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.
To solve this problem, you can use the equation distance = rate * time. Since the distance for both trips is the same, you can set the equations for the two trips equal to each other.
Let's assume the rate for the first trip (going) is y and the rate for the return trip is z. The time for the going trip is 5 hours and the time for the return trip is 4 hours.
For the going trip:
distance = rate * time
d = y * 5
d = 5y
For the return trip:
distance = rate * time
d = z * 4
d = 4z
Since the distance is the same for both trips, we can set the two equations equal to each other:
5y = 4z
We also know that the return rate is 5 mph faster than the going rate:
z = y + 5
Now, we have a system of equations:
5y = 4z
z = y + 5
To solve this system of equations, you can substitute the second equation into the first equation:
5y = 4(y + 5)
Simplify the equation:
5y = 4y + 20
Subtract 4y from both sides:
y = 20
Now, substitute the value of y back into the second equation to find z:
z = 20 + 5
z = 25
Therefore, the speed each way is 20 mi/h for the going trip and 25 mi/h for the return trip.