To solve by substitution, take an equation where one letter is already on its own, then replace that letter in the other equation with the whole expression. This leaves one equation in one unknown.
Substitution is the natural choice whenever a question gives you a rule such as y = 2x − 1. It also becomes essential when one equation is not linear, which you will meet in solving a linear and quadratic pair.
When is substitution the better method?
Scan both equations before you write anything. If either one looks like y = … or x = …, substitute. If neither does, you can either rearrange one of them or use elimination.
A quick test: if isolating a letter would create a fraction, elimination is probably kinder. If the letter is already alone, rearranging is free.
The method, step by step
- Choose the equation that already has a letter alone, or rearrange one so it does.
- Substitute that expression, in brackets, into the other equation.
- Expand and collect like terms.
- Solve for the remaining letter.
- Substitute back into the simple equation to get the other letter.
- Check in the equation you did not use for the final step.
Worked example
Solve: y = 2x − 1 (1) and 3x + 2y = 12 (2)
Step 1, substitute (2x − 1) for y in (2): 3x + 2(2x − 1) = 12.
Step 2, expand: 3x + 4x − 2 = 12.
Step 3, collect: 7x − 2 = 12, so 7x = 14 and x = 2.
Step 4, find y using (1): y = 2(2) − 1 = 3.
Step 5, check in (2): 3(2) + 2(3) = 6 + 6 = 12. This matches.
Answer: x = 2, y = 3.
The mistake to watch for
The usual slip is dropping the brackets when substituting.
Mistaken working: 3x + 2 × 2x − 1 = 12, so 7x − 1 = 12 and x = 13/7.
Only the 2x was multiplied by 2. The −1 sits outside the multiplication, so the equation no longer matches the original.
The fix is to write the brackets every time you substitute, even when the expression looks short. Then expand carefully: 2(2x − 1) means 2 × 2x and 2 × (−1). An answer such as 13/7 is also a signal to pause, since textbook-style questions usually give whole numbers or simple fractions, although not always.
Check yourself
Try these on paper, then open each answer.
1. Solve y = x + 4 and 2x + y = 16.
Show answer
Substitute: 2x + (x + 4) = 16, so 3x + 4 = 16 and x = 4. Then y = 4 + 4 = 8. Check: 2(4) + 8 = 16, which matches.
x = 4, y = 8
2. Solve x = 3y − 2 and 2x − y = 6.
Show answer
Substitute: 2(3y − 2) − y = 6, so 6y − 4 − y = 6 and 5y = 10, giving y = 2. Then x = 3(2) − 2 = 4. Check: 2(4) − 2 = 6, which matches.
x = 4, y = 2
3. Solve 2x + y = 7 and 4x − 3y = −1. (Rearrange first.)
Show answer
From (1), y = 7 − 2x. Substitute: 4x − 3(7 − 2x) = −1, so 4x − 21 + 6x = −1 and 10x = 20, giving x = 2. Then y = 7 − 4 = 3. Check: 4(2) − 3(3) = 8 − 9 = −1, which matches.
x = 2, y = 3
Where this leads next
Once both methods feel comfortable, move to interpreting an intersection in context, where the answers describe a real situation. The simultaneous relationships practice set mixes both methods so you practise choosing. The non-calculator working trainer helps you check each arithmetic step.
If you can do either method when told which one to use but hesitate when you must choose, a teacher in online one-to-one Mathematics tuition can help. They can give you a short test for picking the faster method.