Sec 3 G3 Maths Tutor Bukit Timah | Changing Formula Subject

Secondary 3 G3 Mathematics Tutor Bukit Timah | Changing the Subject of a Formula Without Guesswork — Did you know? A formula is useful in more than one direction. When a question gives area but asks for a width, or gives speed but asks for time, the relationship itself may already be familiar. The missing skill is often changing which variable stands alone. Here is the fast explanation, followed by carefully checked practice for readers who want to become more independent.

Quick answer: how do you change the subject?

Step 1: Circle the requested variable. Write down what the question wants before moving any symbols.
Step 2: Identify which operations surround the variable, including whole brackets and fractions.
Step 3: Undo those operations with equivalent steps on both sides of the equation.
Step 4: Check the new denominator and any relevant restrictions on the variables.
Step 5: Substitute permitted sample values into the original and rearranged formulas to check equivalence.

The 2027 Singapore–Cambridge SEC G3 Mathematics (K310) syllabus includes changing the subject of a formula and finding unknown quantities in given formulas. The content is taught across the secondary course, so your school’s Secondary 3 chapter order may differ. This reading guide is aligned to mathematical principles rather than to a claim that every school reaches the same page in the same week.

Skip to: The big ideas · Worked examples · Common mistakes · Six-week study plan · Questions parents ask.

The important ideas behind rearrangement

Changing the subject changes the viewpoint

A = lw makes area the subject: it tells you what area will be when length and width are given. Dividing the same equality by a nonzero length l gives w = A/l. The formula has not become a different geometric fact; the same relationship has been viewed from the width’s perspective.

Students often confuse rearrangement with moving numbers around until an answer looks convenient. The correct aim is equivalence. Each line must be true whenever the previous line is true under the stated domain conditions. That is what makes the final expression trustworthy.

Reverse the operations in a sensible order

To recover x from y = 3x + 5, subtract five then divide by three. This gives x = (y−5)/3. But y = 3(x+5) is different: dividing by three first gives y/3 = x+5, then subtracting five gives x = y/3−5.

A bracket is a mathematical boundary. It shows what the multiplication applies to. A tutor can help by asking, ‘What was the last operation performed on x to make y?’ Undoing the outer operation first often prevents the most common sign and bracket mistakes.

Denominators can impose restrictions

In y = (ax+b)/c, with a and c nonzero, multiply both sides by c. Now cy=ax+b. Subtract b and divide by a to obtain x=(cy−b)/a. The whole numerator must remain grouped.

A rearrangement is not an excuse to divide by zero. If a symbol used as a divisor could be zero, mention that restriction. A numerical value satisfying a transformed expression may still be invalid in the original formula, so checking the initial relationship remains essential.

Square roots need mathematical and practical checks

If a circular area A=πr² is given and r is required, isolate r²=A/π and then use r=√(A/π) for a positive geometric radius. The algebra of square equations may yield two signs, but a physical radius cannot be negative.

Similarly, y=√(x+c) has nonnegative output. Squaring both sides may produce x=y²−c, but only nonnegative y belongs to the original relationship. The student needs to understand not only the inverse operation but also which answers the starting statement allows.

Units reveal the rearrangement

Suppose distance d = speed v multiplied by time t. Making t the subject gives t=d/v. The units agree: kilometres divided by kilometres per hour leaves hours. Making v the subject gives v=d/t, with units km/h.

Mathematical communication improves when learners state what each letter means. An expression may be symbolically correct while the final number is reported as the wrong physical quantity. A short sentence explaining the answer in context can catch this omission.

Reverse checks save marks

A student can test a rearrangement by undoing the steps or substituting small permitted numbers into both forms. For example, if w=A/l and A=30, l=5, then w=6. Returning to A=lw gives 30=5×6. Agreement supports the result.

The check need not be lengthy. The strongest habit is to know what error it could expose: the missing bracket, an inverted fraction, a lost coefficient or an invalid denominator. The student should be able to explain the checking method rather than treat it as another unexplained button press.

50 checked examples: one question at a time

The following are original constructed teaching examples, not predictions of school examination questions. There are ten types of rearrangement, each repeated with five different numerical settings. Choose one example, cover the method, solve it independently, reveal the check, then return after a short gap. The aim is to choose lawful operations without a tutor giving the first line.

Rectangle width from area

This cluster isolates rectangle width from area. Look for the operation that must be undone first, write one valid line at a time, and remember that the same relationship should remain true after rearrangement. Mixed practice later is more useful than trying to memorise a distinct shortcut for every context.

Worked example 1. Question: Area A = lw. Make w the subject; then find w when A = 15 cm² and l = 3 cm. Method: Divide both sides by l (l is nonzero): w = A/l = 15/3 = 5 cm. Check: Checking the original: 3 × 5 = 15 cm². The relationship is preserved. Watch for: Dividing l by A would reverse the ratio.

Take the idea further: What happens to w if the same area is shared over double the length? Name the subject before calculating. A student who identifies the correct unknown has already completed an essential reasoning step. Try a version with negative values where the context permits them, and explain which signs must be preserved.

