document.write( "Question 169614: A photo is 3 inches longer that it is wide. A 2-inch border is placed around the photo making the total area of the photo and border 108 square inches. What are the dimensions of the photo? \n" ); document.write( "

Algebra.Com's Answer #125115 by solver91311(24713) $\"\"$ $\"About$

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Let the width of just the picture be $\"x\"$ , then the length must be $\"x%2B3\"$ . If the border is 2 inches wide all the way around, the border must add 4 inches to the overall width and 4 inches to the overall length, so the width with the border must be $\"x%2B4\"$ and the length with the border must be $\"x%2B3%2B4\"$ or $\"x%2B7\"$ . Since area is given by length times width, the area of the picture and the border must be $\"%28x%2B4%29%28x%2B7%29\"$ and we know this to be equal to 108 square inches. So:\r
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\n" ); document.write( "\n" ); document.write( " $\"%28x%2B4%29%28x%2B7%29=108\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E2%2B11x%2B28=108\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E2%2B11x-80=0\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"%28x-5%29%28x%2B16%29=0\"$ \r
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\n" ); document.write( "\n" ); document.write( "Therefore, $\"x\"$ , the original width of the picture alone, must be either 5 or -16. -16 for a width is absurd and is therefore an extraneous root introduced by the act of squaring the variable. -16 can be discarded and the width of the picture is therefore 5 inches. The length is 3 inches longer, or 8 inches. \n" ); document.write( "

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