right to left without changing its sign, they might use -1x instead of (-4−3)x. Equally, moving the constant 11 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=4, the left side becomes -5, and the right side becomes -5. Both are exactly -5. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -17 by -24. The resulting solution should be x=5, because the difference between x-coefficients remains -7. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

10. Equation 10: collect variables and constants

Question. Solve 3x − 1 = −x + 27. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -1x from both sides to get 4x − 1=27. Then 4x=28; divide by 4 (which is nonzero) to obtain x=7. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -1x from right to left without changing its sign, they might use 2x instead of (3−-1)x. Equally, moving the constant -1 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=7, the left side becomes 20, and the right side becomes 20. Both are exactly 20. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 27 by 31. The resulting solution should be x=8, because the difference between x-coefficients remains 4. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

11. Equation 11: collect variables and constants

Question. Solve −x + 10 = 4x + 35. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 4x from both sides to get -5x + 10=35. Then -5x=25; divide by -5 (which is nonzero) to obtain x=-5. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 4x from right to left without changing its sign, they might use 3x instead of (-1−4)x. Equally, moving the constant 10 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-5, the left side becomes 15, and the right side becomes 15. Both are exactly 15. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 35 by 30. The resulting solution should be x=-4, because the difference between x-coefficients remains -5. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

12. Equation 12: collect variables and constants

Question. Solve -5x − 2 = 8. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 0x from both sides to get -5x − 2=8. Then -5x=10; divide by -5 (which is nonzero) to obtain x=-2. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 0x from right to left without changing its sign, they might use -5x instead of (-5−0)x. Equally, moving the constant -2 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-2, the left side becomes 8, and the right side becomes 8. Both are exactly 8. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 8 by 3. The resulting solution should be x=-1, because the difference between x-coefficients remains -5. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

13. Equation 13: collect variables and constants

Question. Solve 2x + 9 = -4x + 15. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -4x from both sides to get 6x + 9=15. Then 6x=6; divide by 6 (which is nonzero) to obtain x=1. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -4x from right to left without changing its sign, they might use -2x instead of (2−-4)x. Equally, moving the constant 9 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=1, the left side becomes 11, and the right side becomes 11. Both are exactly 11. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 15 by 21. The resulting solution should be x=2, because the difference between x-coefficients remains 6. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

14. Equation 14: collect variables and constants

Question. Solve -2x − 3 = x − 15. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 1x from both sides to get -3x − 3=-15. Then -3x=-12; divide by -3 (which is nonzero) to obtain x=4. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 1x from right to left without changing its sign, they might use -1x instead of (-2−1)x. Equally, moving the constant -3 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=4, the left side becomes -11, and the right side becomes -11. Both are exactly -11. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -15 by -18. The resulting solution should be x=5, because the difference between x-coefficients remains -3. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

15. Equation 15: collect variables and constants

Question. Solve 5x + 8 = -3x + 64. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -3x from both sides to get 8x + 8=64. Then 8x=56; divide by 8 (which is nonzero) to obtain x=7. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -3x from right to left without changing its sign, they might use 2x instead of (5−-3)x. Equally, moving the constant 8 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=7, the left side becomes 43, and the right side becomes 43. Both are exactly 43. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 64 by 72. The resulting solution should be x=8, because the difference between x-coefficients remains 8. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

16. Equation 16: collect variables and constants

Question. Solve x − 4 = 2x + 1. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 2x from both sides to get -1x − 4=1. Then -1x=5; divide by -1 (which is nonzero) to obtain x=-5. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 2x from right to left without changing its sign, they might use 3x instead of (1−2)x. Equally, moving the constant -4 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-5, the left side becomes -9, and the right side becomes -9. Both are exactly -9. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 1 by 0. The resulting solution should be x=-4, because the difference between x-coefficients remains -1. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

17. Equation 17: collect variables and constants

Question. Solve -3x + 7 = -2x + 9. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -2x from both sides to get -1x + 7=9. Then -1x=2; divide by -1 (which is nonzero) to obtain x=-2. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -2x from right to left without changing its sign, they might use -5x instead of (-3−-2)x. Equally, moving the constant 7 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-2, the left side becomes 13, and the right side becomes 13. Both are exactly 13. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 9 by 8. The resulting solution should be x=-1, because the difference between x-coefficients remains -1. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

18. Equation 18: collect variables and constants

Question. Solve 4x − 5 = 3x − 4. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 3x from both sides to get 1x − 5=-4. Then 1x=1; divide by 1 (which is nonzero) to obtain x=1. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 3x from right to left without changing its sign, they might use 7x instead of (4−3)x. Equally, moving the constant -5 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=1, the left side becomes -1, and the right side becomes -1. Both are exactly -1. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -4 by -3. The resulting solution should be x=2, because the difference between x-coefficients remains 1. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

19. Equation 19: collect variables and constants

Question. Solve 6 = −x + 10. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -1x from both sides to get 1x + 6=10. Then 1x=4; divide by 1 (which is nonzero) to obtain x=4. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -1x from right to left without changing its sign, they might use -1x instead of (0−-1)x. Equally, moving the constant 6 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=4, the left side becomes 6, and the right side becomes 6. Both are exactly 6. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 10 by 11. The resulting solution should be x=5, because the difference between x-coefficients remains 1. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