Perimeter and brackets

This cluster isolates perimeter and brackets. Look for the operation that must be undone first, write one valid line at a time, and remember that the same relationship should remain true after rearrangement. Mixed practice later is more useful than trying to memorise a distinct shortcut for every context.

Worked example 2. Question: In P = 2(l + w), make w the subject. Then use P = 22 cm and l = 8 cm. Method: Divide by 2 first: P/2 = l+w. Subtract l: w = P/2−l = 22/2−8 = 3 cm. Check: Two lots of 8+3 give 22 cm, matching the original perimeter. Watch for: Subtracting l from P before removing the outer factor of two changes the formula.

Take the idea further: Now change the subject to l and show that the steps are symmetrical. After reading the worked method, hide it and begin again. Recognition of a printed step is not the same as being able to retrieve it in a test. Compare with a formula that looks similar but places a bracket in a different position.

Acceleration from a motion relationship

This cluster isolates acceleration from a motion relationship. Look for the operation that must be undone first, write one valid line at a time, and remember that the same relationship should remain true after rearrangement. Mixed practice later is more useful than trying to memorise a distinct shortcut for every context.

Worked example 3. Question: Make a the subject of v = u + at. Evaluate with v = 14 m/s, u = 2 m/s and t = 4 s. Method: Subtract u then divide by t: a = (v−u)/t = (14−2)/4 = 3 m/s². Check: The original gives u+at = 2+3×4 = 14 m/s. Watch for: The entire difference v−u is the numerator; brackets matter.

Take the idea further: Express t in terms of v, u and a, and state the nonzero denominator condition. Trace every change across the equality. A correct algebraic transformation should have the same operation applied to both sides. Explain how a zero denominator would change which inputs are permitted.

Find the x-coordinate from a line equation

This cluster isolates find the x-coordinate from a line equation. Look for the operation that must be undone first, write one valid line at a time, and remember that the same relationship should remain true after rearrangement. Mixed practice later is more useful than trying to memorise a distinct shortcut for every context.

Worked example 4. Question: Change the subject of y = mx+c to x. Calculate x with y=11, m=2 and c=5. Method: Subtract c: y−c=mx. Divide by m: x=(y−c)/m=(11−5)/2=3. Check: Putting x=3 into mx+c produces 2×3+5=11. Watch for: The calculation assumes m is nonzero; an x-independent constant line behaves differently.

Take the idea further: Explain why y−c must remain together above the divisor m. Use the checking line as a separate route to confidence. If it disagrees with the original, locate the first line where equivalence was broken. Generate a new example with different numbers and ask a classmate to verify your rearrangement.

Trapezium area: make a parallel side the subject

This cluster isolates trapezium area: make a parallel side the subject. Look for the operation that must be undone first, write one valid line at a time, and remember that the same relationship should remain true after rearrangement. Mixed practice later is more useful than trying to memorise a distinct shortcut for every context.

Worked example 5. Question: The trapezium area is A=(a+b)h/2. Make b the subject. Use A=18 cm², a=4 cm and h=4 cm. Method: Multiply by 2, divide by nonzero h and subtract a: b = 2A/h−a = 36/4−4 = 5 cm. Check: Area check: (4+5)×4/2 = 18 cm². Watch for: Dividing only A rather than the full doubled expression loses the relationship.

Take the idea further: Rearrange this formula to make h the subject instead. Write the answer’s units. A bare number cannot show whether the result is time, length, acceleration, rate or temperature. Estimate the result’s size before the final numerical division.

Cylinder height from exact volume

This cluster isolates cylinder height from exact volume. Look for the operation that must be undone first, write one valid line at a time, and remember that the same relationship should remain true after rearrangement. Mixed practice later is more useful than trying to memorise a distinct shortcut for every context.

Worked example 6. Question: Cylinder volume V=πr²h. Find h when V=16π cm³ and radius r=2 cm. Method: Divide by πr²: h=V/(πr²)=(16π)/(4π)=4 cm, with nonzero r. Check: Substitute back: π×2²×4=16π cm³. Watch for: Do not divide by r when the original expression contains r squared.

Take the idea further: If volume is constant and radius doubles, how must height change? Change one input and predict the direction of the output’s change before recalculating. This is a test of understanding rather than speed. Start from your answer and work backwards to reconstruct the original equation.

Find an annual simple-interest percentage

This cluster isolates find an annual simple-interest percentage. Look for the operation that must be undone first, write one valid line at a time, and remember that the same relationship should remain true after rearrangement. Mixed practice later is more useful than trying to memorise a distinct shortcut for every context.

Worked example 7. Question: Simple interest I=Prt uses decimal annual rate r. Make r the subject and use I=$40, P=$500, t=2 years. Method: Divide by Pt: r=I/(Pt)=40/(500×2)=0.04. As a percentage this is 4% per year. Check: Checking: 500×0.04×2 gives $40. Watch for: A decimal rate and a percentage rate are not the same printed number.

