4.2 Solve Systems of Equations by Substitution

Domenic Spilotro, MSc

4.2 Solve Systems of Equations by Substitution

Learning Objectives

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

Try It

Before you get started, take this readiness quiz:

1) Simplify $- 5 (3 - x)$ .
2) Simplify $4 - 2 (n + 5)$ .
3) Solve for $y$ : $8 y - 8 = 32 - 2 y$ .
4) Solve for $x$ : $3 x - 9 y = - 3$ .

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.

{\begin{cases} 2 x + y = 7 \\ x - 2 y = 6 \end{cases}

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 Example 4.2.1

Example 4.2.1

How to Solve a System of Equations by Substitution

Solve the system by substitution. ${\begin{cases} 2 x + y = 7 \\ x - 2 y = 6 \end{cases}$

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. — Figure 4.2.1

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. — Figure 4.2.2

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. — Figure 4.2.3

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. — Figure 4.2.4

The fifth row reads, “Step 5: Write the solution as an ordered pair.” Then “The ordered air is (x, y).” Then (4, −1). — Figure 4.2.5

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.” — Figure 4.2.6

Try It

5) Solve the system by substitution.

${\begin{cases} - 2 x + y = - 11 \\ x + 3 y = 9 \end{cases}$

Solution

$(6, 1)$

6) Solve the system by substitution.

${\begin{cases} x + 3 y = 10 \\ 4 x + y = 18 \end{cases}$

Solution

$(4, 2)$

HOW TO

Solve a system of equations by substitution.

Solve one of the equations for either variable.
Substitute the expression from Step 1 into the other equation.
Solve the resulting equation.
Substitute the solution in Step 3 into one of the original equations to find the other variable.
Write the solution as an ordered pair.
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 Example 4.2.2

Example 4.2.2

Solve the system by substitution.

${\begin{cases} x + y = - 1 \\ y = x + 5 \end{cases}$

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.

$\begin{array}{r} {\begin{cases} x + y = - 1 \\ y = x + 5 \end{cases} \end{array}$

Step 1: Replace the $y$ with $x + 5$

$\begin{array}{rcl} y & = & x + 5 \\ x + y & = & - 1 \\ x + x + 5 & = & - 1 \end{array}$

Step 2: Solve the resulting equation for $x$

$\begin{array}{rcl} x + x + 5 & = & - 1 \\ 2 x + 5 & = & - 1 \\ 2 x & = & - 6 \end{array}$

Step 3: Substitute $x = - 3$ into $y = x + 5$ to find $y$

$\begin{aligned} x & = - 3 \\ y & = x + 5 \\ y & = - 3 + 5 \\ y & = 2 \end{aligned}$

The ordered pair is $(- 3, 2)$ .

Step 4: Check the ordered pair in both equations:

$\begin{aligned} x + y & = - 1 \\ - 3 + 2 & \overset{?}{=} - 1 \\ - 1 & = - 1 ✓ \end{aligned}$

\begin{aligned} y & = x + 5 \\ 2 & \overset{?}{=} - 3 + 5 \\ 2 & = 2 ✓ \end{aligned}

The solution is $(- 3, 2)$ .

Try It

7) Solve the system by substitution.

${\begin{cases} x + y = 6 \\ y = 3 x - 2 \end{cases}$

Solution

$(2, 4)$

8) Solve the system by substitution.

${\begin{cases} 2 x - y = 1 \\ y = - 3 x - 6 \end{cases}$

Solution

$(- 1, - 3)$

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 $y$ .

Example 4.2.3

Solve the system by substitution.

${\begin{cases} 3 x + y = 5 \\ 2 x + 4 y = - 10 \end{cases}$

Solution

We need to solve one equation for one variable. Then we will substitute that expression into the other equation.

Solve for $y$

$3 x + y = 5$

Step 1: Substitute into the other equation.

Step 2: Replace the $y$ with $- 3 x + 5$ .

$\begin{aligned} 3 x + y & = 5 \\ y & = - 3 + 5 \\ 2 x + 4 y & = - 10 \\ 2 x + 4 (- 3 + 5) & = - 10 \end{aligned}$

Step 3: Solve the resulting equation for $x$ .

