Solving Systems of Linear Equations by Substitution

March 9, 2011 at 11:04 am Leave a comment

The first algebraic technique for solving systems of two linear equations in two unknowns is the substitution method. I teach my students to look for one of two scenarios for using substitution:

  • One of the equations already has an isolated variable, like y=3x+4 or x=\frac{2}{3}y-9
  • One of the equations can be easily manipulated in order to isolate a variable. What we are really looking for here is an equation that has a “plain” x term or a “plain” y term. That is, there is an equation in which the x term or y term has a coefficient of 1 or -1.

(Later I amend this approach to only using substitution when one of the equations already has x or y isolated.)

Example 1

Solve: \begin{array}{rcl}3x-2y & = & 13\\y & = & 2x-9\end{array}

Note the second equation: y=2x-9. This equation has y isolated on the left side of the equation, and tells us that y is the same as 2x-9, so we can replace y in the equation 3x-2y=13 by the expression 2x-9.

3x-2(2x-9)=13     (Substitute 2x-9 for y.)
3x-4x+18=13       (Distribute.)
-x+18=13              (Combine like terms.)
-x=-5                      (Subtract 18 from both sides.)
x=5                         (Divide both sides by -1.)

Now that we have “half” of our solution, we need to find the y-coordinate of the ordered pair by back-substituting 5 for x in either of the original equations. It will usually be more efficient to substitute into the equation that had the other variable isolated.

y=2(5)-9        (Substitute 5 for x.)
y=1                  (Simplify.)

The solution to this system is (5,1).

We can check our solution by substituting 5 for x and 1 for y in the other equation, 3x-2y=13.

3(5)-2(1)=13              (Substitute 5 for x and 1 for y.)
15-2=13                       (Multiply.)
13=13                          (Subtract.)

The solution checks.

Example 2

Solve: \begin{array}{rcl}x & = & \frac{2}{3}y+5\\6x+7y & = & 96\end{array}

Note the first equation, x=\frac{2}{3}y+5 has x isolated on the left side of the equation, and tells us that x can be replaced by \frac{2}{3}y+5 in the equation 6x+7y=96.

6\left(\frac{2}{3}y+5\right)+7y=96     (Substitute \frac{2}{3}y+5 for x.)
4y+30+7y=96       (Distribute.)
11y+30=96              (Combine like terms.)
11y=66                      (Subtract 30 from both sides.)
y=6                             (Divide both sides by 11.)

To find the x-coordinate of the ordered pair, substitute 6 for y.

x=\frac{2}{3}(6)+5        (Substitute 6 for y.)
x=4+5                                   (Multiply \frac{2}{3}(6).)
x=9                                        (Simplify.)

The solution to this system is (9,6).

We can check our solution by substituting 9 for x and 6 for y in the other equation, 6x+7y=96.

6(9)+7(6)=96              (Substitute 9 for x and 6 for y.)
54+42=96                       (Multiply.)
96=96                               (Subtract.)

The solution checks.


I hope that you find this helpful. If you have any questions, or if you have any other topics you are interested in, you can reach me through the contact page at my web site –


I am a math instructor at College of the Sequoias in Visalia, CA. Each Wednesday I post an article related to General Teaching on my blog. If there’s a particular topic you’d like me to address, or if you have a question or a comment, please let me know. You can reach me through the contact page on my website –


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