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# Applications Using Linear Models

## Solve story problems using linear equations

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Practice Applications Using Linear Models
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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 ### 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: xy=distance in miles=cost of the rental truck Peter pays a flat fee of$40 for the day; this is the y\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: yy0=m(xx0)\begin{align*}y-y_0=m(x-x_0)\end{align*} Plug in the coordinate point: 63y0=m(46x0)\begin{align*}63-y_0=m(46-x_0)\end{align*} Plug in the point (0, 40): 6340=m(460)\begin{align*}63-40=m(46-0)\end{align*} Solve for the slope: 23=46mm=2346=0.5\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 y\begin{align*}y-\end{align*}intercept, the equation of the line is y=0.5x+40\begin{align*}y=0.5x+40\end{align*}.

To find out the cost of driving the truck 220 miles, we plug in x=220\begin{align*}x=220\end{align*} to get y40=0.5(220)y=150\begin{align*}y-40=0.5(220) \Rightarrow y= \ 150\end{align*}. Driving 220 miles would cost150.

#### 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:

xy=number of window shades sold=Anne's earnings

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: yy0=m(xx0)\begin{align*}y-y_0=m(x-x_0)\end{align*}

Plug in the slope: yy0=6(xx0)\begin{align*}y-y_0=6(x-x_0)\end{align*}

Plug in the point (200, 2500): y2500=6(x200)\begin{align*}y-2500=6(x-200)\end{align*}

To find Anne’s base salary, we plug in x=0\begin{align*}x = 0\end{align*} and get y2500=1200y=1300\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: xy=pounds of oranges=pounds of cherries The equation that describes this situation is 2x+3y=12\begin{align*}2x+3y=12\end{align*}. If she buys 4 pounds of oranges, we can plug x=4\begin{align*}x = 4\end{align*} into the equation and solve for y\begin{align*}y\end{align*}: 2(4)+3y=123y=1283y=4y=43 Nadia can buy 113\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: y=mx+b\begin{align*}y=mx+b\end{align*}, where m\begin{align*}m\end{align*} is the slope and the point (0,b)\begin{align*}(0, b)\end{align*} is the y\begin{align*}y-\end{align*}intercept. • Often, we don’t know the value of the y\begin{align*}y-\end{align*}intercept, but we know the value of y\begin{align*}y\end{align*} for a non-zero value of x\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 yy0=m(xx0)\begin{align*}y-y_0=m(x-x_0)\end{align*}, where m\begin{align*}m\end{align*} is the slope and (x0,y0)\begin{align*}(x_0, y_0)\end{align*} is a point on the line. • An equation in standard form is written ax+by=c\begin{align*}ax+by=c\end{align*}, where a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*} are all integers and a\begin{align*}a\end{align*} is positive. (Note that the b\begin{align*}b\end{align*} in the standard form is different than the b\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 12\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: xy=time Peter skateboards=time Peter walks The equation that describes this situation is: 7x+3y=6\begin{align*}7x+3y=6\end{align*} If Peter skateboards 12\begin{align*}\frac{1}{2}\end{align*} an hour, we can plug x=0.5\begin{align*}x = 0.5\end{align*} into the equation and solve for y\begin{align*}y\end{align*}: 7(0.5)+3y=63y=63.53y=2.5y=56 Peter must walk 56\begin{align*}\frac{5}{6}\end{align*} of an hour. ### Explore More For 1-8, write the equation in slope-intercept, point-slope and standard forms. 1. The line has a slope of 23\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 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?