$\begin{array}{rcl} 2 x - 12 x + 20 & = & - 10 \\ - 10 x + 20 & = & - 10 \\ - 10 x & = & - 30 \\ x & = & 3 \end{array}$

Step 4: Substitute $x = 3$ into $3 x + y = 5$ to find $y$ .

$\begin{array}{rcl} x & = & 3 \\ 3 x + y & = & 5 \\ 3 (3) + y & = & 5 \\ 9 + y & = & 5 \\ y & = & - 4 \end{array}$

The ordered pair is $(3, - 4)$ .

Step 5: Check the ordered pair in both equations:

$\begin{aligned} 3 x + y & = 5 \\ - 3 \times 3 + (- 4) & \overset{?}{=} 5 \\ 9 - 4 & \overset{?}{=} 5 \\ 5 & = 5 ✓ \end{aligned}$

$\begin{aligned} 2 x + 4 y & = - 10 \\ 2 \times 3 + 4 (- 4) & = - 10 \\ 6 - 16 & \overset{?}{=} - 10 \\ - 10 & = - 10 ✓ \end{aligned}$

The solution is $(3, - 4)$ .

Try It

9) Solve the system by substitution.

${\begin{cases} 4 x + y = 2 \\ 3 x + 2 y = - 1 \end{cases}$

Solution

$(1, - 2)$

10) Solve the system by substitution.

${\begin{cases} - x + y = 4 \\ 4 x - y = - 2 \end{cases}$

Solution

$(2, 6)$

In Example 4.2.3 it was easiest to solve for $y$ in the first equation because it had a coefficient of $1$ . In Example 4.2.4 it will be easier to solve for $x$ .

Example 4.2.4

Solve the system by substitution.

${\begin{cases} x - 2 y = - 2 \\ 3 x + 2 y = 34 \end{cases}$

Solution

We will solve the first equation for $x$ and then substitute the expression into the second equation.

$x - 2 y = - 2$

Step 1: Solve for $x$ .

Step 2: Substitute into the other equation.

Step 3: Replace the $x$ with $2 y - 2$ .

$\begin{array}{rcl} 3 x + 2 y & = & 34 \\ 3 (2 y - 2) + 2 y & = & 34 \end{array}$

Step 4: Solve the resulting equation for $y$ .

$\begin{array}{rcl} 6 y - 6 + 2 y & = & 34 \\ 8 y - 6 & = & 34 \\ 8 y & = & 40 \\ y & = & 5 \end{array}$

Step 5: Substitute $y = 5$ into $x - 2 y = - 2$ to find $x$ .

$\begin{array}{rcl} x - 2 y & = & - 2 \\ x - 2 \times 5 & = & - 2 \\ x - 10 & = & - 2 \\ x & = & 8 \end{array}$

The ordered pair is $(8, 5)$ .

Step 6: Check the ordered pair in both equations:

$\begin{aligned} x - 2 y & = - 2 \\ 8 - 2 \times 5 & \overset{?}{=} - 2 \\ 8 - 10 & \overset{?}{=} - 2 \\ - 2 & = - 2 ✓ \end{aligned}$

\begin{aligned} 3 x + 2 y & = 34 \\ 3 \times 8 + 2 \times 5 & \overset{?}{=} 34 \\ 24 + 10 & \overset{?}{=} 34 \\ 34 & = 34 ✓ \end{aligned}

The solution is $(8, 5)$ .

Try It

11) Solve the system by substitution.

${\begin{cases} x - 5 y = 13 \\ 4 x - 3 y = 1 \end{cases}$

Solution

$(- 2, - 3)$

12) Solve the system by substitution.

${\begin{cases} x - 6 y = - 6 \\ 2 x - 4 y = 4 \end{cases}$

Solution

$(6, 2)$

When both equations are already solved for the same variable, it is easy to substitute!

Example 4.2.5

Solve the system by substitution.

${\begin{cases} y = - 2 x + 5 \\ y = \frac{1}{2} x \end{cases}$

Solution

Since both equations are solved for $y$ , we can substitute one into the other.

Step 1: Substitute $\frac{1}{2} x$ for $y$ in the first equation.

