Systems of equations with substitution

CCSS Math: 8.EE.C.8, 8.EE.C.8b
Walk through examples of solving systems of equations with substitution.
Let's work to solve this system of equations:
y=2x        Equation 1y = 2x ~~~~~~~~\gray{\text{Equation 1}}
x+y=24        Equation 2x + y = 24 ~~~~~~~~\gray{\text{Equation 2}}
The tricky thing is that there are two variables, xx and yy. If only we could get rid of one of the variables...
Here's an idea! Equation 11 tells us that 2x\goldD{2x} and y\goldD y are equal. So let's plug in 2x\goldD{2x} for y\goldD y in Equation 22 to get rid of the yy variable in that equation:
x+y=24Equation 2x+2x=24Substitute 2x for y\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 xx variable that we know how to solve:
Nice! So we know that xx equals 88. But remember that we are looking for an ordered pair. We need a yy value as well. Let's use the first equation to find yy when xx equals 88:
y=2xEquation 1y=2(8)Substitute 8 for xy=16\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 (8,16)(\blueD8, \greenD{16}). 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:
y=2x16=?2(8)Plug in x = 8 and y = 1616=16Yes!\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:
x+y=248+16=?24Plug in x = 8 and y = 1624=24Yes!\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! (8,16)(\blueD8, \greenD{16}) 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.
4x+y=284x + y = 28
y=3xy = 3x
x=x =
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}
y=y =
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}

Now let's find yy:
The solution:
x=4x = 4
y=12y = 12

Solving for a variable first, then using substitution

Sometimes using substitution is a little bit trickier. Here's another system of equations:
3x+y=9       Equation 1-3x + y = -9~~~~~~~ \gray{\text{Equation 1}}
5x+4y=32       Equation 25x + 4y = 32~~~~~~~ \gray{\text{Equation 2}}
Notice that neither of these equations are already solved for xx or yy. As a result, the first step is to solve for xx or yy 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 yy:
3x+y=9Equation 13x+y+3x=9+3xAdd 3x to each sidey=9+3x\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 xx.
5x+4y=32Equation 25x+4(9+3x)=32Substitute -9 + 3x for y5x36+12x=3217x36=3217x=68x=4Divide each side by 17\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=4x = 4 into one of the original equations, and solve for yy.
3x+y=9The first equation3(4)+y=9Substitute 4 for x12+y=9y=3Add 12 to each side\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 (4,3)(\blueD4, \greenD 3).
Let's check the first equation:
3x+y=93(4)+3=?9Plug in x = 4 and y = 39=9Yes!\begin{aligned} -3x + y &= -9 \\\\ -3(\blueD 4) + \greenD 3 &\stackrel?= -9 &\gray{\text{Plug in x = 4 and y = 3}}\\\\ -9 &= -9 &\gray{\text{Yes!}}\end{aligned}
Let's check the second equation:
5x+4y=325(4)+4(3)=?32Plug in x = 4 and y = 332=32Yes!\begin{aligned} 5x + 4y &= 32 \\\\ 5(\blueD{4}) + 4(\greenD{3}) &\stackrel?= 32 &\gray{\text{Plug in x = 4 and y = 3}}\\\\ 32 &= 32 &\gray{\text{Yes!}}\end{aligned}
Nice! (4,3)(\blueD4, \greenD 3) is indeed a solution. We must not have made any mistakes.

Let's practice!

1) Use substitution to solve the following system of equations.
2x3y=52x - 3y = -5
y=x1y = x - 1
x=x =
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}
y=y =
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}

Now let's find yy:
The solution:
x=8x = 8
y=7y = 7
2) Use substitution to solve the following system of equations.
7x2y=13-7x - 2y = -13
x2y=11x - 2y = 11
x=x =
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}
y=y =
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}

Let's solve the second equation for xx:
x2y=11x=11+2y\begin{aligned} x - 2y &= 11\\\\ x&= 11 + 2y\end{aligned}
Now we can plug this into the first equation:
Now let's find xx:
The solution:
x=3x = 3
y=4y = -4
3) Use substitution to solve the following system of equations.
3x4y=2-3x - 4y = 2
5=5x+5y-5 = 5x + 5y
x=x =
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}
y=y =
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}

Let's solve the second equation for xx:
5=5x+5y1=x+yDivide by 5x=1y\begin{aligned} -5 &= 5x+5y\\\\ -1 &= x+y &\gray{\text{Divide by 5}}\\\\ x&= -1 - y\end{aligned}
Now we can plug this into the first equation:
Now let's find xx:
The solution:
x=2x = -2
y=1y = 1