# 5.3: Applications Using Linear Models

Difficulty Level: At Grade Created by: CK-12
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Practice Applications Using Linear Models

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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 ### 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 of40 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 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*}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 plug in \begin{align*}x=220\end{align*} to get \begin{align*}y-40=0.5(220) \Rightarrow y= \ 150\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 gets6 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 is1300.

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.

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. ### 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. 1. 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*}. 2. The line has a slope of -1 and contains the point \begin{align*}\left(\frac{4}{5}, 0 \right)\end{align*}. 3. The line has a slope of 2 and contains the point \begin{align*}\left(\frac{1}{3}, 10 \right)\end{align*}. 4. The line contains points (2, 6) and (5, 0). 5. The line contains points (5, -2) and (8, 4). 6. The line contains points (-2, -3) and (-5, 1). For 9-10, solve the problem. 1. Andrew has two part time jobs. One pays6 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?
2. 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?

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