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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

## Want to join the conversation?

• This article is filled with errors in explanation. It lacks to explain why this is this and why this is here. It can get confusing and half of the time the user is trying to figure out the whole thing by himself and the article just show the answers. If you get a question wrong and hit show solution it doesn't show you explanations that go with the steps.
• It helps me since I have an algebra test tomorrow and I hope I get an A- or at least an B+
• When do we need to use this in real life?
• Good question. It will likely only matter if you go into a hard science field with lots of math.
• Props to all the homies in the struggle (the comment section) actually asking and answering questions that are relevant to the topic full heartedly. Just a little thank you.
• welp this helped me a little bit
• Thank you so much for this amazing website. I'm Dutch and we don't have 12 grades like the USA. After 6th grade, depending on our "level" we go the some high school which best suits our capabilities I guess you could say. We do some tests at the end of 6th grade and the results, in combination with the school's advice we go to vmbo, havo or vwo (middle, high, preparatory
scientific).

I went to middle. I've never been taught such equations, and now I am able to solve them, which are part of the higher degrees of eduction, which is fantastic!
• I wish we had a similar system here.
• x=y+2,55
2x=5y

• x = y + 2.55
2x = 5y
By substituting the first equation to the second equation we get:
2(y+2.55) = 5y
Now distribute:
2y + 2.55 = 5y
Then we combine like terms:
2.55 = 5y - 2y
which is 2.55 = 3y.
We then finally divide 3 and 2.55 to get y.
2.55/3 = y
y = 0.85. So y equals 0.85. Now we can find x easily.
x = 0.85 + 2.55
x = 3.40
(1 vote)
• -5=5x+5y when you divide it by 5 I don't understand how the equation then becomes x=-1-y when you are dividing by a positive 5 :(
• Two things are happening - first, divide by positive 5, then manipulate the equation to get x by itself.
So take the original equation and divide each term on both sides by positive 5. As long as we divide each term by the same non-zero number, the two sides will remain equal.
    -5 = 5x + 5y    -1  = x + y 

Now to get x by itself, we can subtract y from both sides.
-1      = x + y-1 - y = x + y - y-1 - y = x     

Hope that helps.