If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Systems of equations with elimination

Walk through examples of solving systems of equations with elimination.
In this article, we're going to be solving systems of linear equations using a strategy called elimination. First, we need to understand that it's okay to add equations together.
Key idea: Anytime we have two true equations, we can add or subtract them to create another true equation.
For example, here are two very basic true equations:
2, equals, 2
5, equals, 5
We can add these equations together to create another true equation:
\begin{aligned} 2 &= 2 &\\ +~~~\ 5&=5&\\ \hline\\ 7 &=7 \end{aligned}
Or we can subtract these equations to create another true equation:
\begin{aligned} 2 &= 2 &\\ -\ ~~~5&=5&\\ \hline\\ -3 &=-3 \end{aligned}
Here's another example with more complicated equations:
\begin{aligned} 2x + 3 &= 7 &\\ +~~~\ 4x + 1&=9&\\ \hline\\ 6x+4 &=16 \end{aligned}
Great, now that we see that it's okay to add or subtract equations, we can move onto solving a system of equations using elimination.

## Solving a system of equations using elimination

We'll solve this system of equations as an example:
x, plus, 3, y, equals, 8, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 1, end text, end color gray
4, x, minus, 3, y, equals, 17, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 2, end text, end color gray
The hard thing about solving is that there are two variables x and y. If only we could get rid of one of the variables...
Here's an idea! Let's add the two equations together to cancel out the y variable:
\begin{aligned} x \maroonD{+ 3y} &= 8 &\\ +~~~\ 4x \maroonD{- 3y}&=17&\\ \hline\\ 5x\maroonD{+0} &=25 \end{aligned}
Brilliant! Now we have an equation with just the x variable that we know how to solve:
\begin{aligned} 5x + 0 &= 25\\\\ 5x&= 25\\\\\ \blueD x&\blueD= \blueD5&\gray{\text{Divide each side by 5}}\end{aligned}
Baller! Let's use the first equation to find y when x equals 5:
\begin{aligned} \blueD x + 3y &= 8&\gray{\text{Equation 1}} \\\\ \blueD 5 + 3y &= 8 &\gray{\text{Substitute 5 for x}}\\\\ 3y &= 3 &\gray{\text{Subtract 5 from each side}}\\\\ \greenD y &\greenD = \greenD 1&\gray{\text{Divide each side by 3}}\end{aligned}
Sweet! So the solution to the system of equations is left parenthesis, start color #11accd, 5, end color #11accd, comma, start color #1fab54, 1, end color #1fab54, right parenthesis.
Use elimination to solve the following system of equations.
4, y, minus, 2, x, equals, 4
5, y, plus, 2, x, equals, 23
x, equals
y, equals

## Multiplying one of the equations by a constant, then using elimination

That last example worked out great because the y variable was eliminated when we added the equations. Sometimes it isn't quite that easy.
Take this system of equation as an example:
6, x, plus, 5, y, equals, 28, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 1, end text, end color gray
3, x, minus, 4, y, equals, 1, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 2, end text, end color gray
If we add these equations, neither the x or y variable will be eliminated, so that won't work. Here are the steps for problems like this:
Step 1: Multiply one of the equations by a constant so that when we add it to the other equation, one of the variables is eliminated.
\begin{aligned} \maroonD{-2}(3x -4y) &= \maroonD{-2}(1) &\gray{\text{Multiply the second equation by} -2} \\\\ \blueD{-6x+8y} &\blueD= \blueD{-2}&\gray{\text{Simplify to get a new equation}}\end{aligned}
Step 2: Add the new equation to the equation we didn't use in step 1 in order to eliminate one of the variables.
\begin{aligned} 6x + 5y &= 28 &\gray{\text{Equation 1}}\\ +\ \blueD{-6x + 8y}&\blueD=\blueD{-2}&\gray{\text{The new equation}}\\ \hline\\ 13y &=26 \end{aligned}
Step 3: Solve for y.
\begin{aligned} 13y &= 26\\\\ y&= 2&\gray{\text{Divide each side by 13}}\end{aligned}
Step 4: Substitute y, equals, 2 into one of the original equations, and solve for x.
\begin{aligned} 3x - 4y &= 1 &\gray{\text{Equation 2}} \\\\ 3x -4(2) &= 1 &\gray{\text{Substitute 2 for y}} \\\\ 3x -8 &= 1 \\\\ 3x &= 9 &\gray{\text{Add 8 to each side}} \\\\ x &= 3 &\gray{\text{Divide each side by 3}} \end{aligned}
So our solution is left parenthesis, 3, comma, 2, right parenthesis.
Use elimination to solve the following system of equations.
8, x, plus, 14, y, equals, 12
minus, 6, x, minus, 7, y, equals, minus, 16
x, equals
y, equals

