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Algebra 1

Course: Algebra 1 > Unit 6

Lesson 2: Solving systems of equations with substitution
Math
Algebra 1
Systems of equations
Solving systems of equations with substitution
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Substitution method review (systems of equations)

Google Classroom
The substitution method is a technique for solving a system of equations. This article reviews the technique with multiple examples and some practice problems for you to try on your own.

What is the substitution method?

The substitution method is a technique for solving systems of linear equations. Let's walk through a couple of examples.

Example 1

We're asked to solve this system of equations:
3x+y=−3x=−y+3\begin{aligned} 3x+y &= -3\\\\ x&=-y+3 \end{aligned}
The second equation is solved for x, so we can substitute the expression minus, y, plus, 3 in for x in the first equation:
3x+y=−33(−y+3)+y=−3−3y+9+y=−3−2y=−12y=6 \begin{aligned} 3\blueD{x}+y &= -3\\\\ 3(\blueD{-y+3})+y&=-3\\\\ -3y+9+y&=-3\\\\ -2y&=-12\\\\ y&=6 \end{aligned}
Plugging this value back into one of our original equations, say x, equals, minus, y, plus, 3, we solve for the other variable:
x=−y+3x=−(6)+3x=−3\begin{aligned} x &= -\blueD{y} +3\\\\ x&=-(\blueD{6})+3\\\\ x&=-3 \end{aligned}
The solution to the system of equations is x, equals, minus, 3, y, equals, 6.
We can check our work by plugging these numbers back into the original equations. Let's try 3, x, plus, y, equals, minus, 3.
3x+y=−33(−3)+6=?−3−9+6=?−3−3=−3\begin{aligned} 3x+y &= -3\\\\ 3(-3)+6&\stackrel ?=-3\\\\ -9+6&\stackrel ?=-3\\\\ -3&=-3 \end{aligned}
Yes, our solution checks out.

Example 2

We're asked to solve this system of equations:
7x+10y=36−2x+y=9\begin{aligned} 7x+10y &= 36\\\\ -2x+y&=9 \end{aligned}
In order to use the substitution method, we'll need to solve for either x or y in one of the equations. Let's solve for y in the second equation:
−2x+y=9y=2x+9\begin{aligned} -2x+y&=9 \\\\ y&=2x+9 \end{aligned}
Now we can substitute the expression 2, x, plus, 9 in for y in the first equation of our system:
7x+10y=367x+10(2x+9)=367x+20x+90=3627x+90=363x+10=43x=−6x=−2 \begin{aligned} 7x+10\blueD{y} &= 36\\\\ 7x+10\blueD{(2x+9)}&=36\\\\ 7x+20x+90&=36\\\\ 27x+90&=36\\\\ 3x+10&=4\\\\ 3x&=-6\\\\ x&=-2 \end{aligned}
Plugging this value back into one of our original equations, say y, equals, 2, x, plus, 9, we solve for the other variable:
y=2x+9y=2(−2)+9y=−4+9y=5\begin{aligned} y&=2\blueD{x}+9\\\\ y&=2\blueD{(-2)}+9\\\\ y&=-4+9 \\\\ y&=5 \end{aligned}
The solution to the system of equations is x, equals, minus, 2, y, equals, 5.
Want to learn more about the substitution method? Check out this video.

Practice

Problem 1
  • Current
Solve the following system of equations.
−5x+4y=3x=2y−15\begin{aligned} -5x+4y &= 3\\\\ x&=2y-15 \end{aligned}
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 more practice? Check out this exercise.

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