20. Equation 20: collect variables and constants

Question. Solve -4x − 6 = 4x − 62. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 4x from both sides to get -8x − 6=-62. Then -8x=-56; divide by -8 (which is nonzero) to obtain x=7. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 4x from right to left without changing its sign, they might use 0x instead of (-4−4)x. Equally, moving the constant -6 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=7, the left side becomes -34, and the right side becomes -34. Both are exactly -34. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -62 by -70. The resulting solution should be x=8, because the difference between x-coefficients remains -8. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

21. Equation 21: collect variables and constants

Question. Solve 3x + 5 = -10. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 0x from both sides to get 3x + 5=-10. Then 3x=-15; divide by 3 (which is nonzero) to obtain x=-5. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 0x from right to left without changing its sign, they might use 3x instead of (3−0)x. Equally, moving the constant 5 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-5, the left side becomes -10, and the right side becomes -10. Both are exactly -10. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -10 by -7. The resulting solution should be x=-4, because the difference between x-coefficients remains 3. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

22. Equation 22: collect variables and constants

Question. Solve −x − 7 = -4x − 13. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -4x from both sides to get 3x − 7=-13. Then 3x=-6; divide by 3 (which is nonzero) to obtain x=-2. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -4x from right to left without changing its sign, they might use -5x instead of (-1−-4)x. Equally, moving the constant -7 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-2, the left side becomes -5, and the right side becomes -5. Both are exactly -5. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -13 by -10. The resulting solution should be x=-1, because the difference between x-coefficients remains 3. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

23. Equation 23: collect variables and constants

Question. Solve -5x + 4 = x − 2. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 1x from both sides to get -6x + 4=-2. Then -6x=-6; divide by -6 (which is nonzero) to obtain x=1. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 1x from right to left without changing its sign, they might use -4x instead of (-5−1)x. Equally, moving the constant 4 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=1, the left side becomes -1, and the right side becomes -1. Both are exactly -1. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -2 by -8. The resulting solution should be x=2, because the difference between x-coefficients remains -6. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

24. Equation 24: collect variables and constants

Question. Solve 2x − 8 = -3x + 12. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -3x from both sides to get 5x − 8=12. Then 5x=20; divide by 5 (which is nonzero) to obtain x=4. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -3x from right to left without changing its sign, they might use -1x instead of (2−-3)x. Equally, moving the constant -8 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=4, the left side becomes 0, and the right side becomes 0. Both are exactly 0. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 12 by 17. The resulting solution should be x=5, because the difference between x-coefficients remains 5. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

25. Equation 25: collect variables and constants

Question. Solve -2x + 3 = 2x − 25. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 2x from both sides to get -4x + 3=-25. Then -4x=-28; divide by -4 (which is nonzero) to obtain x=7. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 2x from right to left without changing its sign, they might use 0x instead of (-2−2)x. Equally, moving the constant 3 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=7, the left side becomes -11, and the right side becomes -11. Both are exactly -11. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -25 by -29. The resulting solution should be x=8, because the difference between x-coefficients remains -4. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

26. Equation 26: collect variables and constants

Question. Solve 5x − 9 = -2x − 44. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -2x from both sides to get 7x − 9=-44. Then 7x=-35; divide by 7 (which is nonzero) to obtain x=-5. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -2x from right to left without changing its sign, they might use 3x instead of (5−-2)x. Equally, moving the constant -9 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-5, the left side becomes -34, and the right side becomes -34. Both are exactly -34. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -44 by -37. The resulting solution should be x=-4, because the difference between x-coefficients remains 7. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

27. Equation 27: collect variables and constants

Question. Solve x + 2 = 3x + 6. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 3x from both sides to get -2x + 2=6. Then -2x=4; divide by -2 (which is nonzero) to obtain x=-2. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 3x from right to left without changing its sign, they might use 4x instead of (1−3)x. Equally, moving the constant 2 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=-2, the left side becomes 0, and the right side becomes 0. Both are exactly 0. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant 6 by 4. The resulting solution should be x=-1, because the difference between x-coefficients remains -2. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

28. Equation 28: collect variables and constants

Question. Solve -3x − 10 = −x − 12. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract -1x from both sides to get -2x − 10=-12. Then -2x=-2; divide by -2 (which is nonzero) to obtain x=1. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves -1x from right to left without changing its sign, they might use -4x instead of (-3−-1)x. Equally, moving the constant -10 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=1, the left side becomes -13, and the right side becomes -13. Both are exactly -13. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -12 by -14. The resulting solution should be x=2, because the difference between x-coefficients remains -2. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

29. Equation 29: collect variables and constants

Question. Solve 4x + 1 = 6x − 7. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 6x from both sides to get -2x + 1=-7. Then -2x=-8; divide by -2 (which is nonzero) to obtain x=4. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 6x from right to left without changing its sign, they might use 10x instead of (4−6)x. Equally, moving the constant 1 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=4, the left side becomes 17, and the right side becomes 17. Both are exactly 17. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -7 by -9. The resulting solution should be x=5, because the difference between x-coefficients remains -2. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

30. Equation 30: collect variables and constants

Question. Solve -11 = 2x − 25. Find one value of x that makes both sides equal, without treating the letters as different variables.

