Elimination Method: Steps and Examples

The elimination method solves a system of equations by adding or subtracting the equations so that one variable cancels out. If nothing cancels right away, first multiply one or both equations so a pair of coefficients become opposites. Solve for the variable that’s left, then substitute to find the other one.

Before this lesson: You should know what a system of equations is and how to use the substitution method to find the second variable.

Extra practice: The Algebra 1 Unit 5 Practice Pack has printable problems, answer keys and a cheat sheet for this unit.

How the Elimination Method Works

If two things are equal, and you add the same amount to both, they stay equal. Elimination uses this idea with whole equations.

Look at this system:

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

The y-terms are opposites: $+y$ and $-y$ . When you add the two equations, they cancel, or “eliminate” each other.

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

Now only x is left, so $x = 7$ . Then $7 + y = 10$ , so $y = 3$ . The solution is (7, 3).

Steps for the Elimination Method

Line up the equations so the x-terms, y-terms, equal signs and numbers are in columns.
Make opposites. If no pair of coefficients are opposites, multiply one or both equations by a number so they are. Multiply every term, on both sides.
Add the equations to eliminate one variable.
Solve for the variable that’s left.
Substitute that value into either original equation to find the other variable.
Check the pair in both original equations.

Add or subtract? If the coefficients are opposites, like 3y and −3y, add. If they are the same, like 2y and 2y, subtract, or multiply one equation by −1 and then add.

Worked Examples

Example 1: Add the equations

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

\begin{array}{rrcl} & 2x + y & = & 11 \\ + & 3x - y & = & 9 \\ \hline & 5x & = & 20 \end{array}

Step 1. The y-terms, $y$ and $-y$ , are opposites. Add the equations so the y-terms cancel. (See combining like terms.)

x = 4

Step 2. Divide both sides by 5.

2(4) + y = 11

Step 3. Substitute $x = 4$ into the first equation.

8 + y = 11

Step 4. Multiply.

y = 3

Step 5. Subtract 8 from both sides.

Answer: (4, 3)

Check: $2(4) + 3 = 8 + 3 = 11$ and $3(4) - 3 = 12 - 3 = 9$ .

Example 2: Subtract the equations

Solve the system $3x + 2y = 16$ and $x + 2y = 8$ .

\begin{array}{rrcl} & 3x + 2y & = & 16 \\ - & (x + 2y) & = & 8 \\ \hline & 2x & = & 8 \end{array}

Step 1. The y-terms are the same, $2y$ and $2y$ . Subtract the second equation from the first. Subtract every term, so $2y - 2y = 0$ and $16 - 8 = 8$ .

x = 4

Step 2. Divide both sides by 2.

4 + 2y = 8

Step 3. Substitute $x = 4$ into the second equation.

2y = 4

Step 4. Subtract 4 from both sides.

y = 2

Step 5. Divide both sides by 2.

Answer: (4, 2)

Check: $3(4) + 2(2) = 12 + 4 = 16$ and $4 + 2(2) = 4 + 4 = 8$ .

Example 3: Multiply one equation first

Solve the system $2x + 3y = 12$ and $x - y = 1$ .

3x - 3y = 3

Step 1. Nothing cancels yet. Multiply every term of the second equation by 3, so its y-term becomes $-3y$ . (See distributive property.)

\begin{array}{rrcl} & 2x + 3y & = & 12 \\ + & 3x - 3y & = & 3 \\ \hline & 5x & = & 15 \end{array}

Step 2. Add the new equation to the first equation. The y-terms cancel.

x = 3

Step 3. Divide both sides by 5.

3 - y = 1

Step 4. Substitute $x = 3$ into $x - y = 1$ .

-y = -2

Step 5. Subtract 3 from both sides.

y = 2

Step 6. Divide both sides by $-1$ .

Answer: (3, 2)

Check: $2(3) + 3(2) = 6 + 6 = 12$ and $3 - 2 = 1$ .

Example 4: Multiply both equations

Solve the system $3x + 4y = 10$ and $2x - 3y = 1$ .

9x + 12y = 30

Step 1. To eliminate y, make the y-terms $12y$ and $-12y$ . Multiply every term of the first equation by 3.

