Need help doing this problem please!

Question

Trey will rent a car for the weekend. He can choose one of two plans. The first plan has an initial fee of $55 and cost an additional $0.08 per mile driven. The second plan has an initial fee of $50 and cost an additional $0.10 per mile driven.

for what amount of driving do the two plans cost the same?

what is the cost when the two plans cost the same?

Answer 1

Hello, Leanna,

We can state the two plans in two equations:

Plan 1: Cost1 = 0.08X + 55

Plan 2: Cost2 = 0.10X + 50

where X is miles driven and the initial fees are added to that plan's amount.

"At what amount of driving do the two plans cost the same?" can be stated algebraically by setting:

Cost1 = Cost2

Therefore:

0.08X+55 = 0.10X + 50

Rearranging, we get -0.02X = -5

X is therefore 250 miles.

Check by solving each equation for x = 250. Each results in a total cost of $75 to drive 250 miles.

I hope this helps explain the process,

Bob

Answer 2

Let "C" be the cost of renting a car and let "m" be the number of miles driven.

For the first plan, the cost is 55 plus 0.08 times m or C = 55 + 0.08m

For the second plan, the cost is 50 plus 0.10 times m or C = 50 + 0.10m

Since you want to know about the point when the plans cost the same, then we want to know when "55 + 0.08m" equals "50 + 0.10m" so:

55 + 0.08m = 50 + 0.10m [subtract 50 from both sides to get:

5 + 0.08m = 0.10m [subtract 0.08m from both sides to get:

5 = 0.02m [divide both sides by 0.02 to get:

m = 250 miles

So at 250 miles, both plans will cost the same. The amount they cost is found by plugging in m = 250 to either equation (C = 55 + 0.08m or C = 50 + 0.10m). I'll choose the second plan equation C = 50 + 0.10m

C = 50 + 0.10(250)

C = 50 + 25

C = 75

So the cost is $75