Worked reasoning. Subtract 2x from both sides to get -2x − 11=-25. Then -2x=-14; divide by -2 (which is nonzero) to obtain x=7. The value is the shared balance point, not merely a term that appeared in the question.

First-step trap. If the student moves 2x from right to left without changing its sign, they might use 2x instead of (0−2)x. Equally, moving the constant -11 requires subtracting it from the entire equality. Repeated guessing can land on the right answer without teaching balance.

Check. At x=7, the left side becomes -11, and the right side becomes -11. Both are exactly -11. This direct substitution detects signs or division errors.

Independent next question. Replace the right constant -25 by -27. The resulting solution should be x=8, because the difference between x-coefficients remains -2. This is a changed-number test: the student must understand why the answer shifts rather than copying the old solution.

31. Brackets and both sides 1

Question. Solve -4(x − 2) − 5 = -2(x + 2) + 15. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes -4x + 3, right becomes -2x + 11. Subtract the -2x term from both sides: (-4−-2)x=8. This gives x=-4. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -4(x+-2) but forgetting -4 times the constant, or expanding the right side and losing its outer 15. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-4, the left side is 19; the right side is 19. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=-4 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

32. Brackets and both sides 2

Question. Solve -1(x + 2) + 4 = -4(x − 3) − 7. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes −x + 2, right becomes -4x + 5. Subtract the -4x term from both sides: (-1−-4)x=3. This gives x=1. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -1(x+2) but forgetting -1 times the constant, or expanding the right side and losing its outer -7. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=1, the left side is 1; the right side is 1. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=1 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

33. Brackets and both sides 3

Question. Solve 2(x − 5) = 3(x − 1) − 13. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes 2x − 10, right becomes 3x − 16. Subtract the 3x term from both sides: (2−3)x=-6. This gives x=6. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding 2(x+-5) but forgetting 2 times the constant, or expanding the right side and losing its outer -13. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=6, the left side is 2; the right side is 2. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=6 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

34. Brackets and both sides 4

Question. Solve -4(x − 1) − 4 = 1(x + 1) + 9. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes -4x, right becomes x + 10. Subtract the 1x term from both sides: (-4−1)x=10. This gives x=-2. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -4(x+-1) but forgetting -4 times the constant, or expanding the right side and losing its outer 9. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-2, the left side is 8; the right side is 8. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=-2 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

35. Brackets and both sides 5

Question. Solve -1(x + 3) + 5 = 0(x + 3) − 1. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes −x + 2, right becomes -1. Subtract the 0x term from both sides: (-1−0)x=-3. This gives x=3. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -1(x+3) but forgetting -1 times the constant, or expanding the right side and losing its outer -1. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=3, the left side is -1; the right side is -1. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=3 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

36. Brackets and both sides 6

Question. Solve 2(x − 4) + 1 = -3(x − 2) − 38. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes 2x − 7, right becomes -3x − 32. Subtract the -3x term from both sides: (2−-3)x=-25. This gives x=-5. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding 2(x+-4) but forgetting 2 times the constant, or expanding the right side and losing its outer -38. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-5, the left side is -17; the right side is -17. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=-5 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

37. Brackets and both sides 7

Question. Solve -4(x) − 3 = 4(x) − 3. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes -4x − 3, right becomes 4x − 3. Subtract the 4x term from both sides: (-4−4)x=0. This gives x=0. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -4(x+0) but forgetting -4 times the constant, or expanding the right side and losing its outer -3. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=0, the left side is -3; the right side is -3. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=0 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

38. Brackets and both sides 8

Question. Solve -1(x + 4) + 6 = 2(x + 2) − 17. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes −x + 2, right becomes 2x − 13. Subtract the 2x term from both sides: (-1−2)x=-15. This gives x=5. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -1(x+4) but forgetting -1 times the constant, or expanding the right side and losing its outer -17. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=5, the left side is -3; the right side is -3. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=5 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

39. Brackets and both sides 9

Question. Solve 2(x − 3) + 2 = 0(x − 3) − 10. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes 2x − 4, right becomes -10. Subtract the 0x term from both sides: (2−0)x=-6. This gives x=-3. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding 2(x+-3) but forgetting 2 times the constant, or expanding the right side and losing its outer -10. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-3, the left side is -10; the right side is -10. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=-3 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

40. Brackets and both sides 10

Question. Solve -4(x + 1) − 2 = -2(x − 1) − 12. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes -4x − 6, right becomes -2x − 10. Subtract the -2x term from both sides: (-4−-2)x=-4. This gives x=2. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -4(x+1) but forgetting -4 times the constant, or expanding the right side and losing its outer -12. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=2, the left side is -14; the right side is -14. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=2 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

41. Brackets and both sides 11

Question. Solve -1(x + 5) − 6 = -4(x + 1) − 25. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes −x − 11, right becomes -4x − 29. Subtract the -4x term from both sides: (-1−-4)x=-18. This gives x=-6. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -1(x+5) but forgetting -1 times the constant, or expanding the right side and losing its outer -25. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-6, the left side is -5; the right side is -5. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=-6 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

42. Brackets and both sides 12

Question. Solve 2(x − 2) + 3 = 3(x + 3) − 9. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes 2x − 1, right becomes 3x. Subtract the 3x term from both sides: (2−3)x=1. This gives x=-1. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding 2(x+-2) but forgetting 2 times the constant, or expanding the right side and losing its outer -9. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-1, the left side is -3; the right side is -3. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=-1 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

43. Brackets and both sides 13

Question. Solve -4(x + 2) − 1 = 1(x − 2) − 27. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes -4x − 9, right becomes x − 29. Subtract the 1x term from both sides: (-4−1)x=-20. This gives x=4. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -4(x+2) but forgetting -4 times the constant, or expanding the right side and losing its outer -27. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=4, the left side is -25; the right side is -25. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

Independent next question. Ask the pupil to re-solve the equation after increasing both outer constants by the same amount. The solution stays x=4 because both sides receive the same addition. Explain this property rather than guessing from the numeric result.