8x - 12y = 4

Step 2. Multiply every term of the second equation by 4.

\begin{array}{rrcl} & 9x + 12y & = & 30 \\ + & 8x - 12y & = & 4 \\ \hline & 17x & = & 34 \end{array}

Step 3. Add the new equations. The y-terms cancel.

x = 2

Step 4. Divide both sides by 17.

3(2) + 4y = 10

Step 5. Substitute $x = 2$ into $3x + 4y = 10$ .

6 + 4y = 10

Step 6. Multiply.

4y = 4

Step 7. Subtract 6 from both sides.

y = 1

Step 8. Divide both sides by 4.

Answer: (2, 1)

Check: $3(2) + 4(1) = 6 + 4 = 10$ and $2(2) - 3(1) = 4 - 3 = 1$ .

Example 5: Word problem

Three tacos and two drinks cost $13. Five tacos and two drinks cost $19. How much does one taco cost? How much does one drink cost?

\begin{cases} 3t + 2d = 13 \\ 5t + 2d = 19 \end{cases}

Step 1. Let $t$ be the price of a taco and $d$ the price of a drink. Write an equation for each order.

\begin{array}{rrcl} & 5t + 2d & = & 19 \\ - & (3t + 2d) & = & 13 \\ \hline & 2t & = & 6 \end{array}

Step 2. The d-terms are the same, so subtract the first equation from the second.

t = 3

Step 3. Divide both sides by 2.

3(3) + 2d = 13

Step 4. Substitute $t = 3$ into the first equation.

9 + 2d = 13

Step 5. Multiply.

2d = 4

Step 6. Subtract 9 from both sides.

d = 2

Step 7. Divide both sides by 2.

Answer: A taco costs $3, and a drink costs $2.

Check: Second order: $5(3) + 2(2) = 15 + 4 = 19$ .

Which Method Should You Use?

Method	Best when	Example
Graphing	You want to see the solution, and it has whole numbers	$y = x + 1$ and $y = -x + 5$
Substitution	A variable is already alone or has a coefficient of 1	$y = 2x - 6$ and $4x - 3y = 6$
Elimination	Both equations are in the form $Ax + By = C$	$3x + 4y = 10$ and $2x - 3y = 1$

When Both Variables Cancel

Sometimes x and y both cancel. Look at what is left.

A false statement, like $0 = 6$ , means no solution. The lines are parallel.
A true statement, like $0 = 0$ , means infinitely many solutions. The equations describe the same line.

Learn more in Number of Solutions.

Common Mistakes

Mistake	What it looks like	How to fix it
Multiplying only one side	$3(x - y) = 1$ becomes $3x - 3y = 1$	Multiply both sides: $3x - 3y = 3$ .
Multiplying only one term	$3(x - y)$ becomes $3x - y$	Multiply every term: $3x - 3y$ .
Adding when the terms match	Adding $2y + 2y$ gives 4y, which doesn’t cancel	Same terms: subtract. Opposite terms: add.
Subtracting only part	$(3x + 2y) - (x + 2y) = 2x + 4y$	Subtract every term: $2y - 2y = 0$ .
Stopping after one variable	Answering “ $x = 4$ “	Substitute to find the other variable too.

Practice Problems

Solve each system by elimination. Check your answers.

$x + y = 9$ and $x - y = 3$
$2x + y = 11$ and $3x - y = 9$
$4x + 3y = 25$ and $4x - y = 13$
$x + 2y = 7$ and $3x - 2y = 5$
$2x + 5y = 1$ and $x + 3y = 1$
$3x - 2y = 8$ and $5x + 4y = 6$
$2x + 3y = 7$ and $3x + 4y = 10$
$4x - 6y = 10$ and $2x - 3y = 5$
Two adult tickets and three student tickets cost $36. One adult ticket and three student tickets cost $27. Find the price of each ticket.

Show answers

2x = 12

Add the equations. The y-terms cancel.

x = 6

Divide both sides by 2.

6 + y = 9

Substitute into x + y = 9.

y = 3

Subtract 6 from both sides.

Check: 6 − 3 = 3. Answer: (6, 3).

