Algebra word problem

Question

Twice last​ month, Judy Carter rented a car from a car rental company and traveled around the Southwest on business. The company rents its car for a daily​ fee, plus an additional charge per mile driven. Judy recalls that her first trip lasted 4​ days, she drove 450​ miles, and the rental cost her .$233.50.

On her second business trip she drove 200 miles in 3​ days, and paid $149.00

for the rental. Find the daily​ fee, and find the mileage charge.

Answer 1

Heres an equation that summarizes how to find the total cost of the trip. This helps if you don't know where to start building your equations to solve the problem

days(daily fee) + miles(charge per mile) = total cost

trip 1 = 233.50 = 4(fee) + 450(charge)

trip 2 = 149 = 3(fee) + 200(charge)

You can see that you are missing 2 variables. Let's let the daily fee = x and the charge per mile = y

233.50 = 4x + 450y

149 = 3x+ 200y

To solve for 2 variables, you need to isolate one variable in one of the equations and use that to solve the other equation. Lets use the bottom equation to isolate x.

149 = 3x+ 200y

149 - 200y = 3x

x = (149 - 200y) / 3 -> this can be substituted into the first original equation for the x to solve for y.

233.50 = 4x + 450y

233.50 = 4(149 - 200y) / 3 + 450y

233.50 = 4(149 - 200y) / 3 + 450y

233.50 = (596 - 800y) / 3 + 450y

233.50 = (596/3) - (800y/ 3) + 450y

233.50 = 198.67 - 266.67y + 450y

34.83 = 183.33y

y = 0.19 -> mileage charge

plug this into one of the original equations and solve for x

149 = 3x+ 200y

149 = 3x + 200(0.19)

111=3x

x = $37.00

Solution: the charge per day is $37.00 with a mileage cost of $0.19 per mile.

you can check your solution with the original equations! The tricky part about this problem is being careful with using your calculator and rounding. With money problems it is usually best to wait to round until the end or at least round to two decimal points through the problem.