$\begin{array}{rcl} y & = & \frac{1}{2} x \\ y & = & - 2 x + 5 \end{array}$

Step 2: Replace the $y$ with $\frac{1}{2} x$

$\begin{array}{r} \frac{1}{2} x = - 2 x + 5 \end{array}$

Step 3: Solve the resulting equation. Start by clearing the fraction.

$2 (\frac{1}{2} x) = 2 (- 2 x + 5)$

Step 4: Solve for $x$ .

$\begin{array}{rcl} x & = & - 4 x + 10 \\ 5 x & = & 10 \\ x & = & 2 \end{array}$

Step 5: Substitute $x = 2$ into $y = \frac{1}{2} x$ to find $y$ .

$\begin{array}{rcl} x & = & 2 \\ y & = & \frac{1}{2} x \\ y & = & \frac{1}{2} \times 2 \\ y & = & 1 \end{array}$

The ordered pair is $(2, 1)$ .

Step 6: Check the ordered pair in both equations

$\begin{array}{rcl} y & = & \frac{1}{2} x \\ 1 & \overset{?}{=} & \frac{1}{2} \times 2 \\ 1 & = & 1 ✓ \end{array}$

\begin{array}{rcl} y & = & - 2 x + 5 \\ 1 & \overset{?}{=} & - 2 \times 2 + 5 \\ 1 & = & 1 ✓ \end{array}

The solution is $(2, 1)$ .

Try It

13) Solve the system by substitution.

${\begin{cases} x + y = 6 \\ y = 3 x - 2 \end{cases}$

Solution

$(6, 2)$

14) Solve the system by substitution.

${\begin{cases} 4 x - y = 0 \\ 2 x - 3 y = 5 \end{cases}$

Solution

$(8, 2)$

Be very careful with the signs in the next example.

Example 4.2.6

Solve the system by substitution.

${\begin{cases} 4 x + 2 y = 4 \\ 6 x - y = 8 \end{cases}$

Solution

We need to solve one equation for one variable. We will solve the first equation for $y$ .

$4 x + 2 y = 4$

Step 1: Solve the first equation for $y$ .

$2 y = - 4 x + 4$

Step 2: Substitute $- 2 x + 2$ for $y$ in the second equation.

$\begin{array}{rcl} y & = & - 2 x + 2 \\ 6 x - y & = & 8 \end{array}$

Step 3: Replace the $y$ with $- 2 x + 2$ .

$6 x - (- 2 x + 2) = 8$

Step 4: Solve the equation for $x$ .

$\begin{array}{rcl} 6 x + 2 x - 2 & = & 8 \\ 8 x - 2 & = & 8 \\ 8 x & = & 10 \\ x & = & \frac{5}{2} \end{array}$

Step 5: Substitute $x = \frac{5}{4}$ into $4 x + 2 y = 4$ to find $y$ .

$\begin{array}{rcl} x & = & \frac{5}{2} \\ 4 x + 2 y & = & 4 \\ 4 (\frac{5}{2}) + 2 y & = & 4 \\ 5 + 2 y & = & 4 \\ 2 y & = & - 1 \\ y & = & - \frac{1}{2} \end{array}$

The ordered pair is $(\frac{5}{4}, - \frac{1}{2})$

Step 6: Check the ordered pair in both equations.

$\begin{array}{rc} 4 x + 2 y & = 4 \\ 4 (\frac{5}{4}) + 2 (- \frac{1}{2}) & \overset{?}{=} 4 \\ 5 - 1 & \overset{?}{=} 4 \\ 4 & = 4 ✓ \end{array}$

\begin{array}{rcl} 6 x - y & = & 8 \\ 6 (\frac{5}{4}) - (- \frac{1}{2}) & \overset{?}{=} & 8 \\ \frac{15}{4} - (- \frac{1}{2}) & \overset{?}{=} & 8 \\ \frac{16}{2} & \overset{?}{=} & 8 \\ 8 & = & 8 ✓ \end{array}

The solution is $(\frac{5}{4}, - \frac{1}{2})$

Try It

15) Solve the system by substitution.

