## How to Use the Substitution Method to Solve Systems of Equations

- Isolate a variable in one of the equations.
- Substitute the isolated variable into the other equation.
- Solve for the non-substituted variable.
- Solve for the substituted variable by plugging the answer from Step 3 into either one of the original equations.

## Examples

Use the substitution method to solve this system of equations:

\[y=2x-2\]

\[y=5x-23\]

*Which variable is easiest to isolate and substitute?*

\[y=2x-2\]

\[y=5x-23\]

In this system of equations, the \(y\) is already isolated in both equations.

*What is the result when you substitute \(2x-2\) for \(y\)?*

\[y=2x-2\]

\[y=5x-23\]

The first equation tells me that \(y\) is equal to \(\yellow 2x-2\).

So, I will substitute \(\yellow 2x-2\) everywhere I see a \(y\) in the second equation.

\[({\yellow 2x-2})=5x-23\]

*What does x equal?*

To find the x-coordinate of the solution, I will solve the substituted equation until \(\yellow x\) is by itself.

\[\yellow (2x-2)=5x-23\]

First, I will distribute the invisible coefficient of 1 to the terms inside the parentheses.

\[\yellow 2x-2=5x-23\]

Then I will subtract \(\yellow 2x\) from both sides of the equation to combine like terms.

\[\yellow -2=3x-23\]

Next, I will add \(\yellow 23\) to both sides to undo the subtraction.

\[\yellow 21=3x\]

Then I will divide both sides of the equation by \(\yellow 3\) to undo the multiplication.

\[\yellow 7=x\]

The x-coordinate of my solution is \(\yellow 7\).

*What does y equal?*

Now that I know that \(\yellow x=7\), I can plug \(\yellow 7\) into either one of the original equations to find the value of \(y\).

\[y=2x-2\]

\[y=5x-23\]

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

\[y=2({\yellow 7})-2\]

I will simplify the right side of the equation by multiplying \(2\) and \(\yellow 7\).

\[y=14-2\]

Then I will subtract \(2\) from \(14\).

\[y=12\]

The y-coordinate of my solution is \(\yellow 12\).

\(\yellow (7, 12)\) is the solution to this system of equations.

\[y=2x-2\]

\[y=5x-23\]

I can check my answer by plugging \(\yellow x=7\) and \(\yellow y=12\) into both equations.

**First Equation**

\[{\yellow (12)}=2{\yellow(7)}-2\]

\(2\) times \(7\) is \(14\).

And \(14\) minus \(2\) does equal \(12\).

So, \(\yellow (7, 12)\) is a solution to the first equation.

**Second Equation**

\[{\yellow (12)}=5{\yellow(7)}-23\]

\(5\) times \(7\) is \(35\).

And \(35\) minus \(23\) does equal \(12\).

So, \(\yellow (7, 12)\) is a solution to the second equation.

Use the substitution method to solve this system of equations:

\[x=-3y+19\]

\[-2x+y=18\]

*Which variable is easiest to isolate and substitute?*

\[x=-3y+19\]

\[-2x+y=18\]

In this system of equations, the \(x\) is already isolated in the first equation.

*What is the result when you substitute \(-3y+19\) for \(x\)?*

\[x=-3y+19\]

\[-2x+y=18\]

The first equation tells me that \(x\) is equal to \(\green -3y+19\).

So, I will substitute \(\green -3y+19\) everywhere I see a \(x\) in the second equation.

\[-2({\green -3y+19})+y=18\]

*What does y equal?*

To find the y-coordinate of the solution, I will solve the substituted equation until \(\green y\) is by itself.

\[\green -2(-3y+19)+y=18\]

First, I will distribute the \(\green -2\) coefficient to the terms inside the parentheses.

\[\green 6y-38+y=18\]

Then I will add the \(\green 6y\) and \(\green y\) on the left side of the equation to combine like terms.

\[\green 7y-38=18\]

Next, I will add \(\green 38\) to both sides of the equation to undo the subtraction.

\[\green 7y=56\]

Then I will divide both sides of the equation by \(\green 7\) to undo the multiplication.

\[\green y=8\]

The y-coordinate of my solution is \(\green 8\).

