Making x the subject is straightforward when x appears only once. It becomes less familiar when x occurs in two terms, inside a bracket or in both the numerator and denominator of a fraction.
The central idea stays the same: perform valid operations on the equation until the required variable is isolated. When it appears in several terms, collect those terms and factor out the variable before dividing.
This guide uses original examples for O Level revision. Work alongside the syllabus for your examination year and check which forms are included. For broader planning, read Abeel’s O Level Mathematics guide.
Understand what “make the subject” asks
In v = u + at, v is already isolated on one side, so it is the subject. Making t the subject means expressing t in terms of the other quantities.
Subtract u from both sides to obtain v − u = at. Dividing both sides by a gives t = (v − u)/a, assuming a is non-zero.
The brackets in the numerator matter. Writing v − u/a would divide only u by a instead of dividing the entire difference.
Rearranging is different from substituting values and solving one numerical case. The final formula must express the required relationship symbolically for the permitted values of the other quantities.
Use operations rather than unexplained “moving”
Students often say that a term changes sign when it crosses the equals sign. This shorthand can work, but it hides the actual operation.
In y = 5x − 7, add seven to both sides to get y + 7 = 5x. Then divide by five, giving x = (y + 7)/5.
No symbol moves by itself. The new equation is equivalent because the same valid operation is applied to each side.
When a rearrangement becomes complicated, return to this language. “Subtract 2x from both sides” is less ambiguous than “move x,” particularly when x has a coefficient or appears inside a larger expression.
Collect every term containing the required variable
Consider q = 3x + px. Both terms on the right contain x. Factorising gives q = x(3 + p).
If 3 + p is non-zero, divide to obtain x = q/(3 + p). This is a compact example of the key method: collect, factorise, divide.
Dividing q by three and then subtracting p would not be equivalent. The term px involves multiplication by x, so p cannot be removed as though it were a separate added constant.
Name the common factor explicitly during revision. If every term in a collected group contains x, write the group as x multiplied by a bracket.
Work through an unknown on both sides
Suppose 2x + 3 = kx − 5 and x is the required subject. Subtract kx from both sides and subtract three from both sides, obtaining 2x − kx = −8.
Factorise the left side: x(2 − k) = −8. Dividing gives x = −8/(2 − k), which is equivalent to x = 8/(k − 2), provided k ≠ 2.
Both final forms are valid. Multiplying numerator and denominator by −1 changes neither value nor meaning.
Check a permitted example. If k = 4, the rearranged formula gives x = 4. The original equation then gives 11 on both sides, confirming that numerical case.
Expand brackets only when it helps collection
Consider a(x + b) = c(x − d). Expanding gives ax + ab = cx − cd.
Collecting x terms on the left and the other terms on the right gives ax − cx = −cd − ab. Factorising yields x(a − c) = −cd − ab.
For a ≠ c, x = (−cd − ab)/(a − c). An equivalent form is x = (cd + ab)/(c − a).
Keep track of the minus in c(x − d): it contributes −cd. When moving several terms into a new arrangement, an intermediate expanded line makes sign errors easier to detect.
A variable inside a denominator needs careful clearing
Suppose y = (2x + 3)/(x + 1). The original formula requires x ≠ −1 because its denominator cannot be zero.
Multiply both sides by x + 1: y(x + 1) = 2x + 3. Expanding gives xy + y = 2x + 3.
Collect x terms: xy − 2x = 3 − y. Factorising gives x(y − 2) = 3 − y, so x = (3 − y)/(y − 2), provided y ≠ 2.
Do not multiply y by only x and omit the one in the bracket. Clearing a denominator means multiplying by the entire denominator expression.
Check the restrictions created by division
The condition y ≠ 2 in the previous example is meaningful. Substituting y = 2 into the original relationship gives 2(x + 1) = 2x + 3 after clearing the denominator, which would require 2 = 3.
There is no valid x for that value of y. The excluded denominator therefore reflects an impossible case rather than a cosmetic notation issue.
Other formulas can behave differently when a coefficient becomes zero. In q = x(3 + p), setting p = −3 gives q = 0 regardless of x. If q is non-zero, there is no solution; if q is zero, the equation does not determine a unique x.
You do not need to invent special cases beyond the question’s scope, but avoid dividing by an expression while silently assuming it cannot be zero.
