Let's define:

• $d$ as the distance Brian hiked uphill (in miles),
• $v$ as his speed uphill (in mph).

From the problem, we know the following:

1. Uphill: He took 3 hours to hike uphill. The distance-speed-time relation is: $\text{Distance} = \text{Speed} \times \text{Time}, \quad d = v \times 3$ So, $d = 3v$.

2. Downhill: His speed downhill is 1 mph faster than his speed uphill, i.e., $v + 1$. He took 2 hours to return downhill, so: $d = (v + 1) \times 2$

Now, since both equations represent the same distance $d$, we can set them equal: $3v = 2(v + 1)$

Solving for $v$:

1. Expand the right side: $3v = 2v + 2$
2. Subtract $2v$ from both sides: $v = 2$

So, his speed uphill was $v = 2$ mph.

Finding the distance $d$:

Now that we know his uphill speed, we can find the distance using $d = 3v$: $d = 3 \times 2 = 6 \text{ miles}.$

Therefore, Brian hiked 6 miles uphill.

Would you like further details or have any questions?

Here are some related questions for you to consider:

1. What would the distance be if Brian's downhill speed was 2 mph faster instead of 1 mph?
2. How long would it take if Brian hiked both uphill and downhill at the same speed of 2 mph?
3. What is the average speed for the entire trip (uphill and downhill)?
4. How would the solution change if Brian took 4 hours uphill and 3 hours downhill?
5. If the hill was 10 miles, how long would it take Brian to hike uphill if he maintained the same speed?

Tip: In distance-speed-time problems, equations involving both time and speed are often key to solving for unknowns.