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# Applications Using Linear Models

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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: $x &= \text{distance in miles}\!\\y &= \text{cost of the rental truck}$ Peter pays a flat fee of$40 for the day; this is the $y-$ intercept.

He pays $63 for 46 miles; this is the coordinate point (46,63). Start with the point-slope form of the line: $y-y_0=m(x-x_0)$ Plug in the coordinate point: $63-y_0=m(46-x_0)$ Plug in the point (0, 40): $63-40=m(46-0)$ Solve for the slope: $23=46m \rightarrow m=\frac{23}{46}=0.5$ 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 $y-$ intercept, the equation of the line is $y=0.5x+40$ .

To find out the cost of driving the truck 220 miles, we plug in $x=220$ to get $y-40=0.5(220) \Rightarrow y= \ 150$ .

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: $x &= \text{number of window shades sold}\!\\y &= \text{Anne's earnings}$ 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: $y-y_0=m(x-x_0)$ Plug in the slope: $y-y_0=6(x-x_0)$ Plug in the point (200, 2500): $y-2500=6(x-200)$ To find Anne’s base salary, we plug in $x = 0$ and get $y-2500=-1200 \Rightarrow y=\ 1300$ . 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.

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: $x &= \text{pounds of oranges}\!\\y &= \text{pounds of cherries}$ The equation that describes this situation is $2x+3y=12$ . If she buys 4 pounds of oranges, we can plug $x = 4$ into the equation and solve for $y$ : $2(4)+3y=12 \Rightarrow 3y=12-8 \Rightarrow 3y=4 \Rightarrow y=\frac{4}{3}$ Nadia can buy $1 \frac{1}{3}$ 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: $y=mx+b$ , where $m$ is the slope and the point $(0, b)$ is the $y-$ intercept. • Often, we don’t know the value of the $y-$ intercept, but we know the value of $y$ for a non-zero value of $x$ . 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 $y-y_0=m(x-x_0)$ , where $m$ is the slope and $(x_0, y_0)$ is a point on the line. • An equation in standard form is written $ax+by=c$ , where $a, b$ , and $c$ are all integers and $a$ is positive. (Note that the $b$ in the standard form is different than the $b$ 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 $\frac{1}{2}$ an hour, how long does he need to walk to get to school? Solution Let’s define our variables: $x &= \text{time Peter skateboards}\!\\y &= \text{time Peter walks}$ The equation that describes this situation is: $7x+3y=6$ If Peter skateboards $\frac{1}{2}$ an hour, we can plug $x = 0.5$ into the equation and solve for $y$ : $7(0.5)+3y=6 \Rightarrow 3y=6-3.5 \Rightarrow 3y=2.5 \Rightarrow y=\frac{5}{6}$ Peter must walk $\frac{5}{6}$ 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 $\frac{2}{3}$ and contains the point $\left(\frac{1}{2}, 1 \right)$ . 2. The line has a slope of -1 and contains the point $\left(\frac{4}{5}, 0 \right)$ . 3. The line has a slope of 2 and contains the point $\left(\frac{1}{3}, 10 \right)$ . 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 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?
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?