44. Brackets and both sides 14

Question. Solve -1(x − 5) − 5 = 0(x) + 4. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes −x, right becomes 4. Subtract the 0x term from both sides: (-1−0)x=4. This gives x=-4. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -1(x+-5) but forgetting -1 times the constant, or expanding the right side and losing its outer 4. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-4, the left side is 4; the right side is 4. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

45. Brackets and both sides 15

Question. Solve 2(x − 1) + 4 = -3(x + 2) + 13. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes 2x + 2, right becomes -3x + 7. Subtract the -3x term from both sides: (2−-3)x=5. This gives x=1. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding 2(x+-1) but forgetting 2 times the constant, or expanding the right side and losing its outer 13. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=1, the left side is 4; the right side is 4. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

46. Brackets and both sides 16

Question. Solve -4(x + 3) = 4(x − 3) − 48. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes -4x − 12, right becomes 4x − 60. Subtract the 4x term from both sides: (-4−4)x=-48. This gives x=6. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -4(x+3) but forgetting -4 times the constant, or expanding the right side and losing its outer -48. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=6, the left side is -36; the right side is -36. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

47. Brackets and both sides 17

Question. Solve -1(x − 4) − 4 = 2(x − 1) + 8. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes −x, right becomes 2x + 6. Subtract the 2x term from both sides: (-1−2)x=6. This gives x=-2. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -1(x+-4) but forgetting -1 times the constant, or expanding the right side and losing its outer 8. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-2, the left side is 2; the right side is 2. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

48. Brackets and both sides 18

Question. Solve 2(x) + 5 = 0(x + 1) + 11. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes 2x + 5, right becomes 11. Subtract the 0x term from both sides: (2−0)x=6. This gives x=3. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding 2(x+0) but forgetting 2 times the constant, or expanding the right side and losing its outer 11. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=3, the left side is 11; the right side is 11. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

49. Brackets and both sides 19

Question. Solve -4(x + 4) + 1 = -2(x + 3) + 1. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes -4x − 15, right becomes -2x − 5. Subtract the -2x term from both sides: (-4−-2)x=10. This gives x=-5. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -4(x+4) but forgetting -4 times the constant, or expanding the right side and losing its outer 1. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-5, the left side is 5; the right side is 5. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

50. Brackets and both sides 20

Question. Solve -1(x − 3) − 3 = -4(x − 2) − 8. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes −x, right becomes -4x. Subtract the -4x term from both sides: (-1−-4)x=0. This gives x=0. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -1(x+-3) but forgetting -1 times the constant, or expanding the right side and losing its outer -8. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=0, the left side is 0; the right side is 0. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

51. Brackets and both sides 21

Question. Solve 2(x + 1) + 6 = 3(x) + 3. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes 2x + 8, right becomes 3x + 3. Subtract the 3x term from both sides: (2−3)x=-5. This gives x=5. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding 2(x+1) but forgetting 2 times the constant, or expanding the right side and losing its outer 3. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=5, the left side is 18; the right side is 18. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

52. Brackets and both sides 22

Question. Solve -4(x + 5) + 2 = 1(x + 2) − 5. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes -4x − 18, right becomes x − 3. Subtract the 1x term from both sides: (-4−1)x=15. This gives x=-3. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -4(x+5) but forgetting -4 times the constant, or expanding the right side and losing its outer -5. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-3, the left side is -6; the right side is -6. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

53. Brackets and both sides 23

Question. Solve -1(x − 2) − 2 = 0(x − 3) − 2. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes −x, right becomes -2. Subtract the 0x term from both sides: (-1−0)x=-2. This gives x=2. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -1(x+-2) but forgetting -1 times the constant, or expanding the right side and losing its outer -2. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=2, the left side is -2; the right side is -2. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

54. Brackets and both sides 24

Question. Solve 2(x + 2) − 6 = -3(x − 1) − 35. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes 2x − 2, right becomes -3x − 32. Subtract the -3x term from both sides: (2−-3)x=-30. This gives x=-6. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding 2(x+2) but forgetting 2 times the constant, or expanding the right side and losing its outer -35. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-6, the left side is -14; the right side is -14. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

55. Brackets and both sides 25

Question. Solve -4(x − 5) + 3 = 4(x + 1) + 27. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes -4x + 23, right becomes 4x + 31. Subtract the 4x term from both sides: (-4−4)x=8. This gives x=-1. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -4(x+-5) but forgetting -4 times the constant, or expanding the right side and losing its outer 27. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-1, the left side is 27; the right side is 27. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

