Question
Solve the following pair of linear equations using the substitution method:
2x + 3y = 7
x − y = 1
Solution — Step by Step
The substitution method is ideal here because the second equation has a coefficient of 1 for x — making isolation straightforward.
Step 1: Choose the simpler equation and isolate one variable.
We pick x − y = 1 because both coefficients are 1 (no fractions will appear when isolating).
x − y = 1 x = y + 1
We've expressed x entirely in terms of y.
Step 2: Substitute into the other equation.
Replace x with (y + 1) in the first equation:
2(y + 1) + 3y = 7
This is now one equation with one unknown — we can solve it directly.
Step 3: Expand and simplify.
2y + 2 + 3y = 7 5y + 2 = 7 5y = 5 y = 1
Step 4: Back-substitute to find x.
From Step 1: x = y + 1 = 1 + 1 = 2
Step 5: Verify in BOTH original equations.
Equation 1: 2(2) + 3(1) = 4 + 3 = 7 ✓ Equation 2: 2 − 1 = 1 ✓
Answer
x = 2, y = 1
The solution represents the point (2, 1) — the intersection of the two lines on a coordinate plane.
Why This Works
The substitution method works by reducing a system of two equations with two unknowns to a single equation with one unknown. Once we express x in terms of y, substituting that expression "collapses" one variable, making the problem solvable.
Every pair of simultaneous linear equations with a unique solution can be solved this way. The key is choosing which equation and which variable to isolate first — pick the one with the smallest coefficient to avoid fractions.
Alternative Method: Elimination
We can also verify by elimination:
Multiply equation 2 by 3: 3x − 3y = 3
Add to equation 1: (2x + 3y) + (3x − 3y) = 7 + 3
5x = 10 → x = 2, then y = 1. Same answer.
💡 Expert Tip
The substitution method gets messy when neither equation has a variable with coefficient 1. For example, 3x + 2y = 5 and 4x + 3y = 7 would require expressing x = (5 − 2y)/3, which introduces fractions. In that case, switch to elimination — it's faster and cleaner.
Common Mistake
⚠️ Common Mistake
Mistake: Substituting back into the substitution expression instead of an original equation.
After finding y = 1, some students substitute back into x = y + 1 (the substitution expression) and stop. That's fine here. But make sure you VERIFY in the original equations — not just the derived expression. The derived expression is correct by construction; verification in the original equations catches errors from earlier steps.