Math Core

Substitution works best when some variable has a coefficient of $1$ . For a system like $3x + 4y = 5$ and $2x - 3y = 9$ , isolating any variable drags in fractions. The elimination method avoids that by adding or subtracting whole equations so that one variable cancels.

Adding equations

If $a = b$ and $c = d$ , then $a + c = b + d$ : adding equal amounts to equal amounts keeps things equal. So you may add two equations, left side to left side and right side to right side, and the result is still true for the solution of the system.

That's useful when a variable has opposite coefficients in the two equations:

\begin{array}{rcrcr} 3x & + & 2y & = & 16 \\ +\;\; 5x & - & 2y & = & 16 \\ \hline 8x & & & = & 32 \end{array}

The $y$ -terms cancel, since $2y + (-2y) = 0$ , leaving $8x = 32$ , so $x = 4$ . Substitute into either original equation: $3(4) + 2y = 16$ gives $2y = 4$ , so $y = 2$ . The solution is $(4, 2)$ .

If a variable has the same coefficient in both equations, subtract instead (or multiply one equation by $-1$ and add).

Worked example: Subtracting equations

Solve the system.

\begin{cases} 2x + 5y = 9 \\ 2x + y = -3 \end{cases}

Both equations contain $2x$ . Subtract the second equation from the first, term by term:

(2x - 2x) + (5y - y) = 9 - (-3) \quad\Longrightarrow\quad 4y = 12 \quad\Longrightarrow\quad y = 3.

Substitute into the second equation: $2x + 3 = -3$ , so $2x = -6$ and $x = -3$ .

Check: $2(-3) + 5(3) = -6 + 15 = 9$ and $2(-3) + 3 = -3$ . The solution is $(-3, 3)$ .

Common mistake

When you subtract an equation, subtract every term, including the constant on the right. Above, $9 - (-3) = 12$ , not $6$ . Many students find it safer to multiply the second equation by $-1$ first, turning it into $-2x - y = 3$ , and then add.

Multiplying first

Most systems don't come with matching coefficients. You can create them: multiplying both sides of an equation by the same nonzero number doesn't change its solutions (it's the same line).

The elimination method

Write both equations in standard form, $Ax + By = C$ , with like terms lined up.
Multiply one or both equations by constants so that one variable's coefficients are opposites.
Add the equations to eliminate that variable, and solve for the other.
Substitute back into either original equation to find the eliminated variable.
Check the pair in both original equations.

Worked example: Multiplying one equation

Solve the system.

\begin{cases} x + 3y = 7 \\ 2x - 5y = -8 \end{cases}

The $x$ -coefficients are $1$ and $2$ . Multiply the first equation by $-2$ so they become $-2$ and $2$ . Multiply every term:

\begin{aligned} -2x - 6y &= -14 \\ 2x - 5y &= -8 \end{aligned}

Add: $-11y = -22$ , so $y = 2$ . Back-substitute into $x + 3y = 7$ : $x + 6 = 7$ , so $x = 1$ .

Check: $1 + 3(2) = 7$ and $2(1) - 5(2) = 2 - 10 = -8$ . The solution is $(1, 2)$ .

When neither coefficient divides the other, multiply both equations, aiming for the least common multiple, just as you would find a common denominator.

Worked example: Multiplying both equations

Solve the system.

\begin{cases} 3x + 4y = 5 \\ 2x - 3y = 9 \end{cases}

The $y$ -coefficients $4$ and $-3$ already have opposite signs. Their least common multiple is $12$ , so multiply the first equation by $3$ and the second by $4$ :

\begin{aligned} 9x + 12y &= 15 \\ 8x - 12y &= 36 \end{aligned}

Add: $17x = 51$ , so $x = 3$ . Back-substitute into $3x + 4y = 5$ : $9 + 4y = 5$ , so $4y = -4$ and $y = -1$ .

Check: $3(3) + 4(-1) = 9 - 4 = 5$ and $2(3) - 3(-1) = 6 + 3 = 9$ . The solution is $(3, -1)$ .

You could instead eliminate $x$ by multiplying by $2$ and $-3$ . Either choice gives the same answer; pick whichever keeps the numbers smaller.

Why elimination works

Each time you multiply and add, you create a new equation. The solution of the original system satisfies it too, because you only combined true statements. So the new equation's line passes through the same intersection point. Elimination cleverly chooses the new line to be vertical ( $x = 3$ ) or horizontal ( $y = -1$ ), where you can read the answer directly.

The two original lines and the eliminated equation x = 3 (dashed) all pass through the solution (3, −1).Open in grapher →

Special cases

As with substitution, if both variables cancel, read what's left. A false statement like $0 = 20$ means no solution; a true statement like $0 = 0$ means infinitely many solutions.

Worked example: Both variables cancel

Solve $6x - 4y = 10$ and $-3x + 2y = 5$ .

Multiply the second equation by $2$ : $-6x + 4y = 10$ . Add it to the first:

(6x - 6x) + (-4y + 4y) = 10 + 10 \quad\Longrightarrow\quad 0 = 20.

This is false, so the system has no solution. The lines are parallel: both have slope $\tfrac{3}{2}$ .

Choosing a method

All three methods solve every system; some are just faster for certain systems.

the system looks like	good choice
an equation already solved for $x$ or $y$ , like $y = 4x - 1$	substitution
a variable with coefficient $1$ or $-1$	substitution or elimination
both equations in standard form, no coefficient of $1$	elimination
you want to see the situation or estimate	graphing

Tip

Before eliminating, rearrange each equation so the $x$ -terms, $y$ -terms and constants line up in columns. An equation like $2y = 3x - 1$ becomes $-3x + 2y = -1$ . Lining up columns prevents adding an $x$ -term to a $y$ -term.

Practice

Practice 1

Solve by elimination: $x + y = 10$ and $x - y = 4$ .

Enter a point like (2, -3)

Practice 2

Solve by elimination: $3x + 2y = 12$ and $3x - 4y = -6$ .

Enter a point like (2, -3)

Practice 3

Solve by elimination: $2x + 3y = 1$ and $4x - y = 9$ .

Enter a point like (2, -3)

Practice 4

To eliminate $y$ from the system $2x + 3y = 7$ and $5x - 2y = 8$ , which step works?

Practice 5

Solve by elimination: $5x + 2y = 4$ and $3x + 5y = -9$ .

Enter a point like (2, -3)

Practice 6

Solve by elimination: $4x - 3y = 25$ and $6x + 5y = 9$ .

Enter a point like (2, -3)

Practice 7

Solve: $2y = 3x - 1$ and $4x + y = 16$ .

Enter a point like (2, -3)

Practice 8

Solve the system $6x - 4y = 10$ and $-9x + 6y = -15$ .