56. Brackets and both sides 26

Question. Solve -1(x − 1) − 1 = 2(x + 3) − 18. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes −x, right becomes 2x − 12. Subtract the 2x term from both sides: (-1−2)x=-12. This gives x=4. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -1(x+-1) but forgetting -1 times the constant, or expanding the right side and losing its outer -18. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=4, the left side is -4; the right side is -4. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

57. Brackets and both sides 27

Question. Solve 2(x + 3) − 5 = 0(x − 2) − 7. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes 2x + 1, right becomes -7. Subtract the 0x term from both sides: (2−0)x=-8. This gives x=-4. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding 2(x+3) but forgetting 2 times the constant, or expanding the right side and losing its outer -7. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=-4, the left side is -7; the right side is -7. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

58. Brackets and both sides 28

Question. Solve -4(x − 4) + 4 = -2(x) + 18. This question needs expansion and equation balance; there is still only one unknown.

Worked reasoning. Expand both complete bracket products. Left becomes -4x + 20, right becomes -2x + 18. Subtract the -2x term from both sides: (-4−-2)x=-2. This gives x=1. Keep the constant contributions from both brackets visible before combining.

First-step trap. Common first errors are expanding -4(x+-4) but forgetting -4 times the constant, or expanding the right side and losing its outer 18. When removing a bracket or changing a sign, rewrite both sides so the equality remains intact.

Check. At x=1, the left side is 16; the right side is 16. They match. A correct answer should be checked in the original bracketed form, not only in an intermediate equation.

59. Two fictional pricing plans 1: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $5 and charges $2 for each visit. Plan B has a fixed fee of $8 and charges $1 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 5+2n=8+1n. Subtract 1n from both sides and subtract 5 from both sides to get (2−1)n=3. The model’s equal-cost point is n=3 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

First-step trap. Choosing the plan with the lower per-visit price does not answer when the two totals are equal. The fixed fees matter. A negative or fractional solution would also need interpretation if visits are whole counts, though these examples use feasible positive integers.

Check. At 3 visits, Plan A totals $11; Plan B totals $11. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $2 and Plan B by $1; their totals would differ by $1. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

60. Two fictional pricing plans 2: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $15 and charges $4 for each visit. Plan B has a fixed fee of $3 and charges $6 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 15+4n=3+6n. Subtract 6n from both sides and subtract 15 from both sides to get (4−6)n=-12. The model’s equal-cost point is n=6 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 6 visits, Plan A totals $39; Plan B totals $39. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $4 and Plan B by $6; their totals would differ by $2. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

61. Two fictional pricing plans 3: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $15 and charges $6 for each visit. Plan B has a fixed fee of $42 and charges $3 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 15+6n=42+3n. Subtract 3n from both sides and subtract 15 from both sides to get (6−3)n=27. The model’s equal-cost point is n=9 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 9 visits, Plan A totals $69; Plan B totals $69. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $6 and Plan B by $3; their totals would differ by $3. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

62. Two fictional pricing plans 4: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $15 and charges $8 for each visit. Plan B has a fixed fee of $3 and charges $9 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 15+8n=3+9n. Subtract 9n from both sides and subtract 15 from both sides to get (8−9)n=-12. The model’s equal-cost point is n=12 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 12 visits, Plan A totals $111; Plan B totals $111. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $8 and Plan B by $9; their totals would differ by $1. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

63. Two fictional pricing plans 5: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $8 and charges $10 for each visit. Plan B has a fixed fee of $23 and charges $5 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 8+10n=23+5n. Subtract 5n from both sides and subtract 8 from both sides to get (10−5)n=15. The model’s equal-cost point is n=3 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 3 visits, Plan A totals $38; Plan B totals $38. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $10 and Plan B by $5; their totals would differ by $5. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

64. Two fictional pricing plans 6: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $13 and charges $3 for each visit. Plan B has a fixed fee of $19 and charges $2 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 13+3n=19+2n. Subtract 2n from both sides and subtract 13 from both sides to get (3−2)n=6. The model’s equal-cost point is n=6 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 6 visits, Plan A totals $31; Plan B totals $31. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $3 and Plan B by $2; their totals would differ by $1. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

65. Two fictional pricing plans 7: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $18 and charges $5 for each visit. Plan B has a fixed fee of $0 and charges $7 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 18+5n=0+7n. Subtract 7n from both sides and subtract 18 from both sides to get (5−7)n=-18. The model’s equal-cost point is n=9 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 9 visits, Plan A totals $63; Plan B totals $63. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $5 and Plan B by $7; their totals would differ by $2. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

66. Two fictional pricing plans 8: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $6 and charges $7 for each visit. Plan B has a fixed fee of $42 and charges $4 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 6+7n=42+4n. Subtract 4n from both sides and subtract 6 from both sides to get (7−4)n=36. The model’s equal-cost point is n=12 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 12 visits, Plan A totals $90; Plan B totals $90. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $7 and Plan B by $4; their totals would differ by $3. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

67. Two fictional pricing plans 9: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $11 and charges $9 for each visit. Plan B has a fixed fee of $35 and charges $1 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 11+9n=35+1n. Subtract 1n from both sides and subtract 11 from both sides to get (9−1)n=24. The model’s equal-cost point is n=3 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 3 visits, Plan A totals $38; Plan B totals $38. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $9 and Plan B by $1; their totals would differ by $8. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

