Intermediate Algebra

Question

A family has two cars. The first car has a fuel efficiency of 40

 miles per gallon of gas and the second has a fuel efficiency of 30

 miles per gallon of gas. During one particular week, the two cars went a combined total of 1850

 miles, for a total gas consumption of 50

 gallons. How many gallons were consumed by each of the two cars that week?

Answer 1

40F+30S = 18350(50) F=1st car's gallons, S=second car's gallons=50-F

40F+30(50-F) =1850

4F +3(50-F) = 185

4F +150 -3F = 185

F = 185-150 = 35 gallons

S =50-35= 15 gallons

It's a version of d=rt, distance = rate of speed times time, d=rg

F and S are the gallons (similar to time)

40, 30, 50 are the rates: mpg miles/gallon

total distance = sum of each car's distance

rgf + rgs = d

df + ds = d total

Answer 2

Hi Mary,

To determine the number of miles a car drives, we multiply the fuel efficiency (miles/gallon) by the number of gallons the car consumed. This works because the gallons cancel out: (30 miles/gallon) x (? gallons) = miles traveled

To calculate the combined total miles traveled, we add

(miles traveled by Car 1) + (miles traveled by Car 2) = total miles

Because miles traveled = (fuel efficiency)(gallons consumed), rewrite the equation as

(Car 1 fuel efficiency)(Car 1 gallons) + (Car 2 fuel efficiency)(Car 2 gallons) = total miles

Let's assume that Car 1 consumes x gallons and Car 2 consumes y gallons. Using these variables, we rewrite the equation again as 40x + 30y = 1850. This equation can be simplified to 4x + 3y =185 by dividing each term by 10.

The problem also tells us that the cars' combined gas consumption is 50 gallons. Since x is Car 1's consumption in gallons and y is Car 2's consumption in gallons, x + y = 50.

So now we have a system of 2 equations: 4x + 3y =185

x + y = 50

There are a couple of ways to solve systems of equations. Method 1 is Substitution:

1. Rewrite x + y = 50 as x = 50 - y
2. Plug (50 - y) in for x in the first equation. 4(50-y) + 3y = 185
3. Solve. 4(50-y) + 3y = 185

200 - 4y + 3y = 185

200 - y = 185

-y = - 15

y = 15

1. Plug the value for y into one of the equations to solve for x.

x + y = 50

x + 15 = 50

x = 35

Method 2 is Elimination by Addition. Here, we want to add the equations so we can cancel out one variable.

1. Multiply -3(x + y = 50)

-3x - 3y = -150

1. Add this modified equation to the first equation

(-3x - 3y = -150)

+ (4x + 3y = 185)

x = 35

1. Use the value of x to solve for y.

x + y = 50

35 + y = 50

y = 15

So Car 1 consumes 35 gallons of gas and Car 2 consumes 15 gallons of gas.