SOLUTION: A bicyclist rode into the country for 5 h. In returning, her speed was 5 mi/h faster and the trip took 4 h. What was her speed each way?

Question 70733: A bicyclist rode into the country for 5 h. In returning, her
speed was 5 mi/h faster and the trip took 4 h. What was her speed each way?
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
We'll use the basic equation that says the distance (D) that you travel is equal to the rate
or speed that you are going (R) times the Time (T) that you use. In equation form this is:
.

.
On the trip out this bicyclist rode at some unknown rate (R) and she rode for 5 hours.
Substituting these values into the equation results in:
.

.
On the way back she boosted her rate by 5 mph. Therefore, her new rate was R+5. She rode
for 4 hours. The equation that results is:
.

.
But the distance she traveled out is the same distance that she traveled on the way back.
This means that the left side of our two equations are equal. Because of this we can
site the right side equal also. To be a little more conventional let's write R*5 as 5*R and
(R+5)*4 as 4*(R+5). In equation form the equality of the right sides is:
.

.
Multiplying out the right side results in:
.

.
Subtract 4*R from both sides and the equation becomes:
.

.
This means that she rode out at 20 mph and as a result her return rate, 5 mph faster, was
25 mph.
.
Hope this helps you to understand the problem a little better.
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