Use elimination when two coefficients match or are opposites, and substitution when one variable already has a coefficient of 1 or −1. The right choice keeps the algebra free of fractions.
This lesson is part of systems of equations. It uses only linear equations, but the same habit works for linear and nonlinear pairs.
What is the two-question test?
Before any algebra, ask two questions.
- Do any two coefficients of the same variable match, or are they opposites? If yes, add or subtract the equations (elimination).
- Is there a variable with coefficient 1 or −1 that can be isolated with no fractions? If yes, rearrange and substitute.
If neither applies, multiply one equation to create a matching coefficient, then eliminate.
Worked example 1: elimination is faster
Solve 3x + 2y = 16 and 5x − 2y = 8.
The y-coefficients are +2 and −2, which are opposites. Add the equations: 8x = 24, so x = 3. Then 3(3) + 2y = 16 gives y = 3.5.
Check: 5(3) − 2(3.5) = 15 − 7 = 8.
Now try substitution on the same pair. From the first equation, x = (16 − 2y) ÷ 3, which brings a fraction into the second equation and makes the working heavier. The answer is the same, but the path is longer and the slips are more likely.
Worked example 2: substitution is faster
Solve y = 2x − 1 and 2x + 3y = 13.
The first equation already isolates y, so substitute: 2x + 3(2x − 1) = 13, so 8x − 3 = 13 and x = 2. Then y = 2(2) − 1 = 3.
Check: 2(2) + 3(3) = 4 + 9 = 13.
Elimination here would first rearrange the first equation into 2x − y = 1, then multiply it by 3. That works, but it adds two steps that substitution does not need.
The mistake that costs marks
The common slip is to use whichever method you learnt first and push through the fractions. One sign goes wrong inside a fraction and the answer is lost.
| Step | Wrong | Right |
|---|---|---|
| Read the system | (skipped) | Ask the two questions |
| 3x + 2y = 16, 5x − 2y = 8 | Isolate x with ÷ 3 | Add the equations |
| Result | Fractions and a likely sign slip | 8x = 24, x = 3 |
Ten seconds spent reading the coefficients saves a minute of arithmetic.
Check yourself
Solve 4x − y = 7 and 3x + 2y = 19. Say which method you would pick first and why.
Answer
The first equation has y with coefficient −1, so substitution is quick: y = 4x − 7. Then 3x + 2(4x − 7) = 19 gives 11x − 14 = 19, so x = 3 and y = 5.
Elimination also works: double the first equation to 8x − 2y = 14, and add to get 11x = 33. Both routes give x = 3, y = 5. Check: 4(3) − 5 = 7 and 3(3) + 2(5) = 19.
What to study next
Next, see how answers that are mathematically valid can still be impossible in a story problem, in rejecting invalid solutions in contextual systems. The word-problem structure worksheet helps you set up the equations first.
If you want a teacher to work through method choice with you, see online one-to-one Additional Mathematics tuition.