document.write( "Question 1036139: The area of a rectangle is 91 square centimeters. Its length is x+2 and its width is x-4. Find the perimeter of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #650748 by addingup(3677)![]() ![]() You can put this solution on YOUR website! L*W = 91 \n" ); document.write( "L = x+2 \n" ); document.write( "W = x-4 \n" ); document.write( "(x+2)*(x-4) = 91 \n" ); document.write( "x^2-2x-8 = 91 \n" ); document.write( "x^2-2x = 99 \n" ); document.write( "x^2-2x+1 = 100 \n" ); document.write( "(x-1)^2 = 100 \n" ); document.write( "x-1 = 10 or x-1 = -10 \n" ); document.write( "x = 11 or x = -9 \n" ); document.write( "Toss out the -9, we can't use it, and you have: \n" ); document.write( "x = 11 \n" ); document.write( "The length is 11+2 = 13 \n" ); document.write( "The width is 11-4 = 7 \n" ); document.write( "The perimeter is: \n" ); document.write( "2(13)+2(7)= 40 \n" ); document.write( "````````````````````````````````````````` \n" ); document.write( "Check: \n" ); document.write( "13+7 = 91 Correct \n" ); document.write( " |