The original photo has dimensions of 5 inches wide and 8 inches long, and we need to determine the new length when the width is changed to 10 inches, 15 inches, and 8 inches.

This is a problem of proportional enlargement. The original ratio of width to length is: $\text{Original Ratio} = \frac{5}{8}$

For the new photos, we maintain the same width-to-length ratio.

1. When the width is 10 inches:

We set up the proportion: $\frac{5}{8} = \frac{10}{L}$ Solving for $L$ (the new length): $L = \frac{10 \times 8}{5} = 16 \text{ inches}$

2. When the width is 15 inches:

$\frac{5}{8} = \frac{15}{L}$ Solving for $L$: $L = \frac{15 \times 8}{5} = 24 \text{ inches}$

3. When the width is 8 inches (the same as the original):

$\frac{5}{8} = \frac{8}{L}$ Solving for $L$: $L = \frac{8 \times 8}{5} = 12.8 \text{ inches}$

Final Answers:

• The new photo is 16 inches long if it is 10 inches wide.
• The new photo is 24 inches long if it is 15 inches wide.
• The new photo is 12.8 inches long if it is 8 inches wide.

Would you like any additional details or clarifications on this problem?

Here are 5 related questions to expand on this concept:

1. How would you calculate the enlargement for different aspect ratios?
2. What happens to the photo if the aspect ratio is not preserved during enlargement?
3. Can you find the width if the length is known instead?
4. How does the enlargement affect the area of the photo?
5. How do you apply this proportional method to three-dimensional objects?

Tip: Always make sure to maintain the same ratio between dimensions when enlarging images to avoid distortion.