Distinguish a common denominator from cross multiplication
For a simple equality of fractions, multiplying both sides by valid non-zero denominators can remove them. This is often described as cross multiplication.
The underlying operation is still multiplication of the whole equation. If extra added terms appear, they must also be multiplied by the denominator used.
For example, in y = 2 + 3/x, multiplying by x gives xy = 2x + 3, not xy = 2 + 3. Collecting yields x(y − 2) = 3 and x = 3/(y − 2), with the relevant restrictions.
Write the complete multiplication before simplifying if a formula has several terms. A shortcut that works for one fraction equality may be misapplied when the structure changes.
Keep powers and roots consistent
Suppose A = πr² and the question asks for the radius r. Dividing by π gives r² = A/π. A radius is non-negative, so r = √(A/π) in the geometric setting.
For a purely algebraic equation x² = k with k positive, there are two real solutions, x = √k and x = −√k. The context explains why one or both branches are appropriate.
Do not treat taking a square root as division by two. Squaring and multiplying by two are different operations, so their inverses differ.
Also keep the whole quantity inside the root when required. √(A/π) is not generally the same as √A divided by π.
Check the units of a physical formula
Units can reveal an incorrect rearrangement. In t = (v − u)/a, velocity has units metres per second and acceleration metres per second squared. Their quotient has units seconds.
If a proposed expression instead multiplies a velocity difference by acceleration, its units do not match time. This does not identify every possible algebra error, but it is a useful independent check.
Units cannot prove a formula is correct. Two incorrect formulas can share the same dimensions, and some mathematical formulas have no physical units at all.
Use dimensional reasoning alongside algebraic equivalence, not as a replacement. A correct rearrangement needs both meaningful quantities and valid operations.
Substitute a permitted set of values
Choose values for the original formula, calculate its result and compare the rearranged version. For y = (2x + 3)/(x + 1), take x = 1 to obtain y = 5/2.
Substituting y = 5/2 into x = (3 − y)/(y − 2) gives (1/2)/(1/2) = 1. The test is consistent with the original choice.
Avoid excluded values such as x = −1 or y = 2. A test involving an undefined expression tells you nothing about a valid numerical case.
One successful check is not a proof for all values, but one failed valid check is enough to show that a proposed rearrangement is wrong. Use both numerical checking and clear symbolic steps.
Do not divide a sum term by term incorrectly
From x(a + b) = c, the solution is x = c/(a + b), assuming a + b ≠ 0. It is not generally c/a + c/b.
The whole bracket is one factor multiplying x. Dividing by that factor removes it. Separating the denominator changes the relationship.
A quick numerical example shows the difference. If a = 2, b = 3 and c = 10, the correct x is two. The expression 10/2 + 10/3 gives a completely different value.
Keep a factored bracket intact through the final division. Many otherwise good rearrangements fail on the last line because a denominator is split without a valid rule.
Practise the structure that causes the error
Sort mistakes into categories: incorrect inverse operation, missed bracket term, failure to collect all x terms, incorrect factorisation, divided sum or ignored restriction.
Then practise a fresh formula with the same structure. If the difficulty is x in the denominator, begin by clearing the denominator correctly. If it is repeated x terms, practise collecting and factoring before trying a more elaborate fraction.
Use Abeel’s Cambridge past-paper guide for independent attempts and later reattempts. The O Level revision plan helps place these shorter tasks within a wider schedule.
Explore Abeel Eschool and the course catalogue for current options. Ask through admissions and FAQs, sharing the original formula and the first line you are unsure about.
Frequently asked questions
What do I do when x appears in two terms?
Collect the x terms on one side, factor out x and divide by the remaining factor when it is non-zero.
Must I always expand brackets first?
No. Sometimes a bracket can be isolated directly. Expand when it makes collection or comparison clearer.
Why do restrictions matter?
Division by zero is undefined. An excluded coefficient or denominator can represent no solution or a case where the formula no longer determines a unique value.
Can I split c divided by a plus b into two fractions?
Not as c/a + c/b. The denominator a + b is one complete expression, so keep it grouped unless a valid transformation applies.
Does a numerical check prove the rearrangement?
No, but it can reveal an error. Combine a permitted numerical check with valid symbolic operations.
Have a question about this topic?
Send your query to Abeel Eschool and our admissions team will get back to you.