${\begin{cases} 3 x + y = 5 \\ 2 x + 4 y = - 10 \end{cases}$

Solution

$(2, \frac{3}{2})$

16) Solve the system by substitution.

${\begin{cases} 4 x - y = 0 \\ 2 x - 3 y = 5 \end{cases}$

Solution

$(- \frac{1}{2}, - 2)$

In Example 4.2.7, it will take a little more work to solve one equation for $x$ or $y$ .

Example 4.2.7

Solve the system by substitution.

${\begin{cases} 4 x - 3 y = 6 \\ 15 y - 20 x = - 30 \end{cases}$

Solution

We need to solve one equation for one variable. We will solve the first equation for $x$ .

$\begin{array}{rcl} x & = & \frac{5}{2} \\ 4 x - 3 y & = & 6 \end{array}$

Step 1: Solve the first equation for $x$ .

$4 x = 3 y + 6$

Step 2: Substitute $\frac{3}{4} y + \frac{3}{2}$ for $x$ in the second equation.

$\begin{array}{rcl} x & = & \frac{3}{4} y + \frac{3}{2} \\ 15 y - 20 x & = & - 30 \end{array}$

Step 3: Replace the $x$ with $\frac{3}{4} y + \frac{3}{2}$

$15 y - 20 (\frac{3}{4} y + \frac{3}{2}) = - 30$

Step 4: Solve for $y$ .

$\begin{array}{rcl} 15 y - 15 y - 30 & = & - 30 \\ 0 - 30 & = & - 30 \\ 0 & = & 0 \end{array}$

Since $0 = 0$ is a true statement, the system is consistent. The equations are dependent. The graphs of these two equations would give the same line. The system has infinitely many solutions.

Try It

17) Solve the system by substitution.

${\begin{cases} 2 x - 3 y = 12 \\ - 12 y + 8 x = 48 \end{cases}$

Solution

Infinitely many solutions.

18) Solve the system by substitution.

${\begin{cases} 5 x + 2 y = 12 \\ - 4 y - 10 x = - 24 \end{cases}$

Solution

Infinitely many solutions.

Look back at the equations in Example 4.2.7. Is there any way to recognize that they are the same line?

Let’s see what happens in the next example.

Example 4.2.8

Solve the system by substitution.

${\begin{cases} 5 x - 2 y = - 10 \\ y = \frac{5}{2} x \end{cases}$

Solution

The second equation is already solved for $y$ , so we can substitute for $y$ in the first equation.

Step 1: Substitute $x$ for $y$ in the first equation.

$\begin{array}{rcl} y & = & \frac{5}{2} x \\ 5 x - 2 y & = & - 10 \end{array}$

Step 2: Replace the $y$ with $\frac{5}{2} x$ .

$5 x - 2 (\frac{5}{2} x) = - 10$

Step 3: Solve for $x$ .

$\begin{array}{rcl} 5 x - 5 x & = & - 10 \\ 0 & \neq & - 10 \end{array}$

Since $0 = - 10$ is a false statement the equations are inconsistent. The graphs of the two equation would be parallel lines. The system has no solutions.

Try It

19) Solve the system by substitution.

${\begin{cases} 3 x + 2 y = 9 \\ y = - \frac{3}{2} x + 1 \end{cases}$

Solution

No solution.

20) Solve the system by substitution.

${\begin{cases} 5 x - 3 y = 2 \\ y = \frac{5}{3} x - 4 \end{cases}$

Solution

No solution.

Solve Applications of Systems of Equations by Substitution

We’ll copy here the problem solving strategy we used in the Solving Systems of Equations by Graphing section for solving systems of equations. Now that we know how to solve systems by substitution, that’s what we’ll do in Step 5:

How to use a problem solving strategy for systems of linear equations.

How To:

How to use a problem-solving strategy for systems of linear equations.

Read the problem. Make sure all the words and ideas are understood.
Identify what we are looking for.
Name what we are looking for. Choose variables to represent those quantities.
Translate into a system of equations.
Solve the system of equations using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

Some people find setting up word problems with two variables easier than setting them up with just one variable. Choosing the variable names is easier when all you need to do is write down two letters. Think about this in the next example—how would you have done it with just one variable?

