To solve this problem, let's break it down step by step.

Let:

• $r_1$ = the average speed (in miles per hour) for the first trip,
• $r_2$ = the average speed for the second trip,
• $t_1$ = the time (in hours) taken for the first trip,
• $t_2$ = the time taken for the second trip.

Step 1: Time equation for each trip

• For the first trip: $\text{Distance} = \text{Speed} \times \text{Time} \implies t_1 = \frac{3060}{r_1}$

• For the second trip: $t_2 = \frac{4500}{r_2}$

Step 2: Relationship between the speeds

From the problem, we know that the speed for the second trip is 60 miles per hour slower than the speed for the first trip: $r_2 = r_1 - 60$

Step 3: Relationship between the times

We are also told that the second trip took 4 hours longer than the first trip: $t_2 = t_1 + 4$

Step 4: Set up equations

Substitute the expressions for $t_1$ and $t_2$ into the time relationship equation:

$\frac{4500}{r_2} = \frac{3060}{r_1} + 4$

Now, replace $r_2$ with $r_1 - 60$: $\frac{4500}{r_1 - 60} = \frac{3060}{r_1} + 4$

Step 5: Solve the equation

Now we solve this equation to find $r_1$, the speed of the first trip.

Multiply both sides of the equation by $r_1(r_1 - 60)$ to eliminate the denominators:

$4500r_1 = 3060(r_1 - 60) + 4r_1(r_1 - 60)$

Expanding both sides:

$4500r_1 = 3060r_1 - 3060 \times 60 + 4r_1^2 - 240r_1$

Simplify the equation:

$4500r_1 = 3060r_1 - 183600 + 4r_1^2 - 240r_1$

Combine like terms:

$0 = 4r_1^2 - 240r_1 + 3060r_1 - 4500r_1 - 183600$ $0 = 4r_1^2 - 240r_1 + 3060r_1 - 4500r_1 - 183600$ [