68. Two fictional pricing plans 10: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $27 and charges $2 for each visit. Plan B has a fixed fee of $3 and charges $6 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 27+2n=3+6n. Subtract 6n from both sides and subtract 27 from both sides to get (2−6)n=-24. The model’s equal-cost point is n=6 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 6 visits, Plan A totals $39; Plan B totals $39. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $2 and Plan B by $6; their totals would differ by $4. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

69. Two fictional pricing plans 11: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $4 and charges $4 for each visit. Plan B has a fixed fee of $13 and charges $3 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 4+4n=13+3n. Subtract 3n from both sides and subtract 4 from both sides to get (4−3)n=9. The model’s equal-cost point is n=9 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 9 visits, Plan A totals $40; Plan B totals $40. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $4 and Plan B by $3; their totals would differ by $1. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

70. Two fictional pricing plans 12: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $27 and charges $6 for each visit. Plan B has a fixed fee of $3 and charges $8 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 27+6n=3+8n. Subtract 8n from both sides and subtract 27 from both sides to get (6−8)n=-24. The model’s equal-cost point is n=12 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 12 visits, Plan A totals $99; Plan B totals $99. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $6 and Plan B by $8; their totals would differ by $2. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

71. Two fictional pricing plans 13: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $14 and charges $8 for each visit. Plan B has a fixed fee of $23 and charges $5 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 14+8n=23+5n. Subtract 5n from both sides and subtract 14 from both sides to get (8−5)n=9. The model’s equal-cost point is n=3 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 3 visits, Plan A totals $38; Plan B totals $38. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $8 and Plan B by $5; their totals would differ by $3. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

72. Two fictional pricing plans 14: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $2 and charges $10 for each visit. Plan B has a fixed fee of $50 and charges $2 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 2+10n=50+2n. Subtract 2n from both sides and subtract 2 from both sides to get (10−2)n=48. The model’s equal-cost point is n=6 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 6 visits, Plan A totals $62; Plan B totals $62. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $10 and Plan B by $2; their totals would differ by $8. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

73. Two fictional pricing plans 15: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $39 and charges $3 for each visit. Plan B has a fixed fee of $3 and charges $7 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 39+3n=3+7n. Subtract 7n from both sides and subtract 39 from both sides to get (3−7)n=-36. The model’s equal-cost point is n=9 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 9 visits, Plan A totals $66; Plan B totals $66. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $3 and Plan B by $7; their totals would differ by $4. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

74. Two fictional pricing plans 16: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $12 and charges $5 for each visit. Plan B has a fixed fee of $24 and charges $4 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 12+5n=24+4n. Subtract 4n from both sides and subtract 12 from both sides to get (5−4)n=12. The model’s equal-cost point is n=12 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 12 visits, Plan A totals $72; Plan B totals $72. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $5 and Plan B by $4; their totals would differ by $1. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

75. Two fictional pricing plans 17: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $17 and charges $7 for each visit. Plan B has a fixed fee of $35 and charges $1 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 17+7n=35+1n. Subtract 1n from both sides and subtract 17 from both sides to get (7−1)n=18. The model’s equal-cost point is n=3 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 3 visits, Plan A totals $38; Plan B totals $38. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $7 and Plan B by $1; their totals would differ by $6. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

76. Two fictional pricing plans 18: when are totals equal?

Question. In a made-up comparison, Plan A has a fixed fee of $5 and charges $9 for each visit. Plan B has a fixed fee of $23 and charges $6 for each visit. For what number of visits do their total costs match?

Worked reasoning. Let n count visits. Write 5+9n=23+6n. Subtract 6n from both sides and subtract 5 from both sides to get (9−6)n=18. The model’s equal-cost point is n=6 visits. It is an equation with the same variable appearing on each side, not two unrelated calculations.

Check. At 6 visits, Plan A totals $59; Plan B totals $59. They agree exactly. Real businesses might have caps and discounts, but those are not part of this fictional model.

Independent next question. For one additional visit, Plan A increases by $9 and Plan B by $6; their totals would differ by $3. Ask which plan becomes cheaper beyond the equality point using the difference in rates rather than recomputing a long table.

77. Variables cancel and reveal an identity

Question. Solve 2x+5=2x+5.

Worked reasoning. Subtract 2x from both sides; what remains is 5=5, always true. Every real x is a solution.

First-step trap. An answer x=0 is one valid example but not the complete solution set.

Check. Choose x=−3,0,7: both sides match every time.

Independent next question. Change one of the constants by one and decide whether the equation now has one, none or infinitely many solutions. State why x-coefficients that cancel can leave either a true identity or a contradiction. Explain the first important transformation aloud and retest without the original answer visible.

78. Variables cancel and reveal a contradiction

Question. Solve 2x+5=2x+7.

Worked reasoning. Subtract 2x from both sides: 5=7, impossible. No solution.

First-step trap. Dividing by zero after both variable coefficients cancel is invalid.

Check. No substitution can change a fixed difference of two between the sides.

79. A linear equation need not have one solution

Question. Classify 3(x+2)=3x+6.