Take the idea further: If the principal doubles but the interest and time stay fixed, what happens to r? Ask which bracket or denominator would cause the most trouble under examination pressure. Put a small note beside that risk. Try a version with negative values where the context permits them, and explain which signs must be preserved.

Celsius as the subject of a conversion formula

This cluster isolates celsius as the subject of a conversion formula. Look for the operation that must be undone first, write one valid line at a time, and remember that the same relationship should remain true after rearrangement. Mixed practice later is more useful than trying to memorise a distinct shortcut for every context.

Worked example 8. Question: From F=9C/5+32, make C the subject and convert 50°F. Method: Subtract 32, then multiply by 5/9: C=5(F−32)/9 = 5(50−32)/9=10°C. Check: The forward calculation 9(10)/5+32=50°F agrees. Watch for: The 32 must be subtracted before applying the 5/9 factor.

Take the idea further: Compare the temperature increase when Celsius rises by five degrees. Explain the method aloud using terms such as subtract, divide, square and substitute rather than simply ‘move to the other side’. Compare with a formula that looks similar but places a bracket in a different position.

A fraction containing a linear expression

This cluster isolates a fraction containing a linear expression. Look for the operation that must be undone first, write one valid line at a time, and remember that the same relationship should remain true after rearrangement. Mixed practice later is more useful than trying to memorise a distinct shortcut for every context.

Worked example 9. Question: Make x the subject of y=(ax+b)/2. With a=2, b=4, y=5, find x. Method: Multiply by 2: 2y=ax+b. Hence x=(2y−b)/a = (2×5−4)/2=3. Check: The original evaluates to (2×3+4)/2=5. Watch for: The whole expression 2y−b must be divided by a.

Take the idea further: Replace the denominator 2 with c and rearrange to a general expression for x. Return after two days and try a different numerical version without hints. This tests whether the relationship has been learned. Explain how a zero denominator would change which inputs are permitted.

Square-root relationship and valid values

This cluster isolates square-root relationship and valid values. Look for the operation that must be undone first, write one valid line at a time, and remember that the same relationship should remain true after rearrangement. Mixed practice later is more useful than trying to memorise a distinct shortcut for every context.

Worked example 10. Question: For y=√(x+c), make x the subject and find x when y=3, c=2. Method: Squaring gives y²=x+c. Therefore x=y²−c=3²−2=7. The original square root requires y≥0. Check: √(7+2)=√9=3, so the candidate works in the initial relationship. Watch for: Squaring can hide a sign restriction; always check in the original formula.

Take the idea further: Explain why a negative y is incompatible with the real principal square-root expression. Use the original formula for a reverse calculation. Checks that merely repeat the same mistaken method are less informative. Generate a new example with different numbers and ask a classmate to verify your rearrangement.

Rectangle width from area

Worked example 11. Question: Area A = lw. Make w the subject; then find w when A = 24 cm² and l = 4 cm. Method: Divide both sides by l (l is nonzero): w = A/l = 24/4 = 6 cm. Check: Checking the original: 4 × 6 = 24 cm². The relationship is preserved. Watch for: Dividing l by A would reverse the ratio.

Take the idea further: What happens to w if the same area is shared over double the length? Name the subject before calculating. A student who identifies the correct unknown has already completed an essential reasoning step. Estimate the result’s size before the final numerical division.

Perimeter and brackets

Worked example 12. Question: In P = 2(l + w), make w the subject. Then use P = 26 cm and l = 9 cm. Method: Divide by 2 first: P/2 = l+w. Subtract l: w = P/2−l = 26/2−9 = 4 cm. Check: Two lots of 9+4 give 26 cm, matching the original perimeter. Watch for: Subtracting l from P before removing the outer factor of two changes the formula.

Take the idea further: Now change the subject to l and show that the steps are symmetrical. After reading the worked method, hide it and begin again. Recognition of a printed step is not the same as being able to retrieve it in a test. Start from your answer and work backwards to reconstruct the original equation.

Acceleration from a motion relationship

Worked example 13. Question: Make a the subject of v = u + at. Evaluate with v = 24 m/s, u = 4 m/s and t = 5 s. Method: Subtract u then divide by t: a = (v−u)/t = (24−4)/5 = 4 m/s². Check: The original gives u+at = 4+4×5 = 24 m/s. Watch for: The entire difference v−u is the numerator; brackets matter.

Take the idea further: Express t in terms of v, u and a, and state the nonzero denominator condition. Trace every change across the equality. A correct algebraic transformation should have the same operation applied to both sides. Try a version with negative values where the context permits them, and explain which signs must be preserved.

Find the x-coordinate from a line equation

Worked example 14. Question: Change the subject of y = mx+c to x. Calculate x with y=18, m=3 and c=6. Method: Subtract c: y−c=mx. Divide by m: x=(y−c)/m=(18−6)/3=4. Check: Putting x=4 into mx+c produces 3×4+6=18. Watch for: The calculation assumes m is nonzero; an x-independent constant line behaves differently.

Take the idea further: Explain why y−c must remain together above the divisor m. Use the checking line as a separate route to confidence. If it disagrees with the original, locate the first line where equivalence was broken. Compare with a formula that looks similar but places a bracket in a different position.

