document.write( "Question 965121: A photograph is 12 cm longer than it is wide. It is mounted in a frame 5 cm wide. The area of the frame is 620 cm squared. What is the dimensions of the photo
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Algebra.Com's Answer #589874 by josgarithmetic(39617)\"\" \"About 
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x and y, photograph dimensions.
\n" ); document.write( "a, the area of just the frame, alone\r
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\n" ); document.write( "\n" ); document.write( "\"y=12%2Bx\", assuming y is length.
\n" ); document.write( "Let u be the uniform width of the frame.\r
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\n" ); document.write( "\n" ); document.write( "Total area frame plus photograph,\"%28x%2B2u%29%28y%2B2u%29\"
\n" ); document.write( "area of just the photograph,\"xy\"
\n" ); document.write( "area of just the frame,\"%28x%2B2u%29%28y%2B2u%29-xy\"\r
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\n" ); document.write( "\n" ); document.write( "According to description and assigning,
\n" ); document.write( "\"%28x%2B2u%29%28y%2B2u%29-xy=a\"\r
\n" ); document.write( "\n" ); document.write( "\"xy%2B2yu%2B2xu%2B4u%5E2-xy-a=0\"
\n" ); document.write( "\"2yu%2B2xu%2B4x%5E2-a=0\"\r
\n" ); document.write( "\n" ); document.write( "and through substituting according to description,
\n" ); document.write( "\"2%2812%2Bx%29u%2B2xu%2B4x%5E2-a=0\"
\n" ); document.write( "\"24u%2B2ux%2B2ux%2B4x%5E2-a=0\"
\n" ); document.write( "\"4x%5E2%2B4ux%2B24u-a=0\"
\n" ); document.write( "\"highlight_green%284x%5E2%2B4ux%2B%2824u-a%29=0%29\"\r
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\n" ); document.write( "\n" ); document.write( "Now would be a good step to substitute the given values, compute constant and coefficients and then finish solving for x.
\n" ); document.write( "Use as described:
\n" ); document.write( "x, unknown photograph width
\n" ); document.write( "u=5
\n" ); document.write( "a=620
\n" ); document.write( "y=12+x
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