I am having some trouble working this out . Could you possibily show me how to work it out.Thank You!

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

Could somone walk through the steps to sovle this problem? ASAP Please!!

Could somone walk through the steps to sovle this problem?

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

ASAP Please!!

Sure! Let's break down the problem and find the solution step by step.

Step 1: Understand the problem.
We have a bicyclist who rode into the country for 5 hours. On the return trip, his speed was 5 mph faster and the trip took 4 hours. We need to find the speed of the bicyclist for each leg of the trip.

Step 2: Define the variables.
Let's assume the speed of the bicyclist on the outbound trip (to the country) is "x" mph. Then, the speed on the return trip (back from the country) will be "x + 5" mph.

Step 3: Use the formula for speed.
The formula to calculate speed is speed = distance / time. In this problem, we have time for each leg of the trip, so we can set up two equations using this formula.

For the outbound trip:
Speed = distance / time
x = distance / 5, as the time is given as 5 hours.

For the return trip:
Speed = distance / time
x + 5 = distance / 4, as the time is given as 4 hours.

Step 4: Relate the distances.
The distance traveled for the outbound trip is the same as the distance traveled for the return trip (since the cyclist started and ended at the same point). So we can equate the distances.

distance for outbound trip = distance for return trip

distance = speed * time for each leg

x * 5 = (x + 5) * 4

Step 5: Solve the equation.
Let's solve the equation to find the value of x.

5x = 4(x + 5)
5x = 4x + 20
5x - 4x = 20
x = 20

Step 6: Calculate the speed.
Now that we have the value of x, which is 20 mph, we can calculate the speed for each leg of the trip.

The speed on the outbound trip is x = 20 mph.
The speed on the return trip is x + 5 = 20 + 5 = 25 mph.

Therefore, the speed on the outbound trip is 20 mph and the speed on the return trip is 25 mph.