Substitution & Addition Methods

Graphing gives us intuition, but for exact answers we need algebra. Two fundamental techniques for solving linear systems are the substitution method and the addition (elimination) method.

Substitution Method

⚡Theorem — Substitution Method — Steps

Isolate one variable in one of the equations (choose the one with the simplest coefficient).
Substitute the resulting expression into the other equation.
Solve the single-variable equation that results.
Back-substitute to find the other variable.
Write the solution as an ordered pair and verify it in both original equations.

✎Example — Substitution Method

Solve the system:

$\begin{cases} 2x + y = 5 \\ x - 3y = 1 \end{cases}$

Solution. Equation 2 has $x$ with coefficient 1, so isolate $x$ :

$x = 1 + 3y$

Substitute into equation 1:

$2(1 + 3y) + y = 5$ $2 + 6y + y = 5$ $7y = 3$ $y = \dfrac{3}{7}$

Back-substitute:

$x = 1 + 3 \cdot \dfrac{3}{7} = 1 + \dfrac{9}{7} = \dfrac{16}{7}$

Solution: $\left(\dfrac{16}{7},\; \dfrac{3}{7}\right)$ .

Check: $2 \cdot \dfrac{16}{7} + \dfrac{3}{7} = \dfrac{32 + 3}{7} = 5$ ✓ and $\dfrac{16}{7} - 3 \cdot \dfrac{3}{7} = \dfrac{16 - 9}{7} = 1$ ✓.

Addition (Elimination) Method

The idea is to add the equations so that one variable cancels out.

⚡Theorem — Addition (Elimination) Method — Steps

Multiply one or both equations by suitable constants so that one variable has opposite coefficients in the two equations.
Add the equations — one variable is eliminated.
Solve the resulting single-variable equation.
Substitute back to find the other variable.
Write the solution and verify.

Example: Direct Addition

✎Example — Elimination — Direct Addition

Solve the system:

$\begin{cases} 3x + 2y = 7 \\ x - 2y = 1 \end{cases}$

Solution. The $y$ -coefficients are already opposites ( $+2y$ and $-2y$ ), so add the equations directly:

$(3x + 2y) + (x - 2y) = 7 + 1$ $4x = 8 \implies x = 2$

Substitute $x = 2$ into the first equation:

$3(2) + 2y = 7 \implies 6 + 2y = 7 \implies y = \dfrac{1}{2}$

Solution: $\left(2,\; \dfrac{1}{2}\right)$ .

Example: Multiplication Required

✎Example — Elimination — With Multiplication

Solve the system:

$\begin{cases} 2x + 3y = 8 \\ 3x - 2y = 1 \end{cases}$

Solution. Multiply equation 1 by 2 and equation 2 by 3 to make the $y$ -coefficients opposites:

$4x + 6y = 16$ $9x - 6y = 3$

Add:

$13x = 19 \implies x = \dfrac{19}{13}$

Substitute into equation 1:

$2 \cdot \dfrac{19}{13} + 3y = 8$ $3y = 8 - \dfrac{38}{13} = \dfrac{104 - 38}{13} = \dfrac{66}{13}$ $y = \dfrac{22}{13}$

Solution: $\left(\dfrac{19}{13},\; \dfrac{22}{13}\right)$ .

When to Use Which Method

Situation	Recommended method
One variable has coefficient $1$ or $-1$	Substitution (easy to isolate)
Coefficients are “nice” opposites or multiples	Elimination (direct addition)
Both coefficients are large or complex	Elimination (with multipliers)

In practice, both methods always work — choose whichever looks simpler for the given system.

✏Exercise — Practice

Solve the system using the method of your choice:

$\begin{cases} x + 2y = 7 \\ 3x - y = 7 \end{cases}$

Verify your answer in both equations.