SOLUTION: Automobile companies advertise two rates for fuel mileage. City mileage is the rate fo fuel consumption for driving in stop-and-go traffic. Highway mileage is the rate for drivin

Algebra ->  Equations -> SOLUTION: Automobile companies advertise two rates for fuel mileage. City mileage is the rate fo fuel consumption for driving in stop-and-go traffic. Highway mileage is the rate for drivin      Log On


   

Question 28292: Automobile companies advertise two rates for fuel mileage. City mileage is the rate fo fuel consumption for driving in stop-and-go traffic. Highway mileage is the rate for driving at higher speeds for long periods of time Cynthia's new car gets 17 mi/gal in the city and 25 mi/gal on the highway. On one trip, she drove 220 miles on 11 gallons of gas. A. Define variables and write a system of equations for the gallons burned @ ea. rate.
B. Solve this system and explaing the meaning of your solution C. Find the number of city miles and highway miles Cynthia drove.
Will you please help me with this problem. I am having dificulty setting it up. Thank you.
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
In city her speed = 15mi/gal
On highway her speed = 25mi/gal
Total miles driven = 220 by both the city speed and the highway speed.
Total gallons of gas used (with city and highway speeds) = 11

Since 11 in the highest gallons used you know that the "x" value of which gallons of gas you are trying the find will be less than 11.

We want to find the gallons of gas for city first.
SO Mi driven in city = 17x
With mi driven on highway = 25(11-x)
Total miles driven = 220

Equation:
17x+25(11-x)=220
17x-25x=220-275
-8x=-55
x=6.875 (gallons used for city)

11-6.875=4.125 (gallon used on highway)

6.875(17)=116.875 (miles driven in city)
25(4.125)=103.125 (miles driven on highway)

Hence, she drove 116.875 miles in city and 103.125 miles on the highway.

Paul.