If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Substitution method review (systems of equations)

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=33y+9+y=32y=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=?39+6=?33=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=362x+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
Solve the following system of equations.
5x+4y=3x=2y15\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.

Want to join the conversation?

  • blobby green style avatar for user David
    This is pretty challenging not gonna lie
    (42 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Nancy Crisp
    how do I solve y=2x-1 and y=3x+2 using the substitution method
    (7 votes)
    Default Khan Academy avatar avatar for user
  • leafers seed style avatar for user Theo  Lesko
    are there any easy tips or tricks i can use to remember this?
    (10 votes)
    Default Khan Academy avatar avatar for user
  • mr pants teal style avatar for user Emmery
    In example 2, when it says we have to solve for x or y, how do I get -2x+y=9 into slope intercept form? And also, how did we get rid of the negative once it was in slope intercept form?
    Thanks
    (7 votes)
    Default Khan Academy avatar avatar for user
  • old spice man blue style avatar for user Matt
    I am curious if there are times when either the elimination method or the substitution method would be more appropriate, and or if there would be times when only one way or the other would work. Thank you for the advice in advance!
    (2 votes)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user Matthew Johnson
      Yes, both equations will always work, but yes, at times it is more logical to use one over the other. for instance:
      x-4y=6
      x+4y=12

      we see here that elimination would best fit:
      x+x+4y-4y=6+12


      or:
      x=2y-3
      7x+3y-81=23

      substitute (2y-3) for x.

      Also, once you have a single placeholder, put it ito quadratic form (ax^2+bx+c=0)
      and use the quadratic formula:
      x=(-b±sqrt(b^2-4ac))/2a
      (2 votes)
  • stelly blue style avatar for user kenaniah
    This is so hard I can't.
    (6 votes)
    Default Khan Academy avatar avatar for user
    • stelly blue style avatar for user Kim Seidel
      This is the review. Start at the beginning of the lesson. Go thru each video step by step and make sure you understand before moving to the next one. Ask specific questions when there is some part of the video that you don't understand.

      As you do the practice problems, if you get one wrong, use the hints to learn from your mistakes.

      Then, come back and try the review again.
      (2 votes)
  • blobby green style avatar for user Nicki242424
    how about a problem that has a fraction?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user Humza
      The formatting of the problem never changes. If the fraction is a coefficient, you can multiply it by its own reciprocal (negative flip of the fraction) or you can add or subtract based on whether the fraction is positive or negative. All in all, the way you solve the system stays the same, whether you have fractions, integers or variables.

      Hope this helps:)
      (3 votes)
  • starky sapling style avatar for user CleverDesire
    If i may ask,could anyone help me in this equation:

    3y+2x=7
    y-3x=6

    Please answer the question in with explaination.thanks🙏🏽🥺😊
    (3 votes)
    Default Khan Academy avatar avatar for user
    • stelly blue style avatar for user Kim Seidel
      I'll get you started.
      Take the 2nd equation and add 3x to both sides to isolate "y".
      y = 3x+3
      Now, substitute this value of "y" into the first equation.
      3(3x+6) + 2x = 7
      You can now solve for "x".
      Once you have the value of "x", use it to calculate the value of "y".

      Hope this helps. Comment back if you have questions.
      (3 votes)
  • aqualine sapling style avatar for user Annabel L
    Any tips for turning a wordy problem solving question into simultaneous equations? I can find the answers with working on google and it seems so easy, but I can never get started without being given the equations - I can't find a pattern/rule for determining the equations.
    (4 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user nanice.annmarie
    how would i do y=2(x+3)^2 -5 y=14x+17
    (1 vote)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user GLaneMoore
      To solve using the substitution method, you find what y is, and plug it in to the other equation.
      To do this one: y=14x+17. That means you just plug 14x+17 into the other equation.

      y=2(x+3)^2-5 ---> Now substitute
      14x+17=2(x+3)^2-5 ---> Add 5
      14x+22=2(x+3)^2 ---> Divide 2
      7x+11=(x+3)^2 ---> Multiply out using FOIL
      7x+11=x^2+6x+9 ---> Subtract 7x+11
      0=x^2-x-2 ---> Factor into two binomials and solve for x
      0=(x+1)(x-2) ---> x=-1 or x=2

      Now, you plug in one of the values of x into an original equation to find y:
      x=-1 ---> Substitute
      y=14(-1)+17 ---> Combine
      y=3

      Answer: x=-1, y=3
      (7 votes)