Math Core

Substitution works best when a variable is already alone or easy to isolate. When both equations look like $ax + by = c$ , there's often a faster way: add or subtract the equations so that one variable cancels out. This is called elimination.

Adding equations

If $a = b$ and $c = d$ , then $a + c = b + d$ . Adding equal amounts to both sides keeps things balanced. So you can add two equations, left side to left side and right side to right side, and get a new true equation.

Look at this system:

\begin{aligned} x + y &= 10 \\ x - y &= 4 \end{aligned}

The $y$ -terms are $+y$ and $-y$ . They are opposites, so adding the equations makes them cancel.

Worked example: Opposites cancel

Solve the system $x + y = 10$ and $x - y = 4$ .

Add the equations:

\begin{array}{rcl} x + y &=& 10 \\ x - y &=& 4 \\ \hline 2x &=& 14 \end{array}

So $x = 7$ . Substitute into $x + y = 10$ : $7 + y = 10$ , so $y = 3$ .

Check: $7 + 3 = 10$ ✓ and $7 - 3 = 4$ ✓. The solution is $(7, 3)$ .

Subtracting equations

If a variable has the same coefficient in both equations, subtract instead of add.

Worked example: Same coefficients: subtract

Solve the system $3x + 2y = 17$ and $3x - y = 5$ .

Both equations have $3x$ . Subtract the second equation from the first:

\begin{aligned} (3x + 2y) - (3x - y) &= 17 - 5 \\ 3x + 2y - 3x + y &= 12 \\ 3y &= 12 \\ y &= 4 \end{aligned}

Substitute into $3x - y = 5$ : $3x - 4 = 5$ , so $3x = 9$ and $x = 3$ .

Check: $3(3) + 2(4) = 17$ ✓ and $3(3) - 4 = 5$ ✓. The solution is $(3, 4)$ .

Common mistake

When you subtract an equation, subtract every term, including the right side. $(3x + 2y) - (3x - y)$ is $3y$ , not $y$ , because subtracting $-y$ is the same as adding $y$ . If signs trip you up, multiply the second equation by $-1$ and add instead.

Multiplying first

Often no variable cancels right away. Multiply one equation (or both) by a number so that one variable gets opposite coefficients. Multiplying both sides of an equation by the same number doesn't change its solutions.

Solving by elimination

Write both equations in the form $ax + by = c$ , with like terms lined up.
If needed, multiply one or both equations so the coefficients of one variable are opposites.
Add the equations to eliminate that variable. Solve for the other one.
Substitute back to find the eliminated variable.
Check the pair in both original equations.

Worked example: Multiply one equation

Solve the system $2x + 3y = 13$ and $x - y = 4$ .

Multiply the second equation by $3$ so the $y$ -terms become $+3y$ and $-3y$ :

3(x - y) = 3(4) \quad\Rightarrow\quad 3x - 3y = 12

Add it to the first equation:

\begin{aligned} (2x + 3y) + (3x - 3y) &= 13 + 12 \\ 5x &= 25 \\ x &= 5 \end{aligned}

Substitute into $x - y = 4$ : $5 - y = 4$ , so $y = 1$ .

Check: $2(5) + 3(1) = 13$ ✓ and $5 - 1 = 4$ ✓. The solution is $(5, 1)$ .

Worked example: Multiply both equations

Solve the system $3x + 4y = 10$ and $2x - 3y = -16$ .

To eliminate $y$ , make the $y$ -coefficients $+12$ and $-12$ . Multiply the first equation by $3$ and the second by $4$ :

\begin{aligned} 9x + 12y &= 30 \\ 8x - 12y &= -64 \end{aligned}

Add them: $17x = -34$ , so $x = -2$ .

Substitute into $3x + 4y = 10$ : $3(-2) + 4y = 10$ , so $4y = 16$ and $y = 4$ .

Check: $3(-2) + 4(4) = -6 + 16 = 10$ ✓ and $2(-2) - 3(4) = -4 - 12 = -16$ ✓. The solution is $(-2, 4)$ .

Tip

If both variables cancel, look at what's left. A false statement like $0 = 3$ means no solution. A true statement like $0 = 0$ means infinitely many solutions, just like with substitution.

Practice

Practice 1

Solve the system $x + y = 9$ and $x - y = 3$ .

Enter a point like (2, -3)

Practice 2

Solve the system $2x + y = 11$ and $3x - y = 9$ .

Enter a point like (2, -3)

Practice 3

Solve the system $4x + 3y = 23$ and $4x - y = 3$ .

Enter a point like (2, -3)

Practice 4

Solve the system $x + 2y = 8$ and $3x - 4y = 4$ .

Enter a point like (2, -3)

Practice 5

Solve the system $5x - 2y = 1$ and $3x + 4y = 37$ .

Enter a point like (2, -3)

Practice 6

Solve the system $2x + 3y = -1$ and $3x + 2y = 6$ .

Enter a point like (2, -3)

Practice 7

What is the solution of the system $2x + y = 5$ and $4x + 2y = 7$ ?

Practice 8

Solve the system $3x + 5y = 7$ and $2x - 3y = -8$ .

Enter a point like (2, -3)