SOLUTION: During rush hour, Fernando can drive 35 miles using the side roads in the same time that it takes to travel 30 miles on the freeway. If Fernando’s rate on the side roads is 9 mi/h

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: During rush hour, Fernando can drive 35 miles using the side roads in the same time that it takes to travel 30 miles on the freeway. If Fernando’s rate on the side roads is 9 mi/h       Log On

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Question 73196: During rush hour, Fernando can drive 35 miles using the side roads in the same time that it takes to travel 30 miles on the freeway. If Fernando’s rate on the side roads is 9 mi/h faster than his rate on the freeway, find his rate on the side roads.
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!

Let x=rate on freeway
Then (x+9)=rate on side roads
distance(d)=rate(r) times time(t) or d=rt and t=d/r
time to drive 35 mi using side roads=35%2F%28x%2B9%29
And time to drive 30 miles using freeway=30%2Fx
Now we are told that the above two times are equal, so:
35%2F%28x%2B9%29=30%2Fx multiply both sides by x%28x%2B9%29
35x=30%28x%2B9%29 get rid of parens
35x=30x%2B270 subtract 30 from both sides
35x-30x=30x-30x%2B270 collect like terms
5x=270 divide both sides by 5
x=54 mph----------------------------------rate on freeway
x%2B9=54%2B9=63 mph---------------------------rate on side roads
CK
t=d/r
35/63=30/54
0.55555=0.55555

Hope this helps---ptaylor