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8th grade (Eureka Math/EngageNY)

Unit 4: Lesson 4

Topic D: Systems of linear equations and their solutions

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:
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 x variable that we know how to solve:
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:
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 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:
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! 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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text
y, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

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:
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 x.
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, equals, 4 into one of the original equations, and solve for y.
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 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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text
y, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text
y, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text
y, equals
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Want to join the conversation?

  • mr pants green style avatar for user u
    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.
    (51 votes)
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  • aqualine seedling style avatar for user isse_sam000
    When do we need to use this in real life?
    (31 votes)
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  • spunky sam blue style avatar for user Streety
    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.
    (27 votes)
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  • male robot donald style avatar for user Nicholas P.
    welp this helped me a little bit
    (16 votes)
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  • leafers ultimate style avatar for user Erik Van de Ven
    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!
    (13 votes)
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  • blobby green style avatar for user gabaikangwe321
    x=y+2,55
    2x=5y

    Substitute please
    (9 votes)
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    • duskpin tree style avatar for user abiramirathnav
      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)
  • primosaur sapling style avatar for user sheivakayvan97
    -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 :(
    (5 votes)
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    • leafers ultimate style avatar for user Barbara
      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.
      (13 votes)
  • blobby green style avatar for user Teonna Smith
    i am still stuck is it another way...?
    (6 votes)
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  • mr pink red style avatar for user Iamkalaiah
    this is so confusing i dont know how to do anything
    (6 votes)
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  • duskpin seed style avatar for user samia mays
    In the first question when the ask you 4x+y=28 and the second equation is y=3x what do i do, do i plug in 3x into the first equation please help?
    (3 votes)
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