Let's solve this step-by-step.

Let:

• $d$ be the distance to the place,
• $v_w$ be the walking speed, and
• $v_r$ be the riding speed.

Step 1: Set Up Equations Based on Given Information

1. Equation for walking one way and riding back:
   If it takes him 10 hours to walk one way and ride back, we have: $\frac{d}{v_w} + \frac{d}{v_r} = 10$

2. Equation for riding both ways:
   If it takes him 8 hours to ride both ways, we have: $2 \cdot \frac{d}{v_r} = 8$

Step 2: Solve for $\frac{d}{v_r}$ from the second equation

From the second equation: $2 \cdot \frac{d}{v_r} = 8$ Dividing both sides by 2: $\frac{d}{v_r} = 4$

Step 3: Substitute $\frac{d}{v_r} = 4$ into the first equation

Substitute into the first equation: $\frac{d}{v_w} + 4 = 10$ Subtracting 4 from both sides: $\frac{d}{v_w} = 6$

Step 4: Calculate the time to walk both ways

Since $\frac{d}{v_w} = 6$, the time to walk one way is 6 hours. Therefore, the time to walk both ways is: $2 \cdot 6 = 12 \text{ hours}$

Answer

It will take him 12 hours to walk both ways.

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Would you like more details on any step? Here are some related questions:

1. What if the riding speed was different—how would that affect the outcome?
2. How do we calculate time if given only speeds and not distances?
3. What if he walked and rode in different combinations? How would the times change?
4. How would the solution change if the distance to the place were known?
5. How can this be visualized with a time-distance graph?

Tip: When dealing with rates and times in both directions, setting up equations by separating each direction often simplifies complex distance-rate problems.