Example 4.2.9

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Solution

Step 1: Read the problem.

Step 2: Identify what we are looking for.

We are looking for two numbers.

Step 3: Name what we are looking for.

Let $n =$ the first number

Let $m =$ the second number

Step 4: Translate into a system of equations.

The sum of two numbers is zero.

$n + m = 0$

One number is nine less than the other.

$n = m - 9$

The system is:

${\begin{cases} n + m = 0 \\ n = m - 9 \end{cases}$

Step 5: Solve the system of equations.

We will use substitution since the second equation is solved for $n$ .

Substitute $m - 9$ for $n$ in the first equation.

Solve for $m$ .

$\begin{array}{rcl} m - 9 + m & = & 0 \\ 2 m - 9 & = & 0 \\ 2 m & = & 9 \end{array}$

Substitute $m = \frac{9}{2}$ into the second equation and then solve for $n$ .

$\begin{array}{rcl} m & = & \frac{9}{2} - 9 \\ m & = & \frac{9}{2} - \frac{18}{2} \\ n & = & - \frac{9}{2} \end{array}$

Step 6: Check the answer in the problem.

Do these numbers make sense in the problem? We will leave this to you!

Step 7: Answer the question.

The numbers are $\frac{9}{2}$ and $- \frac{9}{2}$

Try It

21) The sum of two numbers is $10$ . One number is $4$ less than the other. Find the numbers.

Solution

The numbers are $3$ and $7$ .

22) The sum of two number is $- 6$ . One number is $10$ less than the other. Find the numbers.

Solution

The numbers are $2$ and $- 8$ .

In the Example 4.2.10, we’ll use the formula for the perimeter of a rectangle, $P = 2 L + 2 W$ .

Example 4.2.10

The perimeter of a rectangle is $88$ . The length is five more than twice the width. Find the length and the width.

Solution

Step 1: Read the problem.

Horizontal rectangle. Letter W on the left side. Letter L on the bottom side. — Figure 4.2.9

Step 2: Identify what you are looking for.

We are looking for the length and width.

Step 3: Name what we are looking for.

Let $L$ = the length

Let $W$ = the width

Step 4: Translate into a system of equations.

The perimeter of a rectangle is $88$ .

$2 L + 2 W = P$

$2 L + 2 W = 88$

The length is five more than twice the width.

$L = 2 W + 5$

The system is:

${\begin{cases} 2 L + 2 W = 88 \\ L = 2 W + 5 \end{cases}$

Step 5: Solve the system of equations.

We will use substitution since the second equation is solved for $L$ .

Substitute $2 W + 5$ for $L$ in the first equation.

Solve for $W$ .

$\begin{aligned} 2 (2 W + 5) + 2 W & = 88 \\ 4 W + 10 + 2 W & = 88 \\ 6 W + 10 & = 88 \\ 6 W & = 78 \end{aligned}$

Substitute $W = 13$ into the second equation and then solve for $L$ .

$\begin{aligned} L & = 2 \times 13 + 5 \\ L & = 31 \end{aligned}$

Step 6: Check the answer in the problem.

Does a rectangle with length $31$ and width $13$ have perimeter $88$ ? Yes.

Step 7: Answer the equation.

The length is $31$ and the width is $13$ .

Try It

23) The perimeter of a rectangle is $40$ . The length is $4$ more than the width. Find the length and width of the rectangle.

Solution

The length is $12$ and the width is $8$ .

24) The perimeter of a rectangle is $58$ . The length is $5$ more than three times the width. Find the length and width of the rectangle.

Solution

The length is $23$ and the width is $6$ .

For Example 4.2.11, we need to remember that the sum of the measures of the angles of a triangle is $180$ degrees and that a right triangle has one $90$ degree angle.

Example 4.2.11

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

Solution

We will draw and label a figure.

Step 1: Read the problem.

Right Angle Triangle with the letter A in the top-left corner and the letter B in the lower-right corner. Right angle is lower-left. — Figure 4.2.12

Step 2: Identify what you are looking for.

We are looking for the measures of the angles.

Step 3: Name what we are looking for.

Let $a =$ the measure of the 1^st angle.

