# ##\$ Elimination Method

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## How to Use the Elimination Method

1. Add, subtract, or multiply both sides of the equations until the variables are on the same side of the equal sign and the coefficients of one of the variables are equal and opposite.
2. Add the equations together.
3. Solve for the non-eliminated variable.
4. Solve for the eliminated variable by plugging the result from Step 3 into either one of the original equations.

## Examples

Use the elimination method to solve this system of equations:

$3x-5y=25$

$-3x+7y=-29$

Which variable is the easiest to eliminate?

$3x-5y=25$

$-3x+7y=-29$

The $$x$$ variables already have equal and opposite coefficients ($$3$$ and $$-3$$), so it will be easiest to eliminate $$x$$.

All the variables are on the left side of the equation so I don’t need to rearrange the equations at all.

What is the result when you add the equations together?

$\begin{array}{cccc} 3x & -5y & = & 25 \\ -3x & +7y & = & -29 \\ \hline & \yellow 2y & \yellow = & \yellow -4 \end {array}$

When I add the equations together, the $$x$$’s eliminate each other, leaving me with $$\yellow 2y=-4$$.

What does y equal?

$\yellow 2y=-4$

To find the y-coordinate of the solution, I will divide both sides of the eliminated equation by $$\yellow 2$$ to undo the multiplication.

$\yellow y=-2$

The y-coordinate of my solution is $$\yellow -2$$.

What does y equal?

Now that I know that $$\yellow y=-2$$, I can plug $$\yellow -2$$ into either one of the original equations to find the value of $$x$$.

$3x-5y=25$

$-3x+7y=-29$

I think it looks slightly easier to plug $$\yellow y=-2$$ into the first equation, so I will do that.

$3x-5({\yellow -2})=25$

I will simplify the right side of the equation by multiplying $$-5$$ and $$\yellow -2$$.

$3x+10=25$

Next, I will subtract $$10$$ from both sides of the equation to undo the addition.

$3x=15$

Then I will divide both sides of the equation by $$3$$ to undo the multiplication.

$x=5$

The x-coordinate of my solution is $$\yellow 5$$.

$$\yellow (5, -2)$$ is the solution to this system of equations.

$3x-5y=25$

$-3x+7y=-29$

I can check my answer by plugging $$\yellow x=5$$ and $$\yellow y=-2$$ into both equations.

First Equation

$3{\yellow (5)}-5{\yellow(-2)}=25$

$$3$$ times $$5$$ is $$15$$.

And $$5$$ times $$-2$$ is $$-10$$.

$$15$$ minus $$-10$$ does equal $$25$$.

So, $$\yellow (5, -2)$$ is a solution to the first equation.

Second Equation

$-3{\yellow (5)}+7{\yellow(-2)}=-29$

$$-3$$ times $$5$$ is $$-15$$.

And $$7$$ times $$-2$$ is $$-14$$.

$$-15$$ plus $$-14$$ does equal $$-29$$.

So, $$\yellow (5, -2)$$ is a solution to the second equation.

Use the elimination method to solve this system of equations:

$11x-7y=35$

$9x+4y=87$

Which variable is the easiest to eliminate?

$11x-7y=35$

$9x+4y=87$

The coefficients for the $$y$$ variables already have opposite signs, so I think it will be slightly easier to eliminate $$y$$.

The least common multiple of $$7$$ and $$4$$ is $$28$$, so I need to get one of the coefficients of $$y$$ to equal $$28$$ and one of them to equal $$-28$$.

To do that, I am going to multiply the first equation by $$4$$ so that the $$-7y$$ becomes a $$-28y$$.

${\blue 4}[11x-7y=35]$

$\blue 44x-28y=140$

Then I will multiply the second equation by $$7$$ so that the $$4y$$ becomes $$28y$$.

${\blue 7}[9x+4y=87]$

$\blue 63x+28y=609$

All the variables are on the left side of the equations so I don’t need to rearrange the equations at all.

The new versions of the equations are:

$\blue 44x-28y=140$

$\blue63x+28y=609$

The coefficients of the $$y$$ variables are equal and opposite, which is what I want so they eliminate each other in Step 2.

What is the result when you add the equations together?

$\begin{array}{cccc} 44x & -28y & = 140\\ 63x & +28y & = & 609 \\ \hline \blue 107x & & \blue= & \blue 749 \end {array}$

When I add the equations together, the $$y$$’s eliminate each other, leaving me with $$\blue 107x=749$$.

What does y equal?

$\blue 107x=749$

To find the x-coordinate of the solution, I will divide both sides of the eliminated equation by $$\blue 107$$ to undo the multiplication.

$\blue x=7$

The x-coordinate of my solution is $$\blue 7$$.

What does y equal?

Now that I know that $$\blue x-=7$$, I can plug $$\blue 7$$ into either one of the original equations to find the value of $$y$$.

