I need help with this problem

Question

A family has two cars During one particular week the first car consumed 20 gallons of gas and the second consumed 40 gallons of gas.The two cars drove combined total of 2200 miles and the sum of thier fuel efficiencies was 75 miles per gallon.What were the fuel efficiencies of each of the cars that week

Answer 1

x=fuel efficiency of the first car

y=fuel efficiency of the second car

2200=20x+40y

x+y=75

x=75-y

2200=20x+40y

divide both sides by 20

110=x+2y

substitute

110=75-y+2y

110-75=y

35=y

y=35 mpg

x=75-35=40

x=40 mpg

Answer 2

Good Morning DeAsia,

 

Let x be the number of miles the first car traveled during the week.

 

That means the second car traveled 2200 -x

 

The gas mileage (fuel efficiency) is equal to miles traveled divided by gallons used.

 

The fuel efficiency of the first car is x/20 because the problem states that it consumed 20 gallons

 

The fuel efficiency of the second car is (2200 - x)/40

 

When I add these two together, I get 75

 

x/20 + (2200-x)/40 = 75

 

I'm going to multiply each term by 40 to clear the fractions

 

40x/20 + 40 (2200 - x)/40 = 40 * 75

 

2x + (2200 - x) = 3000

 

x + 2200 = 3000

 

x = 800

 

The fuel efficiency of the first car is miles per gallon. So 800 miles / 20 gallons = 40 miles per gallon

 

The second car traveled (2200 - 800) miles which equals 1400 miles

 

So the efficiency of the second car is 1400 miles / 40 gallons = 35 miles per gallon.

 

And to double check, they do add to 75