Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
1. Solving Systems by Graphing
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Exercise 45 Page 369

Practice makes perfect
a Let's find an equation to represent the total cost c of parking h hours in each garage.

Garage A

The hourly fee to park in Garage A is $2.50. Therefore, for h hours of parking, we will pay 2.5h. Furthermore, there is also a flat fee of $5, which is added to the previous expression. We can write our first equation using this information. c= 2.5h+ 5

Garage B

Garage B does not have an hourly fee. When parking here, we pay a flat fee of $20 and nothing else. c= 20

System of Equations

The equations we have written form a system of linear equations. c=2.5h+5 & (I) c=20 & (II)

b To find the solution to the system, we will graph the equations and find the coordinates of the point of intersection, if any. Let's start with Equation (I). For simplicity, we will rewrite 2.5 as 52.

c= 5/2h+ 5Note that the equation above is written in slope-intercept form, where the y-intercept is 5. To find a second point on the graph, we will use the slope 5 2. Starting at (0, 5), we move 2 steps to the right and 5 steps up.

The second equation does not depend on the h-variable. Therefore, it is a horizontal line whose y-intercept is 20. Let's graph it on the same coordinate plane. We will pay close attention to the x-coordinate of the point of intersection, which represents the number of hours for which parking at both garages has the same cost.

We see that the x-coordinate of the point of intersection of the lines is 6. This means that parking for 6 hours has the same cost at either garage.

c Let's consider our graph again. This time, we will pay close attention to the points on both lines where the x-coordinate is 3.

We see above that parking for 3 hours costs $12.50 at Garage A and $20 at Garage B. Therefore, we would choose Garage A.