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The problem asks for the width of a uniform frame that surrounds a printed picture, such that the total area of the picture and the frame is 15 square feet. The picture itself is 3.25 feet tall and 3 feet wide.

Let’s define the width of the frame as \( x \) feet. This width is added to both sides of the picture’s height and width, so the overall dimensions of the picture with the frame will be:

- Height of the framed picture = \( 3.25 + 2x \)
- Width of the framed picture = \( 3 + 2x \)

The total area of the framed picture is given as 15 square feet. Therefore, we can set up the equation:

\[
(3.25 + 2x)(3 + 2x) = 15
\]

Now, I’ll solve this equation to find the value of \( x \).

Let me calculate this.The solution to the equation gives two possible values for the frame width \( x \): -3.5 and 0.375 feet. Since the width of the frame cannot be negative, the valid solution is:

\[
x = 0.375 \text{ feet}
\]

So, the width of the frame should be 0.375 feet to use all of the available wall space.

Would you like any further details or clarifications?

Here are 5 related questions you might consider:
1. How would the solution change if the available wall space was smaller, say 12 square feet?
2. What would be the total perimeter of the framed picture with the current frame width?
3. How would the frame width change if the dimensions of the picture were altered to 4 feet by 2.5 feet?
4. What would be the frame width if only 10 square feet of wall space were available instead of 15 square feet?
5. How can we calculate the area of just the frame itself?

**Tip:** For problems involving geometric shapes, setting up equations based on area or perimeter is a useful way to find unknown dimensions.

You are making a frame of uniform width for a selfie that you took in math class. The picture you printed is 3.25 feet tall and 3 feet wide. You have a wall space of 15 ft². What should the width of the frame be in order to use all of the space?

Solve how to calculate the uniform width of a frame around a selfie, where the picture dimensions are 3.25 feet tall and 3 feet wide, with 15 square feet of wall space. This involves using algebra, quadratic equations, and area calculations. Suitable for Grades 9-10.

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