# How do I solve the following problem? Thank you! A rectangular picture is 9-inches wide and 12-inches long. The picture has a frame of uniform width. If the combined area of picture and frame is...

How do I solve the following problem? Thank you!

A rectangular picture is 9-inches wide and 12-inches long. The picture has a frame of uniform width. If the combined area of picture and frame is 180 in-squared, what is the width of the frame?

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The rectangular picture is 9-inches wide and 12-inches long. It has a frame of uniform width and the combined area of the picture and the frame is 180 in^2.

Let the width of the frame be W. The combined area of the picture and the frame is given by : (9 + 2*W)(12 + 2*W) = 180

=> 108 + 24W + 18W + 4W^2 = 180

=> 4W^2 + 42W - 72 = 0

=> 2W^2 + 21W - 36 = 0

=> 2W^2 + 24W - 3W - 36 = 0

=> 2W(W + 12) - 3(W + 12) = 0

=> (2W - 3)(W + 12) = 0

=> W = 3/2 and W = -12

As the width cannot be negative take the positive root W = 3/2.

**The width of the frame is 3/2 in.**

If w= Width

180 = (2w + 9) * (2w + 12)

180 = 4w^2 + 42w + 108

72 = 4w^2 + 42w

36 = 2w^2 +21w

0 = 2w^2 + 21w - 36

0 = (2w - 3) * (w + 12)

w = 1 1/12 or w = -12

Since it is impossible for any measurement to be a negative number, the width of the frame is 1 1/2 inches.