Bob and Carl each rented the same kind of moving truck from EZ Move. There was a flat rental fee plus a charge per mile that the truck was driven. Bob’s cost for his truck was $112.96 for 138 miles. Carl’s cost for his truck was $142.78 for 209 miles. Which equation can be used to represent the cost of the rental truck?

Question

Round to the nearest hundredth if necessary.
y = 71 x minus 29.82
y = 25 x minus 66
y = 0.42 x + 71
y = 0.42 x + 55

Answer 1

To find the cost equation that represents the rental truck situation, we can set up a system of equations based on the provided information.

Let's denote:

• \( y \) as the total cost of the rental truck,
• \( x \) as the number of miles driven,
• \( f \) as the flat rental fee,
• \( c \) as the charge per mile.

From Bob's rental, we can create the equation: \[ 112.96 = f + 138c \quad (1) \]

From Carl's rental, we have: \[ 142.78 = f + 209c \quad (2) \]

Now, let's subtract equation (1) from equation (2): \[ (142.78 - 112.96) = (f + 209c) - (f + 138c) \] \[ 29.82 = 71c \]

Now, solving for \( c \) (the charge per mile): \[ c = \frac{29.82}{71} \approx 0.42 \]

Now, we can substitute \( c \) back into either equation to find \( f \). Let's use equation (1): \[ 112.96 = f + 138(0.42) \] \[ 112.96 = f + 57.96 \] \[ f = 112.96 - 57.96 = 55 \]

Therefore, we have:

• The flat fee \( f \) is approximately 55,
• The charge per mile \( c \) is approximately 0.42.

Thus, the cost equation can be represented as: \[ y = 0.42x + 55 \]

The correct answer from the options provided is: \[ \boxed{y = 0.42x + 55} \]