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5.2 Solve systems of equations by substitution

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By the end of this section, you will be able to:
  • Solve a system of equations by substitution
  • Solve applications of systems of equations by substitution

Before you get started, take this readiness quiz.

  1. Simplify −5 ( 3 − x ) .
    If you missed this problem, review [link] .
  2. Simplify 4 − 2 ( n + 5 ) .
    If you missed this problem, review [link] .
  3. Solve for y . 8 y − 8 = 32 − 2 y
    If you missed this problem, review [link] .
  4. Solve for x . 3 x − 9 y = −3
    If you missed this problem, review [link] .

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

In this section, we will solve systems of linear equations by the substitution method.

Solve a system of equations by substitution

We will use the same system we used first for graphing.

{ 2 x + y = 7 x − 2 y = 6

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

We’ll fill in all these steps now in [link] .

How to solve a system of equations by substitution

Solve the system by substitution. { 2 x + y = 7 x − 2 y = 6

 

 

 

Solution

This figure has three columns and six rows. The first row says, “Step 1. Solve one of the equations for either variable.” To the right of this, the middl row reads, “We’ll solve the first equation for y.” The third column shows the two equations: 2x + y = 7 and x – 2y = 6. It shows that 2x + y = 7 becomes y = 7 – 2x. The second row reads, “Step 2. Substitute the expression from Step 1 into the other equation.” Then, “We replace y in the second equation with the expression 7 – 2x.” It then shows the x – 2y = 6 becomes x – 2(7 – 2x) = 6. The third row says, “Step 3: Solve the resulting equation.” Then “Now we have an equation with just 1 variable. WE know how to solve this!” It then shows that x – 2(7 – 2x) = 6 becomes x – 14 + 4x = 6 which becomes 5x = 20. Thus x = 4. The fourth row says, “Step 4. Substitute the solution in Step 3 into one of the original quaitons to find the other variable.” Then, “We’ll use the first equation and replace x with 4.” Then it shows that 2x + y = 7 becomes 2(4) + y = 7. This becomes 8 + y = 7, and thus y = −1. The fifth row reads, “Step 5. Write the solution as an ordered pair.” Then “The ordered air is (x, y).” Then (4, −1). The sixth row reads, “Step 6. Check that the order pair is a solution to both original equations.” Then, “Substitute (4, −1) into both equations and make sure they are both true.” It then shows that 2x + y = 7 becomxe 2(4) + −1 = 7, and thus 7 = 7. It also shows that x – 2y = 6 becomes 4 – 2(−1) = 6, and thus 6−6. It also states, “Both equations are ture. (4, −1) is the solution to the system.”
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Solve the system by substitution. { −2 x + y = −11 x + 3 y = 9

( 6 , 1 )

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Solve the system by substitution. { x + 3 y = 10 4 x + y = 18

( 4 , 2 )

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Solve a system of equations by substitution.

  1. Solve one of the equations for either variable.
  2. Substitute the expression from Step 1 into the other equation.
  3. Solve the resulting equation.
  4. Substitute the solution in Step 3 into one of the original equations to find the other variable.
  5. Write the solution as an ordered pair.
  6. Check that the ordered pair is a solution to both original equations.

If one of the equations in the system is given in slope–intercept form, Step 1 is already done! We’ll see this in [link] .

Solve the system by substitution.

{ x + y = −1 y = x + 5

Solution

The second equation is already solved for y . We will substitute the expression in place of y in the first equation.

.
The second equation is already solved for y .
We will substitute into the first equation.
Replace the y with x + 5. .
Solve the resulting equation for x . .
.
.
Substitute x = −3 into y = x + 5 to find y . .
.
The ordered pair is (−3, 2). .
Check the ordered pair in both equations:

x + y = − 1 − 3 + 2 = ? − 1 − 1 = − 1 ✓ y = x + 5 2 = ? − 3 + 5 2 = 2 ✓
The solution is (−3, 2).

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Solve the system by substitution. { x + y = 6 y = 3 x − 2

( 2 , 4 )

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Solve the system by substitution. { 2 x − y = 1 y = −3 x − 6

( −1 , −3 )

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If the equations are given in standard form, we’ll need to start by solving for one of the variables. In this next example, we’ll solve the first equation for y .

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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