A family has two cars. The first car has a fuel efficiency of 20 miles per gallon of gas and the second has a fuel efficiency of 35 miles per gallon of gas. During one particular week, the two cars went a combined total of 1225 miles, for a total gas consumption of 50 gallons. How many gallons were consumed by each of the two cars that week?

first car:
Second:

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alinakincsem alinakincsem · Answer 1 · 2019-06-21T23:35:32+02:00

Answer:

First car: 35 gallons

Second car: 15 gallons

Step-by-step explanation:

Assuming [tex] x [/tex] to be the gas consumed by 1st car and [tex] 50 - x [/tex] to be the gas consumed by 2nd car.

We know that:

Distance traveled = fuel efficiency × gas consumed

[tex]20x+35(50-x)=1225[/tex]

[tex]20x+1750-35x=1225[/tex]

[tex]35x-20x=1750-1225[/tex]

[tex]15x=525[/tex]

[tex]x=35[/tex]

So [tex]50 - x = 50-35=15 [/tex]

Therefore, first car consumed 35 gallons while second car consumed 15 gallons.

luisejr77 luisejr77 · Answer 2 · 2019-06-22T03:00:47+02:00

Answer:

First car: 35 gallons

Second car: 15 gallons

Step-by-step explanation:

Set up a system of equations.Let be "x" the number of gallons consumed by the first car and "y" the number of gallons consumed by the second car.

Then:

[tex]\left \{ {{x+y=50} \atop {20x+35y=1,225}} \right.[/tex]

Applying the Elimination method, multiply the first equation by -20, then add both equations and solve for "y":

[tex]\left \{ {{-20x-20y=1,000} \atop {20x+35y=1,225}} \right. \\.......................\\15y=225\\y=15[/tex]

Substitute this value into the first equation and solve for "x":

[tex]x+15=50\\x=35[/tex]