Answer:
The car uses less gas
They use the same amount of gas after [tex]\frac{640}{7}[/tex] miles
Step-by-step explanation:
Given
The table represents the car mileage
[tex]y = -\frac{1}{5}x + 31[/tex] --- The van
First, calculate the car's slope (m)
[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
From the table, we have:
[tex](x_1,y_1) = (60,13.5);\ \ (x_2,y_2) = (180,10.5)[/tex]
So, we have:
[tex]m = \frac{10.5 - 13.5}{180 - 60}[/tex]
[tex]m = \frac{-3}{120}[/tex]
[tex]m = -\frac{1}{40}[/tex]
Calculate the equation using:
[tex]y = -\frac{1}{40}(x - 60)+13.5[/tex]
[tex]y = -\frac{1}{40}x + 1.5+13.5[/tex]
[tex]y = -\frac{1}{40}x + 15[/tex]
[tex]m = -\frac{1}{40}[/tex] implies that for every mile traveled, the car uses 1/40 gallon of gas
Also:
[tex]y = -\frac{1}{5}x + 31[/tex] --- The van
By comparison to: [tex]y = mx + b[/tex]
[tex]m = -\frac{1}{5}[/tex]
This implies that for every mile traveled, the van uses 1/5 gallon of gas.
By comparison:
[tex]1/40 < 1/5[/tex]
This means that the car uses less gas
Solving (b): Distance traveled for them to use the same amount of gas.
We have:
[tex]y = -\frac{1}{5}x + 31[/tex] --- The van
[tex]y = -\frac{1}{40}x + 15[/tex] --- The car
Equate both
[tex]-\frac{1}{5}x + 31 =-\frac{1}{40}x + 15[/tex]
Collect like terms
[tex]\frac{1}{40}x -\frac{1}{5}x =-31 + 15[/tex]
[tex]\frac{1}{40}x -\frac{1}{5}x =-16[/tex]
Take LCM
[tex]\frac{x - 8x}{40} = -16[/tex]
[tex]\frac{- 7x}{40} = -16[/tex]
Solve for -7x
[tex]-7x = -640[/tex]
Solve for x
[tex]x = \frac{640}{7}[/tex]