Trapezium area: make a parallel side the subject

Worked example 15. Question: The trapezium area is A=(a+b)h/2. Make b the subject. Use A=33 cm², a=5 cm and h=6 cm. Method: Multiply by 2, divide by nonzero h and subtract a: b = 2A/h−a = 66/6−5 = 6 cm. Check: Area check: (5+6)×6/2 = 33 cm². Watch for: Dividing only A rather than the full doubled expression loses the relationship.

Take the idea further: Rearrange this formula to make h the subject instead. Write the answer’s units. A bare number cannot show whether the result is time, length, acceleration, rate or temperature. Explain how a zero denominator would change which inputs are permitted.

Cylinder height from exact volume

Worked example 16. Question: Cylinder volume V=πr²h. Find h when V=45π cm³ and radius r=3 cm. Method: Divide by πr²: h=V/(πr²)=(45π)/(9π)=5 cm, with nonzero r. Check: Substitute back: π×3²×5=45π cm³. Watch for: Do not divide by r when the original expression contains r squared.

Take the idea further: If volume is constant and radius doubles, how must height change? Change one input and predict the direction of the output’s change before recalculating. This is a test of understanding rather than speed. Generate a new example with different numbers and ask a classmate to verify your rearrangement.

Find an annual simple-interest percentage

Worked example 17. Question: Simple interest I=Prt uses decimal annual rate r. Make r the subject and use I=$50, P=$500, t=2 years. Method: Divide by Pt: r=I/(Pt)=50/(500×2)=0.05. As a percentage this is 5% per year. Check: Checking: 500×0.05×2 gives $50. Watch for: A decimal rate and a percentage rate are not the same printed number.

Take the idea further: If the principal doubles but the interest and time stay fixed, what happens to r? Ask which bracket or denominator would cause the most trouble under examination pressure. Put a small note beside that risk. Estimate the result’s size before the final numerical division.

Celsius as the subject of a conversion formula

Worked example 18. Question: From F=9C/5+32, make C the subject and convert 59°F. Method: Subtract 32, then multiply by 5/9: C=5(F−32)/9 = 5(59−32)/9=15°C. Check: The forward calculation 9(15)/5+32=59°F agrees. Watch for: The 32 must be subtracted before applying the 5/9 factor.

Take the idea further: Compare the temperature increase when Celsius rises by five degrees. Explain the method aloud using terms such as subtract, divide, square and substitute rather than simply ‘move to the other side’. Start from your answer and work backwards to reconstruct the original equation.

A fraction containing a linear expression

Worked example 19. Question: Make x the subject of y=(ax+b)/2. With a=3, b=5, y=7, find x. Method: Multiply by 2: 2y=ax+b. Hence x=(2y−b)/a = (2×7−5)/3=3. Check: The original evaluates to (3×3+5)/2=7. Watch for: The whole expression 2y−b must be divided by a.

Take the idea further: Replace the denominator 2 with c and rearrange to a general expression for x. Return after two days and try a different numerical version without hints. This tests whether the relationship has been learned. Try a version with negative values where the context permits them, and explain which signs must be preserved.

Square-root relationship and valid values

Worked example 20. Question: For y=√(x+c), make x the subject and find x when y=4, c=3. Method: Squaring gives y²=x+c. Therefore x=y²−c=4²−3=13. The original square root requires y≥0. Check: √(13+3)=√16=4, so the candidate works in the initial relationship. Watch for: Squaring can hide a sign restriction; always check in the original formula.

Take the idea further: Explain why a negative y is incompatible with the real principal square-root expression. Use the original formula for a reverse calculation. Checks that merely repeat the same mistaken method are less informative. Compare with a formula that looks similar but places a bracket in a different position.

Rectangle width from area

Worked example 21. Question: Area A = lw. Make w the subject; then find w when A = 35 cm² and l = 5 cm. Method: Divide both sides by l (l is nonzero): w = A/l = 35/5 = 7 cm. Check: Checking the original: 5 × 7 = 35 cm². The relationship is preserved. Watch for: Dividing l by A would reverse the ratio.

Take the idea further: What happens to w if the same area is shared over double the length? Name the subject before calculating. A student who identifies the correct unknown has already completed an essential reasoning step. Explain how a zero denominator would change which inputs are permitted.

Perimeter and brackets

Worked example 22. Question: In P = 2(l + w), make w the subject. Then use P = 30 cm and l = 10 cm. Method: Divide by 2 first: P/2 = l+w. Subtract l: w = P/2−l = 30/2−10 = 5 cm. Check: Two lots of 10+5 give 30 cm, matching the original perimeter. Watch for: Subtracting l from P before removing the outer factor of two changes the formula.

Take the idea further: Now change the subject to l and show that the steps are symmetrical. After reading the worked method, hide it and begin again. Recognition of a printed step is not the same as being able to retrieve it in a test. Generate a new example with different numbers and ask a classmate to verify your rearrangement.