$11x-7y=35$

$9x+4y=87$

I think it looks slightly easier to plug $$\blue x=7$$ into the first equation, so I will do that.

$9({\blue 7})+4y=87$

I will simplify the right side of the equation by multiplying $$9$$ and $$\blue 7$$.

$63+4y=87$

Next, I will subtract $$63$$ from both sides of the equation to undo the addition.

$4y=24$

Then I will divide both sides of the equation by $$4$$ to undo the multiplication.

$y=6$

The y-coordinate of my solution is $$\blue 6$$.

$$\blue (7, 6)$$ is the solution to this system of equations.

$11x-7y=35$

$9x+4y=87$

I can check my answer by plugging $$\blue x=7$$ and $$\blue y=6$$ into both equations.

First Equation

$11{\blue (7)}-7{\blue (6)}=35$

$$11$$ times $$7$$ is $$77$$.

And $$7$$ times $$6$$ is $$42$$.

$$77$$ minus $$42$$ does equal $$35$$.

So, $$\blue (7, 6)$$ is a solution to the first equation.

Second Equation

$9{\blue (7)}+4{\blue (6)}=87$

$$9$$ times $$7$$ is $$63$$.

And $$4$$ times $$6$$ is $$24$$.

$$63$$ plus $$24$$ does equal $$87$$.

So, $$\blue (7, 6)$$ is a solution to the second equation.

Use the elimination method to solve this system of equations:

$\frac{2}{3}x-\frac{5}{2}y=-\frac{41}{6}$

$y=4x-1$

Which variable is the easiest to eliminate?

$\frac{2}{3}x-\frac{5}{2}y=-\frac{41}{6}$

$y=4x-1$

The $$y$$ variable in the second equation has a coefficient of 1, so I think it will be easiest to eliminate $$y$$.

First of all, I will multiply the first equation by $$6$$ so I won’t have to deal with the fractions.

${\green 6}[\frac{2}{3}x-\frac{5}{2}y=-\frac{41}{6}]$

$\green 4x-15y=-41$

The coefficient for $$y$$ in the first equation is $$\green -15$$ now, so I need to make the coefficient of $$y$$ in the second equation $$\green 15$$, so they will eliminate each other in Step 2.

To do that, I will multiply the second equation by $$15$$.

${\green 15}[y=4x-1]$

$\green 15y=60x-15$

I will also subtract $$\green 60x$$ from both sides of the equation so that the variables are all on the same side.

$\green -60x+15y=-15$

The new versions of the equations are:

$\green 4x-15y=-41$

$\green -60x+15y=-15$

The coefficients of the $$y$$ variables are equal and opposite, which is what I want so they eliminate each other in Step 2.

What is the result when you add the equations together?

$\begin{array}{cccc} 4x & -15y & = -41\\ -60x & +15y & = & -15 \\ \hline \green -56x & & \green= & \green -56 \end {array}$

When I add the equations together, the $$y$$’s eliminate each other, leaving me with $$\green -56x=-56$$.

What does y equal?

$\green -56x=-56$

To find the x-coordinate of the solution, I will divide both sides of the eliminated equation by $$\green -56$$ to undo the multiplication.

$\green x=1$

The x-coordinate of my solution is $$\green 1$$.

What does y equal?

Now that I know that $$\green x=1$$, I can plug $$\green 1$$ into either one of the original equations to find the value of $$y$$.

$\frac{2}{3}x-\frac{5}{2}y=-\frac{41}{6}$

$y=4x-1$

It looks much easier to plug $$\green x=1$$ into the second equation, so I will do that.

$y=4({\green 1})-1$

I will simplify the right side of the equation by multiplying $$4$$ and $$\green 1$$.

$y=4-1$

Next, I will subtract $$1$$ from $$4$$.

$y=3$

The y-coordinate of my solution is $$\green 3$$.

$$\green (1,3)$$ is the solution to this system of equations.

$\frac{2}{3}x-\frac{5}{2}y=-\frac{41}{6}$

$y=4x-1$

I can check my answer by plugging $$\green x=1$$ and $$\green y=3$$ into both equations.

First Equation

$\frac{2}{3}{\green (1)}-\frac{5}{2}{\green (3)}=-\frac{41}{6}$

$$\frac{2}{3}$$ times $$1$$ is $$\frac{2}{3}$$.

And $$\frac{5}{2}$$ times $$3$$ is $$\frac{15}{2}$$.

$$\frac{2}{3}$$ minus $$\frac{15}{2}$$ does equal $$-\frac{41}{6}$$.

So, $$\green (1, 3)$$ is a solution to the first equation.

Second Equation

${\green (3)}=4{\green (1)}-1$

$$4$$ times $$1$$ is $$4$$.

And $$4$$ minus $$1$$ is $$3$$.

So, $$\green (1, 3)$$ is a solution to the second equation.