Let's break down the problem step-by-step to find the distance Eugenia lives from school.

Step 1: Define Variables and Set Up the Problem

Let:

• $d$ = distance from her house to the school (in miles).
• $t_1$ = time taken to walk to school at 3 mph.
• $t_2$ = time taken to run back home at 7 mph.

Since distance = speed × time, we have:

1. $d = 3 \times t_1$ (for the walk to school),
2. $d = 7 \times t_2$ (for the run back home).

Step 2: Express Total Time

We're given that the total time for both the walk to school and the run back home is 45 minutes, or $\frac{45}{60} = 0.75$ hours.

So, we can write: $t_1 + t_2 = 0.75$

Step 3: Solve the Equations

From equation (1), we have: $t_1 = \frac{d}{3}$

From equation (2), we have: $t_2 = \frac{d}{7}$

Substitute these into the total time equation: $\frac{d}{3} + \frac{d}{7} = 0.75$

To solve for $d$, find a common denominator (21): $\frac{7d}{21} + \frac{3d}{21} = 0.75$

Combine the fractions: $\frac{10d}{21} = 0.75$

Multiply both sides by 21: $10d = 15.75$

Now, divide by 10: $d = 1.575$

Answer

Eugenia lives approximately 1.575 miles from her school.

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Would you like further details on any of these steps, or do you have any other questions?

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Here are some related questions to deepen your understanding:

1. How would the solution change if Eugenia ran back at a different speed?
2. What if the total time were different—how would that impact the distance calculation?
3. How does converting time units affect the accuracy of your solution?
4. Can you generalize this problem for any two speeds and any total time?
5. How might Eugenia reduce the time it takes to retrieve her book?

Tip: When dealing with time problems, always ensure you’re working in the same units throughout the problem to avoid confusion.