Acceleration from a motion relationship

Worked example 23. Question: Make a the subject of v = u + at. Evaluate with v = 36 m/s, u = 6 m/s and t = 6 s. Method: Subtract u then divide by t: a = (v−u)/t = (36−6)/6 = 5 m/s². Check: The original gives u+at = 6+5×6 = 36 m/s. Watch for: The entire difference v−u is the numerator; brackets matter.

Take the idea further: Express t in terms of v, u and a, and state the nonzero denominator condition. Trace every change across the equality. A correct algebraic transformation should have the same operation applied to both sides. Estimate the result’s size before the final numerical division.

Find the x-coordinate from a line equation

Worked example 24. Question: Change the subject of y = mx+c to x. Calculate x with y=27, m=4 and c=7. Method: Subtract c: y−c=mx. Divide by m: x=(y−c)/m=(27−7)/4=5. Check: Putting x=5 into mx+c produces 4×5+7=27. Watch for: The calculation assumes m is nonzero; an x-independent constant line behaves differently.

Take the idea further: Explain why y−c must remain together above the divisor m. Use the checking line as a separate route to confidence. If it disagrees with the original, locate the first line where equivalence was broken. Start from your answer and work backwards to reconstruct the original equation.

Trapezium area: make a parallel side the subject

Worked example 25. Question: The trapezium area is A=(a+b)h/2. Make b the subject. Use A=52 cm², a=6 cm and h=8 cm. Method: Multiply by 2, divide by nonzero h and subtract a: b = 2A/h−a = 104/8−6 = 7 cm. Check: Area check: (6+7)×8/2 = 52 cm². Watch for: Dividing only A rather than the full doubled expression loses the relationship.

Take the idea further: Rearrange this formula to make h the subject instead. Write the answer’s units. A bare number cannot show whether the result is time, length, acceleration, rate or temperature. Try a version with negative values where the context permits them, and explain which signs must be preserved.

Cylinder height from exact volume

Worked example 26. Question: Cylinder volume V=πr²h. Find h when V=96π cm³ and radius r=4 cm. Method: Divide by πr²: h=V/(πr²)=(96π)/(16π)=6 cm, with nonzero r. Check: Substitute back: π×4²×6=96π cm³. Watch for: Do not divide by r when the original expression contains r squared.

Take the idea further: If volume is constant and radius doubles, how must height change? Change one input and predict the direction of the output’s change before recalculating. This is a test of understanding rather than speed. Compare with a formula that looks similar but places a bracket in a different position.

Find an annual simple-interest percentage

Worked example 27. Question: Simple interest I=Prt uses decimal annual rate r. Make r the subject and use I=$60, P=$500, t=2 years. Method: Divide by Pt: r=I/(Pt)=60/(500×2)=0.06. As a percentage this is 6% per year. Check: Checking: 500×0.06×2 gives $60. Watch for: A decimal rate and a percentage rate are not the same printed number.

Take the idea further: If the principal doubles but the interest and time stay fixed, what happens to r? Ask which bracket or denominator would cause the most trouble under examination pressure. Put a small note beside that risk. Explain how a zero denominator would change which inputs are permitted.

Celsius as the subject of a conversion formula

Worked example 28. Question: From F=9C/5+32, make C the subject and convert 68°F. Method: Subtract 32, then multiply by 5/9: C=5(F−32)/9 = 5(68−32)/9=20°C. Check: The forward calculation 9(20)/5+32=68°F agrees. Watch for: The 32 must be subtracted before applying the 5/9 factor.

Take the idea further: Compare the temperature increase when Celsius rises by five degrees. Explain the method aloud using terms such as subtract, divide, square and substitute rather than simply ‘move to the other side’. Generate a new example with different numbers and ask a classmate to verify your rearrangement.

A fraction containing a linear expression

Worked example 29. Question: Make x the subject of y=(ax+b)/2. With a=4, b=6, y=9, find x. Method: Multiply by 2: 2y=ax+b. Hence x=(2y−b)/a = (2×9−6)/4=3. Check: The original evaluates to (4×3+6)/2=9. Watch for: The whole expression 2y−b must be divided by a.

Take the idea further: Replace the denominator 2 with c and rearrange to a general expression for x. Return after two days and try a different numerical version without hints. This tests whether the relationship has been learned. Estimate the result’s size before the final numerical division.

Square-root relationship and valid values

Worked example 30. Question: For y=√(x+c), make x the subject and find x when y=5, c=4. Method: Squaring gives y²=x+c. Therefore x=y²−c=5²−4=21. The original square root requires y≥0. Check: √(21+4)=√25=5, so the candidate works in the initial relationship. Watch for: Squaring can hide a sign restriction; always check in the original formula.

Take the idea further: Explain why a negative y is incompatible with the real principal square-root expression. Use the original formula for a reverse calculation. Checks that merely repeat the same mistaken method are less informative. Start from your answer and work backwards to reconstruct the original equation.

Rectangle width from area

Worked example 31. Question: Area A = lw. Make w the subject; then find w when A = 48 cm² and l = 6 cm. Method: Divide both sides by l (l is nonzero): w = A/l = 48/6 = 8 cm. Check: Checking the original: 6 × 8 = 48 cm². The relationship is preserved. Watch for: Dividing l by A would reverse the ratio.