Worked reasoning. Expansion gives 3x+6=3x+6, so every real x satisfies the equation.

First-step trap. Guessing just x=6 because it appears as a constant.

Check. Subtract the identical sides; the result is 0=0.

80. A contradiction can hide in brackets

Question. Classify 2(x+3)=2x+5.

Worked reasoning. Expand to 2x+6=2x+5; subtract 2x and get 6=5. No solution.

First-step trap. Moving the x-terms and trying to divide by zero creates an undefined step.

Check. The left side always exceeds the right by one.

81. Negative coefficients still permit a unique balance

Question. Solve −3x+7=x−9.

Worked reasoning. Subtract x and seven: −4x=−16, so x=4.

First-step trap. Changing −3x to positive 3x when moving between sides.

Check. Left is −12+7=−5; right is 4−9=−5.

82. A fraction coefficient on both sides

Question. Solve x/2+3=x/4+7.

Worked reasoning. Subtract x/4 from both sides: x/4+3=7, so x/4=4 and x=16.

First-step trap. Combining 1/2 and 1/4 coefficients by adding denominators incorrectly.

Check. At 16, left 8+3=11 and right 4+7=11.

83. An inequality is not an equation

Question. The question is 3x+1>2x+6. What answer type is required?

Worked reasoning. Subtract 2x then one: x>5. This gives a range, not the single value x=5.

First-step trap. Treating the > symbol as an equals sign would incorrectly include the boundary.

Check. At x=6, 19>18; at x=5, 16>16 is false.

84. Units in a context need care

Question. A hypothetical same-cost model solves to n=3.5 visits, but visits are whole numbers. What does that mean?

Worked reasoning. There is no integer visit count where costs are exactly equal under that model. The algebraic intersection is 3.5 in a continuous extension.

First-step trap. Rounding 3.5 to four changes the claim from exact equality to a different comparison.

Check. Compute both plans at three and four whole visits instead, if a decision is needed.

85. Equation balance can be checked with a scale analogy

Question. Explain why subtracting x from each side of 3x+5=2x+9 is valid.

Worked reasoning. Subtracting the same quantity from two equal totals preserves equality. The equation becomes 2x+5=x+9 and finally x=4.

First-step trap. Moving x across the equals sign without understanding equivalent operations can lose a sign.

Check. At x=4, both original sides equal seventeen.

86. A minus outside a bracket affects every term

Question. Solve 4x−(x+2)=10.

Worked reasoning. Expand as 4x−x−2=10, so 3x=12 and x=4.

First-step trap. Writing 4x−x+2=10 forgets that the negative multiplies +2 as well.

Check. Original at four: 16−6=10.

87. Subtracting a negative variable term

Question. Solve 5x+2=−x+14.

Worked reasoning. Add x to both sides to get 6x+2=14, so x=2.

First-step trap. Subtracting x when the right coefficient is −1 may produce 4x incorrectly.

Check. Both original sides equal twelve when x=2.

88. The equation may include a zero coefficient

Question. Solve 0x+7=3x+1.

Worked reasoning. This is 7=3x+1, so x=2. Zero times x disappears without making the whole equation meaningless.

First-step trap. Assuming any zero coefficient creates infinitely many solutions.

Check. At x=2 both sides equal seven.

89. Different paths can solve the same equation

Question. Solve 2(x+4)=x+13 by expansion or by simple balance.

Worked reasoning. Expand left to 2x+8=x+13, subtract x then eight: x=5. Both approaches preserve equality.

First-step trap. Believing a memorised sequence is the only valid method hides the invariant.

Check. At x=5, 2(9)=18 and 5+13=18.

90. One example cannot prove identity

Question. A classmate says 2x+5=3x+3 is true for every x because it works at x=2. Is that correct?

Worked reasoning. No. It works at x=2 only; solving yields x=2. At x=0, the sides are five and three.

First-step trap. One successful substitution demonstrates a candidate, not an identity.

Check. Distinguish a unique solution from all real numbers.

91. No calculator needed for a final sign check

Question. Solve −2x+3=4x−9.

Worked reasoning. Subtract 4x and three to get −6x=−12, so x=2. The two negatives divide to positive.

First-step trap. The student may divide −12 by 6 and report −2.

Check. At x=2, both sides equal −1.

Five recurring errors and their repairs

A variable term is moved without changing its operation

If 5x+2=3x+10 becomes 8x+2=10, the pupil has added the coefficients instead of subtracting 3x from both sides. Show the equal operation written on both sides of the equation, then simplify. A different-number check will show whether the reasoning is understood rather than just copied.

A negative bracket is partially expanded

The expression 4x−(x+2) becomes 4x−x−2, not 4x−x+2. Give the student a numerical test at x=1 to expose the wrong expansion before proceeding. A tutor can focus on multiplication by −1 without re-teaching all algebra.

The pupil divides by zero after x cancels

An equation like 2x+5=2x+7 reduces to 5=7. It is a contradiction, not an invitation to divide by 2−2=0. State no solution. Compare with 5=5 after cancellation, which means every real x satisfies the identity.

The answer is found but not checked

A result x=4 is only convincing if substituting four restores the original balance. For 3x+5=2x+9, each side must equal seventeen. If they differ, the sign, bracket or constant was changed incorrectly. Check the original question rather than an intermediate rearranged version.

