# 5.3: Applications Using Linear Models

**At Grade**Created by: CK-12

**Practice**Applications Using Linear Models

What if your car rental company charges $25 per day plus $0.25 per mile? When the car is returned to you the trip odometer reads 324 miles and the customer's bill totals $156. How could you determine the number of days the customer rented the car? In this Concept, you'll be able to solve real-world problems like this one.

### Watch This

CK-12 Foundation: 0503S Solving Real-World Problems with Linear Equations (H264)

### Guidance

Let’s solve some word problems where we need to write the equation of a straight line in point-slope form.

#### Example A

*Marciel rented a moving truck for the day. Marciel only remembers that the rental truck company charges $40 per day and some number of cents per mile. Marciel drives 46 miles and the final amount of the bill (before tax) is $63. What is the amount per mile the truck rental company charges? Write an equation in point-slope form that describes this situation. How much would it cost to rent this truck if Marciel drove 220 miles?*

**Solution**

Let’s define our variables:

\begin{align*}x = \text{distance in miles}\!\\
y = \text{cost of the rental truck}\end{align*}

Peter pays a flat fee of $40 for the day; this is the \begin{align*}y-\end{align*}

He pays $63 for 46 miles; this is the coordinate point (46,63).

Start with the point-slope form of the line: \begin{align*}y-y_0=m(x-x_0)\end{align*}

Plug in the coordinate point: \begin{align*}63-y_0=m(46-x_0)\end{align*}

Plug in the point (0, 40): \begin{align*}63-40=m(46-0)\end{align*}

Solve for the slope: \begin{align*}23=46m \rightarrow m=\frac{23}{46}=0.5\end{align*}

The slope is 0.5 dollars per mile, so the truck company charges 50 cents per mile ($0.5 = 50 cents). Plugging in the slope and the \begin{align*}y-\end{align*}

To find out the cost of driving the truck 220 miles, we plug in \begin{align*}x=220\end{align*}

**Driving 220 miles would cost $150.**

#### Example B

*Anne got a job selling window shades. She receives a monthly base salary and a $6 commission for each window shade she sells. At the end of the month she adds up sales and she figures out that she sold 200 window shades and made $2500. Write an equation in point-slope form that describes this situation. How much is Anne’s monthly base salary?*

**Solution**

Let’s define our variables:

\begin{align*}x = \text{number of window shades sold}\!\\ y = \text{Anne’s earnings}\end{align*}

We see that we are given the slope and a point on the line:

Nadia gets $6 for each shade, so the slope is 6.

She made $2500 when she sold 200 shades, so the point is (200, 2500).

Start with the point-slope form of the line: \begin{align*}y-y_0=m(x-x_0)\end{align*}

Plug in the slope: \begin{align*}y-y_0=6(x-x_0)\end{align*}

Plug in the point (200, 2500): \begin{align*}y-2500=6(x-200)\end{align*}

To find Anne’s base salary, we plug in \begin{align*}x = 0\end{align*} and get \begin{align*}y-2500=-1200 \Rightarrow y=\$ 1300\end{align*}.

**Anne’s monthly base salary is $1300.**

**Solving Real-World Problems Using Linear Models in Standard Form**

Here are two examples of real-world problems where the standard form of an equation is useful.

#### Example C

*Nadia buys fruit at her local farmer’s market. This Saturday, oranges cost $2 per pound and cherries cost $3 per pound. She has $12 to spend on fruit. Write an equation in standard form that describes this situation. If she buys 4 pounds of oranges, how many pounds of cherries can she buy?*

**Solution**

Let’s define our variables:

\begin{align*}x = \text{pounds of oranges}\!\\ y = \text{pounds of cherries}\end{align*}

The equation that describes this situation is \begin{align*}2x+3y=12\end{align*}.

If she buys 4 pounds of oranges, we can plug \begin{align*}x = 4\end{align*} into the equation and solve for \begin{align*}y\end{align*}:

\begin{align*}2(4)+3y=12 \Rightarrow 3y=12-8 \Rightarrow 3y=4 \Rightarrow y=\frac{4}{3}\end{align*}

**Nadia can buy \begin{align*}1 \frac{1}{3}\end{align*} pounds of cherries.**

Watch this video for help with the Examples above.

