Solving Linear Systems by Elimination

This summary covers why some linear systems are awkward to solve by graphing or by substitution, and how to solve a linear system by adding or subtracting its equations.

When other methods are awkward

Consider the following linear system:

$$\begin{array}{rlrl}& \text{A}:& 5x+2y& =10\\ & \text{B}:& 3x+2y& =2\end{array}$$

Solving this linear system by graphing, the solution appears to be $(4,-5)$.

The lines 5x + 2y = 10 and 3x + 2y = 2 cross at the marked point (4, -5)

Now consider solving the linear system by substitution. Does it look like it will be easy to solve using this method?

Not really. No matter which variable we isolate, we end up with a fraction for a coefficient. For example, from A:

$$\begin{array}{rl}5x+2y& =10\\ 2y& =-5x+10\\ y& =-\frac{5}{2}x+5\end{array}$$

Solving by elimination

There is another method. It is called solving linear systems by elimination. Here is how it works:

1. Look for terms that have matching coefficients.
2. Add or subtract the equations to eliminate a variable.
3. Continue to solve for the variable that is left.
4. Substitute the value you found into either equation, and solve for the other variable.

Example 1

Solve the linear system above by elimination.

The $y$-terms have matching coefficients, so subtract B from A.

$$\begin{array}{crcrcrl}& 5x& +& 2y& =& 10& \text{A}\\ -& 3x& +& 2y& =& 2& \text{B}\\ & 2x& & & =& 8& \end{array}$$ $$\begin{array}{rl}\frac{2x}{2}& =\frac{8}{2}\\ x& =4\end{array}$$

Substitute $x=4$ into B.

$$\begin{array}{rl}3x+2y& =2\\ 3(4)+2y& =2\\ 12+2y-12& =2-12\\ 2y& =-10\\ \frac{2y}{2}& =\frac{-10}{2}\\ y& =-5\end{array}$$

$\therefore$ the solution is $(4,-5)$, as the graph suggested.

Example 2

Solve by elimination.

The $x$-terms have opposite coefficients, so add the equations.

$$\begin{array}{crcrcrl}& 4x& +& 5y& =& 7& \text{A}\\ +& -4x& -& 2y& =& -1& \text{B}\\ & & & 3y& =& 6& \end{array}$$ $$\begin{array}{rl}\frac{3y}{3}& =\frac{6}{3}\\ y& =2\end{array}$$

Substitute $y=2$ into A.

$$\begin{array}{rl}4x+5y& =7\\ 4x+5(2)& =7\\ 4x+10-10& =7-10\\ 4x& =-3\\ \frac{4x}{4}& =\frac{-3}{4}\\ x& =-\frac{3}{4}\end{array}$$

$\therefore$ the solution is $\left(-\frac{3}{4},2\right)$. Fractions are OK!

Example 3

Solve by elimination.

$$\begin{array}{rlrl}& \text{A}:& 2x-3y& =2\\ & \text{B}:& 5x+6y& =5\end{array}$$

No coefficients match. But what if we multiply both sides of A by $2$? As long as we multiply both sides by the same value, the equation stays equivalent.

Now we can add the equations to eliminate $y$.

$$\begin{array}{crcrcrl}& 4x& -& 6y& =& 4& \text{A}\times 2\\ +& 5x& +& 6y& =& 5& \text{B}\\ & 9x& & & =& 9& \end{array}$$ $$\begin{array}{rl}\frac{9x}{9}& =\frac{9}{9}\\ x& =1\end{array}$$

Substitute $x=1$ into B.

$$\begin{array}{rl}5x+6y& =5\\ 5(1)+6y& =5\\ 5+6y-5& =5-5\\ 6y& =0\\ \frac{6y}{6}& =\frac{0}{6}\\ y& =0\end{array}$$

$\therefore$ the solution is $(1,0)$.

Example 4

Solve by elimination.

$$\begin{array}{rlrl}& \text{A}:& 3x-7y& =13\\ & \text{B}:& 4x-5y& =13\end{array}$$

If needed to get matching coefficients, we can multiply both equations.

Now we can subtract to eliminate $x$.

$$\begin{array}{crcrcrl}& 12x& -& 28y& =& 52& \text{A}\times 4\\ -& 12x& -& 15y& =& 39& \text{B}\times 3\\ & & & -13y& =& 13& \end{array}$$ $$\begin{array}{rl}\frac{-13y}{-13}& =\frac{13}{-13}\\ y& =-1\end{array}$$

Substitute $y=-1$ into A.

$$\begin{array}{rl}3x-7y& =13\\ 3x-7(-1)& =13\\ 3x+7-7& =13-7\\ 3x& =6\\ \frac{3x}{3}& =\frac{6}{3}\\ x& =2\end{array}$$

$\therefore$ the solution is $(2,-1)$.

TIP

If you’d like – this is completely optional – you can watch any or all of these (vintage) videos where I explain how to solve linear systems by elimination:

Video Showcase – Solving Linear Systems by Elimination

Curriculum connection

B1.1

solve systems of two linear equations involving two variables, using the algebraic method of substitution or elimination (Sample problem: Solve $y=\frac{1}{2}x-5$, $3x+2y=-2$ for $x$ and $y$ algebraically, and verify algebraically and graphically);

Link to original