document.write( "Question 528814: the area of the rectangle shown is 40in^2. find the length and width. Length of one side is x+7, and the width of the other side is 2x+3. Thank you for taking the time to read this. \n" ); document.write( "
Algebra.Com's Answer #349418 by nerdybill(7384)\"\" \"About 
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the area of the rectangle shown is 40in^2. find the length and width. Length of one side is x+7, and the width of the other side is 2x+3.
\n" ); document.write( "(x+7)(2x+3) = 40
\n" ); document.write( "FOIL the left to get:
\n" ); document.write( "2x^2+3x+14x+21 = 40
\n" ); document.write( "2x^2+17x+21 = 40
\n" ); document.write( "2x^2+17x-19 = 0
\n" ); document.write( "2x^2-2x+19x-19 = 0
\n" ); document.write( "(2x^2-2x)+(19x-19) = 0
\n" ); document.write( "2x(x-1)+19(x-1) = 0
\n" ); document.write( "(x-1)(2x+19) = 0
\n" ); document.write( "x = {-19, 1}
\n" ); document.write( "throw out the negative solution (extraneous) leaving:
\n" ); document.write( "x = 1
\n" ); document.write( ".
\n" ); document.write( "Length:
\n" ); document.write( "x+7 = 1+7 = 8 inches
\n" ); document.write( ".
\n" ); document.write( "Width:
\n" ); document.write( "2x+3 = 2(1)+3 = 2+3 = 5 inches \n" ); document.write( "
\n" );