Question

Helppop!!A car and van are driving on a highway. The table shows the amount y (in gallons) of gas in the cars gas tank after driving x miles. The amount of gas in the van’s gas tank after driving x miles is represented by the equation y=- 1/5x + 31. Which vehicle uses less gasoline per mile? How many miles must the vehicles travel for the amount of gas in each tank to be the same?

Answer 1

Answer:

The car uses less gas

They use the same amount of gas after $\frac{640}{7}$ miles

Step-by-step explanation:

Given

The table represents the car mileage

$y = -\frac{1}{5}x + 31$ --- The van

First, calculate the car's slope (m)

$m = \frac{y_2 - y_1}{x_2 - x_1}$

From the table, we have:

$(x_1,y_1) = (60,13.5);\ \ (x_2,y_2) = (180,10.5)$

So, we have:

$m = \frac{10.5 - 13.5}{180 - 60}$

$m = \frac{-3}{120}$

$m = -\frac{1}{40}$

Calculate the equation using:

$y = -\frac{1}{40}(x - 60)+13.5$

$y = -\frac{1}{40}x + 1.5+13.5$

$y = -\frac{1}{40}x + 15$

$m = -\frac{1}{40}$ implies that for every mile traveled, the car uses 1/40 gallon of gas

Also:

$y = -\frac{1}{5}x + 31$ --- The van

By comparison to: $y = mx + b$

$m = -\frac{1}{5}$

This implies that for every mile traveled, the van uses 1/5 gallon of gas.

By comparison:

$1/40 < 1/5$

This means that the car uses less gas

Solving (b): Distance traveled for them to use the same amount of gas.

We have:

$y = -\frac{1}{5}x + 31$ --- The van

$y = -\frac{1}{40}x + 15$ --- The car

Equate both

$-\frac{1}{5}x + 31 =-\frac{1}{40}x + 15$

Collect like terms

$\frac{1}{40}x -\frac{1}{5}x =-31 + 15$

$\frac{1}{40}x -\frac{1}{5}x =-16$

Take LCM

$\frac{x - 8x}{40} = -16$

$\frac{- 7x}{40} = -16$

Solve for -7x

$-7x = -640$

Solve for x

$x = \frac{640}{7}$