The substitution method solves a system of two equations by turning it into one equation with one unknown. You make one variable the subject of one equation, then substitute that expression into the other equation.
The method
- Isolate a variable in one of the equations, such as y = 2x − 1.
- Substitute that expression for the same variable in the other equation.
- Solve the new equation, which has only one variable.
- Substitute back to find the other variable.
- Check the pair of values in both original equations.
Worked examples
Example 1: one equation already gives y
Solve y = 2x − 1 and 3x + y = 9.
Replace y in the second equation with 2x − 1: 3x + (2x − 1) = 9.
Simplify: 5x − 1 = 9, so 5x = 10 and x = 2.
Substitute back: y = 2 × 2 − 1 = 3. The solution is (2, 3).
Check: 3 × 2 + 3 = 9 ✓ and 2 × 2 − 1 = 3 ✓.
See Example 1 in the Intersection Finder →
Example 2: isolate a variable first
Solve x + 2y = 7 and 2x − y = 4.
The second equation has −y, which is easy to isolate: y = 2x − 4.
Substitute into the first: x + 2(2x − 4) = 7, so x + 4x − 8 = 7, 5x = 15, and x = 3.
Then y = 2 × 3 − 4 = 2. The solution is (3, 2).
Check: 3 + 2 × 2 = 7 ✓ and 2 × 3 − 2 = 4 ✓.
Example 3: both equations give y
Solve y = ½x + 1 and y = −x + 4.
Both right-hand sides equal y, so they equal each other: ½x + 1 = −x + 4.
Add x to both sides and subtract 1: 3/2 x = 3, so x = 2.
Then y = −2 + 4 = 2. The solution is (2, 2).
Tips
- Choose the easiest variable to isolate. Look for one with a coefficient of 1 or −1, so you don’t create fractions.
- Use brackets when you substitute. Writing 2(2x − 4) rather than 2 × 2x − 4 stops you losing a term.
- Substitute into the other equation. Putting the expression back into the equation it came from just gives 0 = 0.
- If x disappears completely, you will be left with a statement like 1 = −3 (no solution) or 4 = 4 (infinitely many solutions). See No solution or infinitely many solutions.
When neither equation has a variable that is easy to isolate, elimination is usually quicker.