Solving by Substitution

≈ 30 min

Substitution turns two equations into one by replacing a variable with what it equals.

Worked example. Solve:

y = x + 1 \qquad (1)

x + y = 7 \qquad (2)

Equation (1) hands us $y$ on a plate. Substitute $x + 1$ for $y$ in (2):

x + (x + 1) = 7

2x + 1 = 7 \;\Rightarrow\; 2x = 6 \;\Rightarrow\; x = 3

Back-substitute into (1): $y = 3 + 1 = 4$ . Solution: $(3, 4)$ .

Check in (2): $3 + 4 = 7$ ✓ — always check in the equation you did not use for back-substitution.

Solve by substitution and give the value of $x$ :

y = x + 1 \qquad x + y = 7

Solve by substitution and give the value of $y$ :

y = 2x \qquad x + y = 12

For the system $y = x - 2$ and $3x + y = 10$ , which substitution is correct?

$3x + (x - 2) = 10$

$3(x - 2) + y = 10$

$3x + x = 10 - 2$

$3(y + 2) + y = 10$ only