The Substitution Method
Best when one variable is already alone, or easy to isolate.
The explanation
Substitution turns two equations into one.
1. Isolate a variable in whichever equation makes it easiest.
2. Substitute that expression into the *other* equation.
3. Solve the single-variable equation.
4. Substitute back to find the other variable.
For y = 3x − 4 and 2x + y = 11:
2x + (3x − 4) = 11 → 5x − 4 = 11 → x = 3.
Then y = 3(3) − 4 = 5. Solution: (3, 5).
Substituting back into the *same* equation you started from is the classic error. It collapses to something true like 11 = 11 and tells you nothing.
Substitution replaces one variable with an equivalent expression, reducing a 2×2 system to a single equation in one unknown. It is valid because the substituted expression is equal to the variable on the solution set.
It is the method of choice when a variable has coefficient ±1, since isolating avoids fractions. When all coefficients are larger, elimination is usually cleaner.
Substituting back into the equation used for isolation produces an identity because that step is not independent information — the second equation is what pins down the value.
Degenerate systems announce themselves the same way as single equations: if both variables vanish and leave a contradiction, the system is inconsistent; if they leave an identity, it is dependent and the solution set is the whole line, written parametrically as (t, mt + b).
Substitution generalises beyond linear systems, and is the standard approach for solving a linear equation together with a quadratic one.
Worked example
Solve x − 2y = 1 and 3x + y = 17 by substitution.
- Isolate x in the first: x = 2y + 1.
- Substitute into the second: 3(2y + 1) + y = 17.
- Expand: 6y + 3 + y = 17, so 7y = 14, y = 2.
- Back-substitute: x = 2(2) + 1 = 5.
Answer: (5, 2)
Common mistakes
- Substituting back into the equation you rearranged.
- Forgetting to distribute the coefficient across the substituted expression.