A real-world x has the wrong interpretation

A fictional equal-cost equation may produce n=3.5 when n counts whole visits. That is a valid intersection in a continuous linear model but not a feasible exact whole-visit equality. Parents and pupils should report the mathematical result, then discuss the physical domain if the question asks for an actual decision.

A focused three-pupil Mathematics lesson

The immutable eduKateSG Clementi Maths tutorial reference describes 90-minute lessons for three pupils near Sixth Avenue MRT. A Secondary 2 G1 lesson can give one pupil simple coefficient movement, another negative bracket expansion and a third interpretation of no-solution identities. This is an illustrative differentiated teaching approach, not confirmation of a current class opening.

10 minutes: retrieve signed numbers, basic balancing and bracket expansion.
20 minutes: compare one-variable equations with and without x on both sides.
25 minutes: work individual equations with feedback on the first invalid line.
15 minutes: classify one-solution, no-solution and identity cases.
10 minutes: attempt an unseen equal-cost story independently.
10 minutes: substitute final results and set a short delayed retest.

A short five-day home plan

Day 1: solve three positive-coefficient equations with x on both sides.
Day 2: include one negative coefficient and check by substitution.
Day 3: expand one bracket on each side before collecting variables.
Day 4: compare a unique solution, a contradiction and an identity.
Day 5: form an equal-cost equation from a small made-up table and interpret the result.

A useful parent prompt is ‘What did you do to both sides on this line?’ Ask it at the point where the signs change. If the pupil can explain why equality survives the operation and solve a changed example without hints, the underlying skill is strengthening.

Frequently asked questions

Are variables on both sides still one-variable equations?

Yes. The symbol x represents the same unknown wherever it appears. There are not two unknown quantities merely because there are two x terms.

What if the x terms cancel?

If a true statement remains, all real x are solutions. If a false statement remains, there is no solution.

Can I always divide by the remaining x coefficient?

Only if it is nonzero. Dividing by zero is undefined.

Which sign should I change when moving x across?

Rather than memorising a sign-switch slogan, subtract or add the same term to both sides, then simplify.

Is this appropriate for Secondary 2 G1?

Solving linear equations in one variable is in the official Secondary Two G1 outline. The teacher’s current chapter and problem complexity determine the immediate practice target.

How can a tutor check genuine improvement?

Ask the learner to solve a changed equation independently, explain one equality-preserving transformation and substitute the answer into both sides.

Official references and reader notes

Reader note: All equations and fictional pricing scenarios are original educational examples, not official school questions or actual commercial prices. A correct solution should satisfy the original relation; a real-world model must also respect its domain. Verify the student’s current subject level, tutorial timing, fees and teacher placement directly.

Bring one marked x-on-both-sides equation to eduKate Singapore and ask how a tutor would identify whether the first issue is sign movement, bracket expansion or interpreting the solution.

Secondary 2 G1 Mathematics Tutor Bukit Timah | Equations with Variables on Both Sides

Quick answer: solving x on both sides

Jump to the part that fixes the mistake

The one-minute diagnostic

The equals sign means the two totals match

Collect the variable terms without changing only one side

Collect constants separately

A negative coefficient is still part of x

Expand the full bracket before collecting x

Why an equation can have one, no or many solutions

91 original linear-equation clinics

1. Equation 1: collect variables and constants

2. Equation 2: collect variables and constants

3. Equation 3: collect variables and constants

4. Equation 4: collect variables and constants

5. Equation 5: collect variables and constants

6. Equation 6: collect variables and constants

7. Equation 7: collect variables and constants

8. Equation 8: collect variables and constants

9. Equation 9: collect variables and constants

10. Equation 10: collect variables and constants

11. Equation 11: collect variables and constants

12. Equation 12: collect variables and constants

13. Equation 13: collect variables and constants

14. Equation 14: collect variables and constants

15. Equation 15: collect variables and constants

16. Equation 16: collect variables and constants

17. Equation 17: collect variables and constants

18. Equation 18: collect variables and constants

19. Equation 19: collect variables and constants

20. Equation 20: collect variables and constants

21. Equation 21: collect variables and constants

22. Equation 22: collect variables and constants

23. Equation 23: collect variables and constants

24. Equation 24: collect variables and constants

25. Equation 25: collect variables and constants

26. Equation 26: collect variables and constants

27. Equation 27: collect variables and constants

28. Equation 28: collect variables and constants

29. Equation 29: collect variables and constants

30. Equation 30: collect variables and constants

31. Brackets and both sides 1

32. Brackets and both sides 2

33. Brackets and both sides 3

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58. Brackets and both sides 28

59. Two fictional pricing plans 1: when are totals equal?

60. Two fictional pricing plans 2: when are totals equal?

61. Two fictional pricing plans 3: when are totals equal?

62. Two fictional pricing plans 4: when are totals equal?

63. Two fictional pricing plans 5: when are totals equal?

64. Two fictional pricing plans 6: when are totals equal?

65. Two fictional pricing plans 7: when are totals equal?

66. Two fictional pricing plans 8: when are totals equal?

67. Two fictional pricing plans 9: when are totals equal?

68. Two fictional pricing plans 10: when are totals equal?

69. Two fictional pricing plans 11: when are totals equal?