Perimeter and brackets

Worked example 32. Question: In P = 2(l + w), make w the subject. Then use P = 34 cm and l = 11 cm. Method: Divide by 2 first: P/2 = l+w. Subtract l: w = P/2−l = 34/2−11 = 6 cm. Check: Two lots of 11+6 give 34 cm, matching the original perimeter. Watch for: Subtracting l from P before removing the outer factor of two changes the formula.

Acceleration from a motion relationship

Worked example 33. Question: Make a the subject of v = u + at. Evaluate with v = 50 m/s, u = 8 m/s and t = 7 s. Method: Subtract u then divide by t: a = (v−u)/t = (50−8)/7 = 6 m/s². Check: The original gives u+at = 8+6×7 = 50 m/s. Watch for: The entire difference v−u is the numerator; brackets matter.

Find the x-coordinate from a line equation

Worked example 34. Question: Change the subject of y = mx+c to x. Calculate x with y=38, m=5 and c=8. Method: Subtract c: y−c=mx. Divide by m: x=(y−c)/m=(38−8)/5=6. Check: Putting x=6 into mx+c produces 5×6+8=38. Watch for: The calculation assumes m is nonzero; an x-independent constant line behaves differently.

Trapezium area: make a parallel side the subject

Worked example 35. Question: The trapezium area is A=(a+b)h/2. Make b the subject. Use A=75 cm², a=7 cm and h=10 cm. Method: Multiply by 2, divide by nonzero h and subtract a: b = 2A/h−a = 150/10−7 = 8 cm. Check: Area check: (7+8)×10/2 = 75 cm². Watch for: Dividing only A rather than the full doubled expression loses the relationship.

Cylinder height from exact volume

Worked example 36. Question: Cylinder volume V=πr²h. Find h when V=175π cm³ and radius r=5 cm. Method: Divide by πr²: h=V/(πr²)=(175π)/(25π)=7 cm, with nonzero r. Check: Substitute back: π×5²×7=175π cm³. Watch for: Do not divide by r when the original expression contains r squared.

Find an annual simple-interest percentage

Worked example 37. Question: Simple interest I=Prt uses decimal annual rate r. Make r the subject and use I=$70, P=$500, t=2 years. Method: Divide by Pt: r=I/(Pt)=70/(500×2)=0.07. As a percentage this is 7% per year. Check: Checking: 500×0.07×2 gives $70. Watch for: A decimal rate and a percentage rate are not the same printed number.

Celsius as the subject of a conversion formula

Worked example 38. Question: From F=9C/5+32, make C the subject and convert 77°F. Method: Subtract 32, then multiply by 5/9: C=5(F−32)/9 = 5(77−32)/9=25°C. Check: The forward calculation 9(25)/5+32=77°F agrees. Watch for: The 32 must be subtracted before applying the 5/9 factor.

A fraction containing a linear expression

Worked example 39. Question: Make x the subject of y=(ax+b)/2. With a=5, b=7, y=11, find x. Method: Multiply by 2: 2y=ax+b. Hence x=(2y−b)/a = (2×11−7)/5=3. Check: The original evaluates to (5×3+7)/2=11. Watch for: The whole expression 2y−b must be divided by a.

Square-root relationship and valid values

Worked example 40. Question: For y=√(x+c), make x the subject and find x when y=6, c=5. Method: Squaring gives y²=x+c. Therefore x=y²−c=6²−5=31. The original square root requires y≥0. Check: √(31+5)=√36=6, so the candidate works in the initial relationship. Watch for: Squaring can hide a sign restriction; always check in the original formula.

Rectangle width from area

Worked example 41. Question: Area A = lw. Make w the subject; then find w when A = 63 cm² and l = 7 cm. Method: Divide both sides by l (l is nonzero): w = A/l = 63/7 = 9 cm. Check: Checking the original: 7 × 9 = 63 cm². The relationship is preserved. Watch for: Dividing l by A would reverse the ratio.

Take the idea further: What happens to w if the same area is shared over double the length? Name the subject before calculating. A student who identifies the correct unknown has already completed an essential reasoning step. Estimate the result’s size before the final numerical division.

Perimeter and brackets

Worked example 42. Question: In P = 2(l + w), make w the subject. Then use P = 38 cm and l = 12 cm. Method: Divide by 2 first: P/2 = l+w. Subtract l: w = P/2−l = 38/2−12 = 7 cm. Check: Two lots of 12+7 give 38 cm, matching the original perimeter. Watch for: Subtracting l from P before removing the outer factor of two changes the formula.

Take the idea further: Now change the subject to l and show that the steps are symmetrical. After reading the worked method, hide it and begin again. Recognition of a printed step is not the same as being able to retrieve it in a test. Start from your answer and work backwards to reconstruct the original equation.

Acceleration from a motion relationship

Worked example 43. Question: Make a the subject of v = u + at. Evaluate with v = 66 m/s, u = 10 m/s and t = 8 s. Method: Subtract u then divide by t: a = (v−u)/t = (66−10)/8 = 7 m/s². Check: The original gives u+at = 10+7×8 = 66 m/s. Watch for: The entire difference v−u is the numerator; brackets matter.