Let $b =$ the measure of the 2^nd angle.

Step 4: Translate into a system of equations.

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle.

$a = 3 b + 10$

The sum of the measures of the angles of a triangle is $180$ .

$a + b + 90 = 180$

The system is:

${\begin{cases} a = 3 b + 10 \\ a + b + 90 = 180 \end{cases}$

Step 5: Solve the system of equations.

We will use substitution since the first equation is solved for $a$ .

Substitute $3 b + 10$ for $a$ in the second equation.

$(3 b + 10) + b + 90 = 180$

Solve for $b$ .

$\begin{aligned} 4 b + 100 & = 180 \\ 4 b & = 80 \end{aligned}$

Substitute $b = 20$ into the first equation and then solve for $a$ .

$\begin{aligned} a & = 3 \times 20 + 10 \\ a & = 70 \end{aligned}$

Step 6: Check the answer in the problem.

We will leave this to you!

Step 7: Answer the question.

The measures of the small angles are $20$ and $70$ .

Try It

25) The measure of one of the small angles of a right triangle is $2$ more than $3$ times the measure of the other small angle. Find the measure of both angles.

Solution

The measure of the angles are $22$ degrees and $68$ degrees.

26) The measure of one of the small angles of a right triangle is $18$ less than twice the measure of the other small angle. Find the measure of both angles.

Solution

The measure of the angles are $36$ degrees and $54$ degrees.

Example 4.2.12

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $$ 25, 000$ plus $$ 15$ for each training session. Option B would pay her $$ 10, 000 + $ 40$ for each training session. How many training sessions would make the salary options equal?

Solution

Step 1: Read the problem.

Step 2: Identify what you are looking for.

We are looking for the number of training sessions that would make the pay equal.

Step 3: Name what we are looking for.

Let $s =$ Heather’s salary.

Let $n =$ the number of training sessions

Step 4: Translate into a system of equations.

Option A would pay her $$ 25, 000$ plus $15$ for each training session.

$s = 25, 000 + 15 n$

Option B would pay her $$ 10, 000 + $ 40$ for each training session

$s = 10, 000 + 40 n$

The system is:

${\begin{cases} s & = 25, 000 + 15 n \\ s & = 10, 000 + 40 n \end{cases}$

Step 5: Solve the system of equations.

We will use substitution.

s=25,000+15n s=10,000+40n — Figure 4.2.15

Substitute $25, 000 + 15 n$ for $s$ in the second equation.

$25, 000 + 15 n = 10, 000 + 40 n$

Solve for $n$ .

$\begin{aligned} 25, 000 & = 10, 000 + 25 n \\ 15, 000 & = 25 n \\ 600 & = n \end{aligned}$

Step 6: Check the answer.

Are $600$ training sessions a year reasonable?

Are the two options equal when $n = 600$ ?

Step 7: Answer the question.

The salary options would be equal for $600$ training sessions.

Try It

27) Geraldine has been offered positions by two insurance companies. The first company pays a salary of $$ 12, 000$ plus a commission of $$ 100$ for each policy sold. The second pays a salary of $$ 20, 000$ plus a commission of $$ 50$ for each policy sold. How many policies would need to be sold to make the total pay the same?

Solution

There would need to be $160$ policies sold to make the total pay the same.

28) Kenneth currently sells suits for company A at a salary of $$ 22, 000$ plus a $$ 10$ commission for each suit sold. Company B offers him a position with a salary of $$ 28, 000$ plus a $$ 4$ commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

Solution

Kenneth would need to sell $1, 000$ suits.

Access these online resources for additional instruction and practice with solving systems of equations by substitution.

Instructional Video-Solve Linear Systems by Substitution

Instructional Video-Solve by Substitution

Key Concepts

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.

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

After reviewing this checklist, what will you do to become confident for all objectives?

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Fanshawe Pre-Health Sciences Mathematics 1 Copyright © 2022 by Domenic Spilotro, MSc is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted.

Solve a System of Equations by Substitution

HOW TO

Solve a system of equations by substitution.

Solve Applications of Systems of Equations by Substitution

How To:

How to use a problem-solving strategy for systems of linear equations.

Self Check

License

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