substitution/elimination

Question

The outdoors club held a car wash to fundraise. They washed cars for $5 and vans for $7 each. They washed 90 vehicles and earned $510. How many of each type of vehicle did they wash? (please solve algebraically using substitution or elimination)

Answer 1

Let C = #cars washed

V = #vans washed

Then:

C + V = 90

If cars are washed for $5, then 5C represents the money brought if for each car that is washed.

Similarly, 7V represents the money brought in for van washings.

That means:

5C + 7V = 510

Probably easiest to use substitution because it is easy to isolate one of the variables without introducing fractions, Solve the first equation for V.

V = 90 - C (substitute that equivalent expression for V into the other equation:

5C + 7(90 - C) = 510

5C + 630 - 7C = 510

-2C = -120

C = 60

V = 90 - 60 = 30

So, 60 cars and 30 vans.

5(60) + 7(30) = 300 + 210 = 510 (check)