CK-12 Foundation: Solving Real-World Problems with Linear Equations

### Vocabulary

- A common form of a line (linear equation) is
**slope-intercept form:**\begin{align*}y=mx+b\end{align*}, where \begin{align*}m\end{align*} is the slope and the point \begin{align*}(0, b)\end{align*} is the \begin{align*}y-\end{align*}intercept .

- Often, we don’t know the value of the \begin{align*}y-\end{align*}intercept, but we know the value of \begin{align*}y\end{align*} for a non-zero value of \begin{align*}x\end{align*}. In this case, it’s often easier to write an equation of the line in
**point-slope form.**An equation in point-slope form is written as \begin{align*}y-y_0=m(x-x_0)\end{align*}, where \begin{align*}m\end{align*} is the slope and \begin{align*}(x_0, y_0)\end{align*} is a point on the line.

- An equation in
**standard form**is written \begin{align*}ax+by=c\end{align*}, where \begin{align*}a, b\end{align*}, and \begin{align*}c\end{align*} are all integers and \begin{align*}a\end{align*} is positive. (Note that the \begin{align*}b\end{align*} in the standard form is different than the \begin{align*}b\end{align*} in the slope-intercept form.)

### Guided Practice

*Peter skateboards part of the way to school and walks the rest of the way. He can skateboard at 7 miles per hour and he can walk at 3 miles per hour. The distance to school is 6 miles. Write an equation in standard form that describes this situation. If he skateboards for \begin{align*}\frac{1}{2}\end{align*} an hour, how long does he need to walk to get to school?*

**Solution**

Let’s define our variables:

\begin{align*}x = \text{time Peter skateboards}\!\\ y = \text{time Peter walks}\end{align*}

The equation that describes this situation is: \begin{align*}7x+3y=6\end{align*}

If Peter skateboards \begin{align*}\frac{1}{2}\end{align*} an hour, we can plug \begin{align*}x = 0.5\end{align*} into the equation and solve for \begin{align*}y\end{align*}:

\begin{align*}7(0.5)+3y=6 \Rightarrow 3y=6-3.5 \Rightarrow 3y=2.5 \Rightarrow y=\frac{5}{6}\end{align*}

**Peter must walk \begin{align*}\frac{5}{6}\end{align*} of an hour.**

### Practice

For 1-8, write the equation in slope-intercept, point-slope and standard forms.

- The line has a slope of \begin{align*}\frac{2}{3}\end{align*} and contains the point \begin{align*}\left(\frac{1}{2}, 1 \right)\end{align*}.
- The line has a slope of -1 and contains the point \begin{align*}\left(\frac{4}{5}, 0 \right)\end{align*}.
- The line has a slope of 2 and contains the point \begin{align*}\left(\frac{1}{3}, 10 \right)\end{align*}.
- The line contains points (2, 6) and (5, 0).
- The line contains points (5, -2) and (8, 4).
- The line contains points (-2, -3) and (-5, 1).

For 9-10, solve the problem.

- Andrew has two part time jobs. One pays $6 per hour and the other pays $10 per hour. He wants to make $366 per week. Write an equation in standard form that describes this situation. If he is only allowed to work 15 hours per week at the $10 per hour job, how many hours does he need to work per week in his $6 per hour job in order to achieve his goal?
- Anne invests money in two accounts. One account returns 5% annual interest and the other returns 7% annual interest. In order not to incur a tax penalty, she can make no more than $400 in interest per year. Write an equation in standard form that describes this problem. If she invests $5000 in the 5% interest account, how much money does she need to invest in the other account?

### Notes/Highlights Having trouble? Report an issue.

Color | Highlighted Text | Notes | |
---|---|---|---|

Please Sign In to create your own Highlights / Notes | |||

Show More |

### Image Attributions

Here you'll learn how to solve real-world problems whose equations are straight lines in either point-slope, slope-intercept, or standard form.