*What does x equal?*

Now that I know that \(\green y=8\), I can plug \(\green 8\) into either one of the original equations to find the value of \(x\).

\[x=-3y+19\]

\[-2x+y=18\]

I think it looks easier to plug \(\green y=8\) into the first equation, so I will do that.

\[x=-3({\green 8})+19\]

I will simplify the right side of the equation by multiplying \(-3\) and \(\green 8\).

\[x=-24+19\]

Then I will add \(-24\) and \(19\).

\[x=-5\]

The x-coordinate of my solution is \(\green -5\).

\(\green (-5,8)\) is the solution to this system of equations.

\[x=-3y+19\]

\[-2x+y=18\]

I can check my answer by plugging \(\green x=-5\) and \(\green y=8\) into both equations.

**First Equation**

\[{\green (-5)}=-3{\green (8)}+19\]

\(-3\) times \(8\) is \(-24\).

And \(-24\) plus \(19\) does equal \(-5\).

So, \(\green (-5, 8)\) is a solution to the first equation.

**Second Equation**

\[-2{\green (-5)}+{\green (8)}=18\]

\(-2\) times \(-5\) is \(10\).

And \(10\) plus \(8\) does equal \(18\).

So, \(\green (-5, 8)\) is a solution to the second equation.

Use the substitution method to solve this system of equations:

\[7x-2y=-19\]

\[2x+y=15\]

*Which variable is easiest to isolate and substitute?*

\[7x-2y=-19\]

\[2x+y=15\]

I think that it would be easiest to isolate the \(y\) in the second equation.

\[2x+y=15\]

To get \(y\) by itself, I need to subtract \(2x\) from both sides of the equation.

\[y=-2x+15\]

*What is the result when you substitute \(-2x+15\) for \(y\)?*

\[7x-2y=-19\]

\[y=-2x+15\]

The rearranged version of the second equation tells me that \(y\) is equal to \(\purple -2x+15\).

So, I will substitute \(\purple -2x+15\) everywhere I see a \(y\) in the first equation.

\[7x-2({\purple -2x+15})=-19\]

*What does x equal?*

To find the x-coordinate of the solution, I will solve the substituted equation until \(\purple x\) is by itself.

\[\purple 7x-2(-2x+15)=-19\]

First, I will distribute the \(\purple -2\) coefficient to the terms inside the parentheses.

\[\purple 7x+4x-30=-19\]

Then I will add the \(\purple 7x\) and \(\purple 4x\) on the left side of the equation to combine like terms.

\[\purple 11x-30=-19\]

Next, I will add \(\purple 30\) to both sides of the equation to undo the subtraction.

\[\purple 11x=11\]

Then I will divide both sides of the equation by \(\purple 11\) to undo the multiplication.

\[\purple x=1\]

The x-coordinate of my solution is \(\purple 1\).

*What does y equal?*

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

\[7x-2y=-19\]

\[2x+y=15\]

I think it looks easiest to plug \(\purple x=1\) into the second equation, so I will do that.

\[2({\purple 1})+y=15\]

I will simplify the right side of the equation by multiplying \(2\) and \(\purple 1\).

\[2+y=15\]

Then I will subtract \(2\) from both sides to undo the addition.

\[y=13\]

The y-coordinate of my solution is \(\purple 13\).

\(\purple (1, 13)\) is the solution to this system of equations.

\[7x-2y=-19\]

\[2x+y=15\]

I can check my answer by plugging \(\purple x=1\) and \(\purple y=13\) into both equations.

**First Equation**

\[7{\purple (1)}-2{\purple (13)}=-19\]

\(7\) times \(1\) is \(7\).

And \(2\) times \(13\) is \(26\).

\(7\) minus\(26\) does equal \(-19\).

So, \(\purple (1, 13)\) is a solution to the first equation.

**Second Equation**

\[2{\purple (1)}+{\purple (13)}=15\]

\(2\) times \(1\) is \(2\).

And \(2\) plus \(13\) does equal \(15\).

So, \(\purple (1, 13)\) is a solution to the second equation.