### Applications using Linear Models

Solve word problems using the equation of a straight line.

#### Real-World Application: Moving Trucks

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?

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*}-intercept.

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, so the truck company charges 0.5 dollars, or 50 cents, per mile ($0.5 = 50 cents). Plugging in the slope and the \begin{align*}y\end{align*}-intercept, the equation of the line is \begin{align*}y=0.5x+40\end{align*}.

To find out the cost of driving the truck 220 miles, we substitute 220 for \begin{align*}x\end{align*} to get \begin{align*}y-40=0.5(220) \Rightarrow y=\$150\end{align*}.

Driving 220 miles would cost $150.

#### Real-World Application: Sales Commission

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. Use the equation to determine Anne’s monthly base salary.

First define the variables:

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

You 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 there is a point at (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.

#### Real-World Application: Buying Fruit

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?

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\\ 3y &=12-8\\ 3y &=4\\ y &=\frac{4}{3}\end{align*}

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

### Example

#### Example 1

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*}\tfrac{1}{2}\end{align*} an hour, how long does he need to walk to get to school?

Define the 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*}

Peter skateboards \begin{align*}\tfrac{1}{2}\end{align*} an hour, so substitute \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\\ 3y &=6-3.5\\ 3y &=2.5\\ y &=\frac{5}{6}\end{align*}

Peter must walk \begin{align*}\tfrac{5}{6}\end{align*} of an hour to get to school.

### Review

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

- The line has a slope of \begin{align*}\tfrac{2}{3}\end{align*} and contains the point \begin{align*}\left(\tfrac{1}{2}, 1 \right)\end{align*}
- The line has a slope of -1 and contains the point \begin{align*}\left(\tfrac{4}{5}, 0 \right)\end{align*}
- The line has a slope of 2 and contains the point \begin{align*}\left(\tfrac{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 how much she should invest to earn the maximum interest without penalty. If she invests $5000 in the 5% interest account, how much money can she invest in the other account?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 5.3.