Solution

The area of the picture is the area of the inner rectangle, and the area of the frame is the area of the outer rectangle minus the area of the inner rectangle.
Since the area of a rectangle is the length times the width, you need to know the length and width of the inner and outer rectangles.

In the figure, you are given that the length of the inner rectangle is x inches, but the width is not given. However, since you know that the width of the frame is 1 inch, it follows that the width of the inner rectangle is equal to the width of the outer rectangle minus 2 inches, or x−2 inches. Thus, the area of the inner rectangle is x(x−2) square inches. In the figure, you are given that the width of the outer rectangle is x inches, but the length is not given. However, since you know that the width of the frame is 1 inch, it follows that the length of the outer rectangle is equal to the length of the inner rectangle plus 2 inches, or x+2 inches. Thus, the area of the outer rectangle is x(x+2) square inches. Since the area of the frame is the area of the outer rectangle minus the area of the inner rectangle, the area of the frame is \(x(x+2) - x(x-2)=x^2+2x-x^2+2x=4x\) square inches. Now you are ready to set up the equation. Set the expression for the area of the picture equal to the expression for the area of the frame and solve the resulting equation for x, as follows.

\(x(x-2)=4x\)

\(x^2-2x=4x\)

\(x^2-6x=0\)

\(x(x-6)=0\)

There are two solutions to the equation, x=0 and x=6. Since x represents the length of a picture, in inches, the solution \(x=0\) does not make sense in this context. Therefore, when \(x=6\), the area of the picture equals the area of the frame.

The correct answer is \(C.\)

Carcass' solution is perfect.
However, if you didn't come up with that approach, another approach is to check each answer choice to see which one yields the correct dimensions.
It's a tedious approach, but it works.

Cheers,
Brent
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We see that the area of the picture (without the frame) is x(x - 2) and the total area of the picture and the frame is x(x + 2). Since the area of the picture is equal to the area of the frame, the total area of the picture and the frame is thus twice the area of the picture. That is, we have:

x(x + 2) = 2x(x - 2)

x^2 + 2x = 2x^2 - 4x

x^2 - 6x = 0

x(x - 6) = 0

x = 0 or x = 6

Since x can’t be 0, x must be 6.

Answer: C
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