5x = 20

Add the equations. The y-terms cancel.

x = 4

Divide both sides by 5.

2(4) + y = 11

Substitute into 2x + y = 11.

8 + y = 11

Multiply.

y = 3

Subtract 8 from both sides.

Check: 3(4) − 3 = 9. Answer: (4, 3).

4y = 12

Subtract the second equation from the first. The x-terms cancel, and 3y − (−y) = 4y.

y = 3

Divide both sides by 4.

4x - 3 = 13

Substitute into 4x − y = 13.

4x = 16

Add 3 to both sides.

x = 4

Divide both sides by 4.

Check: 4(4) + 3(3) = 16 + 9 = 25. Answer: (4, 3).

4x = 12

Add the equations. The y-terms cancel.

x = 3

Divide both sides by 4.

3 + 2y = 7

Substitute into x + 2y = 7.

2y = 4

Subtract 3 from both sides.

y = 2

Divide both sides by 2.

Check: 3(3) − 2(2) = 9 − 4 = 5. Answer: (3, 2).

2x + 6y = 2

Multiply every term of the second equation by 2.

-y = -1

Subtract it from the first equation. The x-terms cancel.

y = 1

Divide both sides by −1.

x + 3 = 1

Substitute into x + 3y = 1.

x = -2

Subtract 3 from both sides.

Check: 2(−2) + 5(1) = −4 + 5 = 1. Answer: (−2, 1).

6x - 4y = 16

Multiply every term of the first equation by 2.

11x = 22

Add it to the second equation. The y-terms cancel.

x = 2

Divide both sides by 11.

3(2) - 2y = 8

Substitute into 3x − 2y = 8.

6 - 2y = 8

Multiply.

-2y = 2

Subtract 6 from both sides.

y = -1

Divide both sides by −2.

Check: 5(2) + 4(−1) = 10 − 4 = 6. Answer: (2, −1).

6x + 9y = 21

Multiply every term of the first equation by 3.

6x + 8y = 20

Multiply every term of the second equation by 2.

y = 1

Subtract the second new equation from the first. The x-terms cancel.

2x + 3 = 7

Substitute into 2x + 3y = 7.

2x = 4

Subtract 3 from both sides.

x = 2

Divide both sides by 2.

Check: 3(2) + 4(1) = 6 + 4 = 10. Answer: (2, 1).

4x - 6y = 10

Multiply every term of the second equation by 2.

0 = 0

Subtract it from the first equation. Every term cancels.

0 = 0 is always true. The equations describe the same line, so there are infinitely many solutions.

2a + 3s = 36

Let a be the adult price and s the student price. Write the first equation.

a + 3s = 27

Write the second equation.

a = 9

Subtract the second from the first. The s-terms cancel.

9 + 3s = 27

Substitute into a + 3s = 27.

3s = 18

Subtract 9 from both sides.

s = 6

Divide both sides by 3.

Check: 2(9) + 3(6) = 18 + 18 = 36. An adult ticket costs $9, and a student ticket costs $6.

Want more practice? The Algebra 1 Unit 5 Practice Pack has 55 pages of problems on this and the other lessons in the unit, with step-by-step answers and a cheat sheet.

Elimination Method FAQ

What is the elimination method?

The elimination method solves a system by adding or subtracting the equations so one variable cancels. You may need to multiply an equation first to make opposite coefficients.

When is elimination better than substitution?

Elimination is usually faster when both equations are in standard form, $Ax + By = C$ , and no variable is already alone.

Should you eliminate x or y?

Either one works and gives the same answer. Choose the variable that needs the least multiplying, or one that already has opposite coefficients.

What does it mean if both variables cancel?

If you get a false statement like $0 = 6$ , there is no solution. If you get a true statement like $0 = 0$ , there are infinitely many solutions.

Summary

Line up the equations by x-terms, y-terms and numbers.
Make one pair of coefficients opposites, multiplying every term if you need to.
Add the equations to eliminate a variable. Subtract if the terms are the same.
Solve, substitute to find the other variable, and check in both equations.
$0 = 0$ means infinitely many solutions. A false statement means no solution.

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Next lesson: Number of Solutions

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