Question

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

distance =rate*time
The distance is the same.
y = rate going
z = rate returning
Going: d = y*t = y*5 = 5y
return: d = z*t = z*4 = 4z
we know the return rate was 5 mi/hr faster; therefore, the return rate of z = y+5
set d = d
5y=4z
z=y+5
=========
Solve simultaneously
Post your work if you need further assistance.

many students find distance-rate-time problems really easy if they make a chart

..........│..D..│..R..│..T..│
----------------------------
1st trip..│.5x..│..x..│..5..│
----------------------------
2nd trip..│4(x+5)│x+5│..4..│

clearly the two trips are the same
so 5x=4(x+5).........etc, piece of cake

The chart really IS a good idea.