Take the idea further: Express t in terms of v, u and a, and state the nonzero denominator condition. Trace every change across the equality. A correct algebraic transformation should have the same operation applied to both sides. Try a version with negative values where the context permits them, and explain which signs must be preserved.

Find the x-coordinate from a line equation

Worked example 44. Question: Change the subject of y = mx+c to x. Calculate x with y=51, m=6 and c=9. Method: Subtract c: y−c=mx. Divide by m: x=(y−c)/m=(51−9)/6=7. Check: Putting x=7 into mx+c produces 6×7+9=51. Watch for: The calculation assumes m is nonzero; an x-independent constant line behaves differently.

Take the idea further: Explain why y−c must remain together above the divisor m. Use the checking line as a separate route to confidence. If it disagrees with the original, locate the first line where equivalence was broken. Compare with a formula that looks similar but places a bracket in a different position.

Trapezium area: make a parallel side the subject

Worked example 45. Question: The trapezium area is A=(a+b)h/2. Make b the subject. Use A=102 cm², a=8 cm and h=12 cm. Method: Multiply by 2, divide by nonzero h and subtract a: b = 2A/h−a = 204/12−8 = 9 cm. Check: Area check: (8+9)×12/2 = 102 cm². Watch for: Dividing only A rather than the full doubled expression loses the relationship.

Take the idea further: Rearrange this formula to make h the subject instead. Write the answer’s units. A bare number cannot show whether the result is time, length, acceleration, rate or temperature. Explain how a zero denominator would change which inputs are permitted.

Cylinder height from exact volume

Worked example 46. Question: Cylinder volume V=πr²h. Find h when V=288π cm³ and radius r=6 cm. Method: Divide by πr²: h=V/(πr²)=(288π)/(36π)=8 cm, with nonzero r. Check: Substitute back: π×6²×8=288π cm³. Watch for: Do not divide by r when the original expression contains r squared.

Take the idea further: If volume is constant and radius doubles, how must height change? Change one input and predict the direction of the output’s change before recalculating. This is a test of understanding rather than speed. Generate a new example with different numbers and ask a classmate to verify your rearrangement.

Find an annual simple-interest percentage

Worked example 47. Question: Simple interest I=Prt uses decimal annual rate r. Make r the subject and use I=$80, P=$500, t=2 years. Method: Divide by Pt: r=I/(Pt)=80/(500×2)=0.08. As a percentage this is 8% per year. Check: Checking: 500×0.08×2 gives $80. Watch for: A decimal rate and a percentage rate are not the same printed number.

Take the idea further: If the principal doubles but the interest and time stay fixed, what happens to r? Ask which bracket or denominator would cause the most trouble under examination pressure. Put a small note beside that risk. Estimate the result’s size before the final numerical division.

Celsius as the subject of a conversion formula

Worked example 48. Question: From F=9C/5+32, make C the subject and convert 86°F. Method: Subtract 32, then multiply by 5/9: C=5(F−32)/9 = 5(86−32)/9=30°C. Check: The forward calculation 9(30)/5+32=86°F agrees. Watch for: The 32 must be subtracted before applying the 5/9 factor.

Take the idea further: Compare the temperature increase when Celsius rises by five degrees. Explain the method aloud using terms such as subtract, divide, square and substitute rather than simply ‘move to the other side’. Start from your answer and work backwards to reconstruct the original equation.

A fraction containing a linear expression

Worked example 49. Question: Make x the subject of y=(ax+b)/2. With a=6, b=8, y=13, find x. Method: Multiply by 2: 2y=ax+b. Hence x=(2y−b)/a = (2×13−8)/6=3. Check: The original evaluates to (6×3+8)/2=13. Watch for: The whole expression 2y−b must be divided by a.

Take the idea further: Replace the denominator 2 with c and rearrange to a general expression for x. Return after two days and try a different numerical version without hints. This tests whether the relationship has been learned. Try a version with negative values where the context permits them, and explain which signs must be preserved.

Square-root relationship and valid values

Worked example 50. Question: For y=√(x+c), make x the subject and find x when y=7, c=6. Method: Squaring gives y²=x+c. Therefore x=y²−c=7²−6=43. The original square root requires y≥0. Check: √(43+6)=√49=7, so the candidate works in the initial relationship. Watch for: Squaring can hide a sign restriction; always check in the original formula.

Take the idea further: Explain why a negative y is incompatible with the real principal square-root expression. Use the original formula for a reverse calculation. Checks that merely repeat the same mistaken method are less informative. Compare with a formula that looks similar but places a bracket in a different position.

The five mistakes to fix first

Dividing only one term rather than a whole expression

Bracket a complete numerator or denominator when several terms belong together. A single missing pair of brackets may change the entire value. Then use an unfamiliar numerical version to check that the student has repaired the underlying decision rather than memorised the corrected line.

