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# Systems of equations with substitution

Walk through examples of solving systems of equations with substitution.
Let's work to solve this system of equations:
y, equals, 2, x, 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
x, plus, y, equals, 24, 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 tricky thing is that there are two variables, x and y. If only we could get rid of one of the variables...
Here's an idea! Equation 1 tells us that start color #e07d10, 2, x, end color #e07d10 and start color #e07d10, y, end color #e07d10 are equal. So let's plug in start color #e07d10, 2, x, end color #e07d10 for start color #e07d10, y, end color #e07d10 in Equation 2 to get rid of the y variable in that equation:
\begin{aligned} x + \goldD y &= 24 &\gray{\text{Equation 2}} \\\\ x + \goldD{2x} &= 24 &\gray{\text{Substitute 2x for y}}\end{aligned}
Brilliant! Now we have an equation with just the x variable that we know how to solve:
\begin{aligned} x + \goldD{2x} &= 24\\\\ 3x&= 24\\\\\ \dfrac{3x}{\maroonD3}&= \dfrac{24}{\maroonD3}&\gray{\text{Divide each side by 3}}\\\\ \blueD{x}&\blueD= \blueD8&\gray{\text{}}\end{aligned}
Nice! So we know that x equals 8. But remember that we are looking for an ordered pair. We need a y value as well. Let's use the first equation to find y when x equals 8:
\begin{aligned} y &= 2\blueD x &\gray{\text{Equation 1}} \\\\ y &= 2(\blueD8) &\gray{\text{Substitute 8 for x}}\\\\ \greenD y &\greenD= \greenD{16}\end{aligned}
Sweet! So the solution to the system of equations is left parenthesis, start color #11accd, 8, end color #11accd, comma, start color #1fab54, 16, end color #1fab54, right parenthesis. It's always a good idea to check the solution back in the original equations just to be sure.
Let's check the first equation:
\begin{aligned} y &= 2x \\\\ \greenD{16} &\stackrel?= 2(\blueD{8}) &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 16 &= 16 &\gray{\text{Yes!}}\end{aligned}
Let's check the second equation:
\begin{aligned} x +y &= 24 \\\\ \blueD{8} + \greenD{16} &\stackrel?= 24 &\gray{\text{Plug in x = 8 and y = 16}}\\\\ 24 &= 24 &\gray{\text{Yes!}}\end{aligned}
Great! left parenthesis, start color #11accd, 8, end color #11accd, comma, start color #1fab54, 16, end color #1fab54, right parenthesis is indeed a solution. We must not have made any mistakes.
Your turn to solve a system of equations using substitution.
Use substitution to solve the following system of equations.
4, x, plus, y, equals, 28
y, equals, 3, x
x, equals
y, equals

## Solving for a variable first, then using substitution

Sometimes using substitution is a little bit trickier. Here's another system of equations:
minus, 3, x, plus, y, equals, minus, 9, 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
5, x, plus, 4, y, equals, 32, 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
Notice that neither of these equations are already solved for x or y. As a result, the first step is to solve for x or y first. Here's how it goes:
Step 1: Solve one of the equations for one of the variables.
Let's solve the first equation for y:
\begin{aligned} -3x + y &= -9 &\gray{\text{Equation 1}} \\\\ -3x + y + \maroonD{3x} &= -9 +\maroonD{3x} &\gray{\text{Add 3x to each side}} \\\\ y &= {-9 +3x} &\gray{\text{}}\end{aligned}
Step 2: Substitute that equation into the other equation, and solve for x.
\begin{aligned} 5x + 4\goldD y &= 32 &\gray{\text{Equation 2}} \\\\ 5x +4(\goldD{-9 + 3x}) &= 32 &\gray{\text{Substitute -9 + 3x for y}} \\\\ 5x -36 +12x &= 32 &\gray{\text{}} \\\\ 17x - 36 &= 32 &\gray{\text{}} \\\\ 17x &= 68 &\gray{\text{}} \\\\ \blueD x &\blueD= \blueD4 &\gray{\text{Divide each side by 17}}\end{aligned}
Step 3: Substitute x, equals, 4 into one of the original equations, and solve for y.
\begin{aligned} -3\blueD x + y &= -9 &\gray{\text{The first equation}} \\\\ -3(\blueD{4}) +y &= -9 &\gray{\text{Substitute 4 for x}} \\\\ -12 + y &= -9 &\gray{\text{}} \\\\ \greenD y &\greenD= \greenD3 &\gray{\text{Add 12 to each side}} \end{aligned}
So our solution is left parenthesis, start color #11accd, 4, end color #11accd, comma, start color #1fab54, 3, end color #1fab54, right parenthesis.

## Let's practice!

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

2) Use substitution to solve the following system of equations.
minus, 7, x, minus, 2, y, equals, minus, 13
x, minus, 2, y, equals, 11
x, equals
y, equals

3) Use substitution to solve the following system of equations.
minus, 3, x, minus, 4, y, equals, 2
minus, 5, equals, 5, x, plus, 5, y
x, equals
y, equals