document.write( "Question 861703: a photo is cut so that its length is 2 cm longer than its width. it is then centered on a background to form a 3 cm border around the photo. the area of the background paper is 168. what are the dimensions of the photo \n" ); document.write( "
Algebra.Com's Answer #519213 by mananth(16946)\"\" \"About 
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let width of picture be x
\n" ); document.write( "length = x+2\r
\n" ); document.write( "\n" ); document.write( "3 cm border on all sides\r
\n" ); document.write( "\n" ); document.write( "width including the border = x+6
\n" ); document.write( "length including border = x+2+6=x+8\r
\n" ); document.write( "\n" ); document.write( "Area of background paper = 168 cm^2\r
\n" ); document.write( "\n" ); document.write( "Area = L*W\r
\n" ); document.write( "\n" ); document.write( "(x+6)(x+8)=168\r
\n" ); document.write( "\n" ); document.write( "x(x+8)+6(x+8)=168\r
\n" ); document.write( "\n" ); document.write( "x^2+8x+6x+48=168\r
\n" ); document.write( "\n" ); document.write( "x^2+14x+48=168\r
\n" ); document.write( "\n" ); document.write( "x^2+!4x-120=0\r
\n" ); document.write( "\n" ); document.write( "x^2+20x-6x-120=0\r
\n" ); document.write( "\n" ); document.write( "x(x+20)-6(x+20)=0\r
\n" ); document.write( "\n" ); document.write( "(x+20)(x-6)=0
\n" ); document.write( "x cannot be negative\r
\n" ); document.write( "\n" ); document.write( "x= 6 , width\r
\n" ); document.write( "\n" ); document.write( "length = x+2= 8\r
\n" ); document.write( "\n" ); document.write( "6 by 8 cm\r
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