document.write( "Question 378785: please help me solve this problem: the dimensions of a rectangle are such that its length is 3 inches more than its width. if the length were doubled and if the with decreased by 1 inch the area would be increased by 50 inches^2. what are the length and width of the rectangle? \n" ); document.write( "
Algebra.Com's Answer #269037 by mananth(16946)\"\" \"About 
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the dimensions of a rectangle are such that its length is 3 inches more than its width. if the length were doubled and if the with decreased by 1 inch the area would be increased by 50 inches^2. what are the length and width of the rectangle?
\n" ); document.write( "...
\n" ); document.write( "width be x
\n" ); document.write( "length = x+3
\n" ); document.write( "..
\n" ); document.write( "Area = x(x+3)
\n" ); document.write( "Area = x^2+3x
\n" ); document.write( "...
\n" ); document.write( "length doubled = 2(x+3)
\n" ); document.write( "width x-1
\n" ); document.write( "Area = 2(x+3)*(x-1)
\n" ); document.write( "Area = 2(x^2+2x-3)
\n" ); document.write( "...
\n" ); document.write( "2(x^2+2x-3)-(x^2+3x)=50
\n" ); document.write( "2x^2+4x-6-x^2-3x=50
\n" ); document.write( "x^2+x-6=50
\n" ); document.write( "x^2+x-56=0
\n" ); document.write( "x^2+8x-7x-56=0
\n" ); document.write( "x(x+8)-7(x+8)=0
\n" ); document.write( "(x+8)(x-7)=0
\n" ); document.write( "x=7 ignore -8
\n" ); document.write( "width = 7 inches
\n" ); document.write( "length = x+3 = 10 inches
\n" ); document.write( "...
\n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( "
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