Moving a term without preserving equality

Say what operation acts on both sides of an equation instead of simply claiming that a sign moves across an equals sign. Then use an unfamiliar numerical version to check that the student has repaired the underlying decision rather than memorised the corrected line.

Losing a squared radius or other exponent

Read every exponent before rearranging. If r² is isolated, taking an appropriate square root is still necessary. Then use an unfamiliar numerical version to check that the student has repaired the underlying decision rather than memorised the corrected line.

Forgetting conditions for division and roots

Check that a divisor cannot be zero and that a physical length or principal square root has a sensible sign. Then use an unfamiliar numerical version to check that the student has repaired the underlying decision rather than memorised the corrected line.

Stopping without using the original relationship

Substitute the answer back into the starting formula to check that it produces the supplied quantity. Then use an unfamiliar numerical version to check that the student has repaired the underlying decision rather than memorised the corrected line.

How a small-group tutor can build independence

For families seeking a Secondary 3 G3 Mathematics tutor in Bukit Timah, ask how a lesson distinguishes misunderstanding from careless arithmetic. A tutor can use a rectangle, a straight line and a speed relationship that all require the same inverse-operation reasoning. Once the pupil recognises that common structure, later practice becomes more flexible.

Small-group teaching is effective only when each student writes independently, explains a decision and receives feedback specific to their attempt. A lesson can begin with a short retrieval task, move to one unfamiliar example, then withdraw hints and retest after a delay. Class size by itself is not a guarantee of understanding, so parents should ask how independent progress is assessed.

The unchanged Clementi Secondary 1 Mathematics small-group guide describes eduKateSG’s earlier-level teaching approach, while this Secondary 3 G3 Bukit Timah overview relates directly to the upper-secondary year level. They serve different purposes and should not be confused.

A six-week plan for formula fluency

Week 1 — identify the obstacle

Collect two original school questions. Check whether the student selected the wrong subject, reversed an operation incorrectly or struggled with fractions. Repair the earliest incorrect decision. Keep one dated first attempt and one later unprompted correction to see whether the learner can apply the same idea after a break.

Week 2 — practise one inverse operation at a time

Rearrange simple multiplication, addition and division equations, then explain every equivalent step aloud before writing a final expression. Keep one dated first attempt and one later unprompted correction to see whether the learner can apply the same idea after a break.

Week 3 — add brackets and restrictions

Contrast y=3x+5 with y=3(x+5), and practise denominators that require nonzero conditions. Record which change in notation changes the correct method. Keep one dated first attempt and one later unprompted correction to see whether the learner can apply the same idea after a break.

Week 4 — connect formulas to physical quantities

Use perimeter, area, speed, temperature and interest. Name what the subject means and check the output units. Keep one dated first attempt and one later unprompted correction to see whether the learner can apply the same idea after a break.

Week 5 — mix contexts after a delay

Mix rearrangement with another algebra chapter and remove the method label from questions. The learner must choose the first step unaided. Keep one dated first attempt and one later unprompted correction to see whether the learner can apply the same idea after a break.

Week 6 — verify transfer with unseen items

Review a fresh set involving brackets, fractions and squares. Compare the student’s independent explanation and checking ability with Week 1. Keep one dated first attempt and one later unprompted correction to see whether the learner can apply the same idea after a break.

Frequently asked questions

What does ‘make x the subject’ actually mean?

Express x by itself on one side so it is described in terms of the other variables. The practical test is a fresh question solved without the first step being supplied by a tutor.

Should I always move symbols across the equals sign?

It is safer to name and perform an equivalent operation on both sides, preserving the original equality. The practical test is a fresh question solved without the first step being supplied by a tutor.

When do I need brackets?

When several terms together form a numerator, difference, or expression multiplied by one outside factor. The practical test is a fresh question solved without the first step being supplied by a tutor.

What if the unknown is squared?

Isolate the square, then take an appropriate square root and apply any restrictions from the question’s context. The practical test is a fresh question solved without the first step being supplied by a tutor.

How can I tell whether a denominator is allowed?

A denominator must not be zero. State the condition and check answers in the original formula. The practical test is a fresh question solved without the first step being supplied by a tutor.

Is this in the 2027 G3 syllabus?

Yes. The K310 algebraic expressions and formulae section includes changing the subject of a formula. The practical test is a fresh question solved without the first step being supplied by a tutor.

What is an effective 10-minute home exercise?

Choose two formulas, identify a different required subject in each, rearrange and verify with allowed sample numbers. The practical test is a fresh question solved without the first step being supplied by a tutor.

What would progress look like?

The pupil can choose legal operations, use brackets correctly, explain the final quantity and check independently. The practical test is a fresh question solved without the first step being supplied by a tutor.

The official 2027 SEAB G3 syllabus index and K310 Mathematics syllabus remain the authoritative references for examination instructions and topic scope. Different schools may reach chapters at different times.

The takeaway

Did you know? A student who learns to change a formula’s subject learns a more general habit: identify what is wanted, undo operations lawfully and check the result in the original relationship. A few carefully corrected examples repeated after a delay can make that skill more dependable than a long page of copied algebra.