To make a letter the subject when it appears more than once, collect every term containing that letter on one side, factorise it out, then divide. This is a standard Extended-style skill and it also appears inside longer formula and geometry questions.
It uses the balancing from solving with brackets and the factorising from algebraic structure.
What is the idea behind it?
Solving for x with numbers and making x the subject with letters are the same process. You are getting x alone using inverse operations. The only new difficulty is that the answer contains other letters.
When x appears in two terms, you first group them. Then x is a common factor, and brackets let you treat it as one piece.
How to do it, step by step
- Clear any fractions or brackets by multiplying out.
- Move every term containing the wanted letter to one side, and everything else to the other.
- Factorise the wanted letter out of the group.
- Divide both sides by the bracket.
- Test with simple numbers.
Worked example
Make x the subject of 5(x − p) = 2x + q
Step 1, expand: 5x − 5p = 2x + q.
Step 2, x terms together: subtract 2x from both sides: 3x − 5p = q.
Step 3, other letters across: add 5p to both sides: 3x = q + 5p.
Step 4, divide: x = (q + 5p)/3.
Test: let p = 2 and q = 4. Then x = (4 + 10)/3 = 14/3. Left side 5(14/3 − 2) = 5 × 8/3 = 40/3. Right side 2 × 14/3 + 4 = 28/3 + 12/3 = 40/3. ✓
Now a case where factorising is needed. Make x the subject of y = (x + 3)/(x − 2).
Multiply both sides by (x − 2): y(x − 2) = x + 3. Expand: yx − 2y = x + 3.
Bring x terms together: yx − x = 3 + 2y. Factorise: x(y − 1) = 2y + 3.
Divide: x = (2y + 3)/(y − 1).
Test with y = 5: x = 13/4, and (13/4 + 3)/(13/4 − 2) = (25/4)/(5/4) = 5. ✓
The mistake to watch for
The common slip is moving a term across without changing its sign.
Mistaken working: 5x − 5p = 2x + q becomes 5x + 2x = q + 5p, so x = (q + 5p)/7.
The student added 2x to the left instead of subtracting it from both sides.
Test with p = 2 and q = 4: the formula gives x = 2. Then the left side 5(2 − 2) = 0, but the right side is 2 × 2 + 4 = 8. They do not match.
The correction is to write ”− 2x” under both sides each time, so the sign change is visible instead of mental.
Check yourself
1. Make t the subject of 4(t + m) = 7t − n
Show answer
Expand: 4t + 4m = 7t − n. Subtract 4t: 4m = 3t − n. Add n: 4m + n = 3t. So t = (4m + n)/3.
Test with m = 1 and n = 2: t = 2. Left side 4 × 3 = 12, right side 14 − 2 = 12. ✓
2. Make x the subject of px + q = rx + s
Show answer
Subtract rx and q: px − rx = s − q. Factorise: x(p − r) = s − q. Divide: x = (s − q)/(p − r).
Test with p = 5, r = 2, q = 1, s = 10: x = 9/3 = 3. Then 15 + 1 = 16 and 6 + 10 = 16. ✓
3. Make b the subject of a = (2b + 1)/(b − 3)
Show answer
Multiply: a(b − 3) = 2b + 1. Expand: ab − 3a = 2b + 1. Collect b terms: ab − 2b = 1 + 3a. Factorise: b(a − 2) = 3a + 1. So b = (3a + 1)/(a − 2).
Test with a = 5: b = 16/3. Then (32/3 + 1)/(16/3 − 3) = (35/3)/(7/3) = 5. ✓
Where this leads next
With the balancing skill in place, try forming an equation from a verbal constraint, which turns sentences into equations you can solve. Then use the equations and formulas practice set.
If the routine works when you follow an example but not on a fresh formula, online one-to-one Mathematics tuition lets a teacher watch that transition and adjust the explanation.