A car gets 21 mi/gal in city driving and 28 mi/gal highway driving. If 18 gal of gas are used in traveling 448 mi, how many miles were driven in the city, how many driven on the highway (assuming that only the given rates of usage were actually used)?

Question

A car gets 21 mi/gal in city driving and 28 mi/gal highway driving. If 18 gal of gas are used in traveling 448 mi, how many miles were driven in the city, how many driven on the highway (assuming that only the given rates of usage were actually used)?

Answer 1

A car gets 21 mi/gal in city driving and 28 mi/gal highway driving. If 18 gal of gas are used in traveling 448 mi, how many miles were driven in the city, how many driven on the highway (assuming that only the given rates of usage were actually used)?

H = highway gallons and C = city gallons.
Then,
H + C = 18 and

21C + 28H = 448.

I suspect you can take it from here.

Answer 2

To solve this problem, we can set up a system of equations using the given information.

Let's assume that x miles were driven in the city and y miles were driven on the highway.

We need to find x and y.

Given that the car gets 21 miles per gallon in city driving, and 18 gallons were used to travel a total of 448 miles, we can write the equation:

x miles / 21 miles per gallon + y miles / 28 miles per gallon = 448 miles / 18 gallons

Now we can solve this equation to find the values of x and y.

First, let's simplify the equation by multiplying through by the least common multiple of the denominators of the fractions, which is 588:

(588x) / 21 + (588y) / 28 = (588*448) / 18

Simplifying further:

28x + 21y = 32944 / 3

Now, let's simplify the equation further by dividing the right side by 3:

28x + 21y = 10981.33

To find integer solutions, we can multiply the equation by 3 to eliminate the decimal:

84x + 63y = 32944

Now, we need to solve this linear system of equations consisting of:
28x + 21y = 10981.33
84x + 63y = 32944

I will use the method of elimination to solve this system of equations:

Multiply the first equation by 3:

3 * (28x + 21y) = 3 * 10981.33
84x + 63y = 32944

84x + 63y = 32943.99 -> 84x + 63y = 32944

As we can see, both equations are the same.

Therefore, there are infinitely many solutions to this system of equations.

In this case, we cannot determine the specific values of x and y. We can only conclude that the values of x and y can vary, as long as they satisfy the equation 84x + 63y = 32944.