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The problem involves a 5 inch by 7 inch photograph placed inside a picture frame, where both the length and width of the frame are \(2x\) inches larger than the photograph’s dimensions. We are asked to find an expression representing the perimeter of the frame.

### Step-by-step breakdown:

1. **Dimensions of the photograph:**
   - Width of the photograph: 5 inches
   - Length of the photograph: 7 inches

2. **Dimensions of the frame:**
   - Since the frame's width and length are \(2x\) inches larger than the photograph, the new dimensions of the frame are:
     - Width of the frame: \(5 + 2x\) inches
     - Length of the frame: \(7 + 2x\) inches

3. **Perimeter of the frame:**
   The formula for the perimeter \(P\) of a rectangle is:
   \[
   P = 2 \times (\text{Length} + \text{Width})
   \]
   Substituting the values:
   \[
   P = 2 \times \left((7 + 2x) + (5 + 2x)\right)
   \]
   Simplify the expression inside the parentheses:
   \[
   P = 2 \times \left(12 + 4x\right)
   \]
   Distribute the 2:
   \[
   P = 24 + 8x
   \]

So, the expression for the perimeter of the frame is \(8x + 24\).

### The correct answer is:
\[
\boxed{8x + 24}
\]

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Do you want further details or have any questions?

Here are 5 related questions to expand on this topic:
1. How would the perimeter change if the frame’s dimensions increased by \(3x\) instead of \(2x\)?
2. Can you express the area of the frame in terms of \(x\)?
3. What would the perimeter be if \(x = 3\)?
4. How would you calculate the perimeter if the photo were circular instead of rectangular?
5. How do you factor the expression \(8x + 24\) to make it more simplified?

**Tip:** When dealing with perimeter problems, always remember that you are adding twice the length and twice the width!

A 5 inch × 7 inch photograph is placed inside a picture frame. Both the length and width of the frame are 2x inches larger than the width and length of the photograph. Which expression represents the perimeter of the frame?

Learn how to find the perimeter of a picture frame when the dimensions of a 5 inch × 7 inch photograph are increased by 2x inches. This problem covers key algebraic and geometric concepts, including perimeter calculation and algebraic expressions. Suitable for students in Grades 6-8.

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