document.write( "Question 443324: I can't figure this out: The dimensions of a rectangle are the length is 3 in more than its width, if the length were doubled and the width decreased by 1 in., the area would be increased by 126^2, what are the length and width of the rectangle? Please HELP! \n" ); document.write( "
Algebra.Com's Answer #305793 by swincher4391(1107)![]() ![]() You can put this solution on YOUR website! You say the area would be increased by 126^2, do you mean 126in^2? \n" ); document.write( "======================== \n" ); document.write( "You will have a system of equations: \n" ); document.write( "A = lw \n" ); document.write( "[l = w+3] \n" ); document.write( "A = (w+3)*w \n" ); document.write( "=============== \n" ); document.write( "2l * w-1 = A + 126 \n" ); document.write( "2(w+3) * w-1 = A + 126 \n" ); document.write( "(2w+6)*(w-1) = A +126 \n" ); document.write( "2w^2 +6w -2w -6 = A +126 \n" ); document.write( "======================== \n" ); document.write( "2w^2 +4w -132 = A \n" ); document.write( "===================== \n" ); document.write( "2w^2 + 4w -132 = w(w+3) \n" ); document.write( "2w^2 + 4w + -132= w^2 +3w \n" ); document.write( "w^2 + w + -132 = 0 \n" ); document.write( "(w-11)(w+12) \n" ); document.write( "================ \n" ); document.write( "w = 11 or w =-12\r \n" ); document.write( "\n" ); document.write( "-12 inches doesn't make any sense, so 11 inches is the only one that makes sense.\r \n" ); document.write( "\n" ); document.write( "So the width is 11 inches. \n" ); document.write( "======================== \n" ); document.write( "Plug this back in for \n" ); document.write( "l = w+3 \n" ); document.write( "l = 11 +3 \n" ); document.write( "l = 14 \n" ); document.write( "======================== \n" ); document.write( "The dimensions are 14 x 11 or \n" ); document.write( " \n" ); document.write( " |