Let's define the distance from Tom's house to Easton as $d$ miles. We'll use the relationship between distance, speed, and time, which is given by:

$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$

Step 1: Set up the equation for Tom's trip to Easton

• For the trip to Easton, Tom's speed was 50 mph.
• The time for this trip is:

$\text{Time to Easton} = \frac{d}{50}$

Step 2: Set up the equation for Tom's return trip

• For the return trip, Tom's speed was 45 mph.
• The time for this trip is:

$\text{Time to return} = \frac{d}{45}$

Step 3: Use the information about the time difference

We know that the return trip took 30 minutes (or $\frac{1}{2}$ hour) longer than the trip to Easton. So, we can write the following equation:

$\frac{d}{45} = \frac{d}{50} + \frac{1}{2}$

Step 4: Solve the equation

To solve for $d$, first eliminate the fractions by multiplying through by a common denominator. The least common denominator of 45, 50, and 2 is 450.

Multiply the entire equation by 450:

$450 \times \frac{d}{45} = 450 \times \frac{d}{50} + 450 \times \frac{1}{2}$

This simplifies to:

$10d = 9d + 225$

Now subtract $9d$ from both sides:

$d = 225$

Step 5: Conclusion

The distance from Tom's house to Easton is $\boxed{225}$ miles.

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Would you like further clarification or additional details about the solution?

Here are some related questions for further exploration:

1. How do we interpret the relationship between speed, time, and distance in real-life situations?
2. What would happen if Tom's return trip was faster? How would that change the equation?
3. How can this problem be generalized for other situations involving delays in travel?
4. How does changing the time difference affect the solution for the distance?
5. What are the impacts of increasing the speed on the total time for a trip?

Tip: When working with problems involving rates, always express the time difference clearly, and use common denominators to simplify fractions!