SOLUTION: Eric drives to a town 350 miles away, then returns driving 20 miles per hour faster. If the return trip takes him 2 hours less, how fast was he driving during the drive out?

Question 963399: Eric drives to a town 350 miles away, then returns driving 20 miles per hour faster. If the return trip takes him 2 hours less, how fast was he driving during the drive out?

Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
.
x=original speed
(350mi/x)-2hrs=350mi/(x+20mph) Find common denominator on left side
(350/x)-(2x/x)=350/(x+20) Simplify left side.
(350-2x)/x=350/(x+20) Cross multiply.
(350-2x)(x+20)=350x Multiply out left side.
350x+7000-2x^2-40x=350x Subtract 310x from each side.
7000-2x^2=40x Subtract 7000-2x^2 from each side.
0=2x^2+40x-7000 Divide by 2.
0=x^2+20x-3500 Factor right side.
0=(x-50)(x+70)
x-50=0 or x+70=0
x=50 or x=-70
ANSWER: The original speed (x) was 50 mph.
CHECK:
(350mi/50mph)-2hrs=350mi/(50mph+20mph)
7hrs-2hrs=350mi/70mph
5hrs=5hrs
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