## Multiplying both of the equations by a constant, then using elimination

Sometimes we'll need to multiply both equations by a constant when using elimination.
For example, consider this system of equations:
5, x, plus, 3, y, equals, 14, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 1, end text, end color gray
3, x, plus, 2, y, equals, 8, space, space, space, space, space, space, space, space, start color gray, start text, E, q, u, a, t, i, o, n, space, 2, end text, end color gray
Here are the steps to solve a system of equations like this one:
Step 1: Multiply each equation by a constant so we can eliminate one variable.
5, x, plus, 3, y, equals, 14, right arrow, start color #ca337c, start text, M, u, l, t, i, p, l, y, space, b, y, space, 2, end text, end color #ca337c, right arrow, 10, x, plus, 6, y, equals, 28
3, x, plus, 2, y, equals, 8, right arrow, start color #ca337c, start text, M, u, l, t, i, p, l, y, space, b, y, space, 3, end text, end color #ca337c, right arrow, 9, x, plus, 6, y, equals, 24
Step 2: Combine the new equations to eliminate one variable.
\begin{aligned} 10x + 6y &= 28 \\ 9x + 6y&=24\\ \hline\\ x + 0 &=4 &\gray{\text{Subtract the equations}} \end{aligned}
Step 3: Substitute x, equals, 4 into one of the original equations, and solve for y.
\begin{aligned} 3x + 2y &= 8 &\gray{\text{Equation 2}} \\\\ 3(4) + 2y &= 8 &\gray{\text{Substitute 4 for x}} \\\\ 12 +2y &= 8 \\\\ 2y &= -4 &\gray{\text{Subtract 12 from each side}} \\\\ y &= -2 &\gray{\text{Divide each side by 2}} \end{aligned}
So our solution is left parenthesis, 4, comma, minus, 2, right parenthesis.
Use elimination to solve the following system of equations.
5, x, plus, 4, y, equals, minus, 14
3, x, plus, 6, y, equals, 6
x, equals
y, equals

## Let's practice!

1) Use elimination to solve the following system of equations.
3, y, plus, x, equals, 7
2, y, minus, x, equals, minus, 2
x, equals
y, equals

2) Use elimination to solve the following system of equations.
minus, 7, y, minus, 4, x, equals, 1
7, y, minus, 2, x, equals, 53
x, equals
y, equals

3) Use elimination to solve the following system of equations.
minus, 9, y, plus, 4, x, minus, 20, equals, 0
minus, 7, y, plus, 16, x, minus, 80, equals, 0
x, equals
y, equals

4) Use elimination to solve the following system of equations.
3, x, minus, 11, y, equals, minus, 1
2, x, minus, 5, y, equals, minus, 3
x, equals
y, equals

## Challenge problem

A school is selling tickets to a play. On the first day of ticket sales, the school sold 6 adult tickets and 10 student tickets for a total of dollar sign, 140. On the second day of ticket sales, the school sold 7 adult tickets and 3 student tickets for a total of dollar sign, 94.
Write and solve a system of equations to find the cost of an adult ticket and the cost of a student ticket.
The cost of an adult ticket is dollar sign
.
The cost of a student ticket is dollar sign
.