SOLUTION: Help a mother teaching her son Algebra The story problem is Jose was taking a trip from point a to point b, on the way there, he drove 45mph. On the way back, he drove the same di

Question 87658: Help a mother teaching her son Algebra
The story problem is Jose was taking a trip from point a to point b, on the way there, he drove 45mph. On the way back, he drove the same distance at 50mph and got back 20 minutes faster. What was the distance for point a to point b. (The equations that I came up with was 45x=50x-20 to get the time but I know that it is not totally correct. I came up with 180 miles but the answer in the teachers guide is 150. Don't know how they got it.)
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
The equation is that Distance going = Distance returning.

Let t = time (in hours) of the return trip
t+1/3 = time going
50 MPH = rate of the return
45 MPH = rate going

Distance going = Distance returning
RT = RT
45(t+1/3) = 50(t)
45t + 15= 50t
15= 50t-45t
15=5t
t= 3 hours

Distance = 3*50= 150 miles

R^2 Retired from SCC
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