document.write( "Question 682017: The dimensions of a rectangle are such that it's length is 3 in. more than its width. If the length were doubled and if the width were decreased by 1 in. the area would be increased by 204 in squared. What are the length and width of the rectangle? \n" ); document.write( "
Algebra.Com's Answer #423007 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! The dimensions of a rectangle are such that it's length is 3 in. more than its width. \n" ); document.write( "L = W+3 \n" ); document.write( ": \n" ); document.write( "If the length were doubled and if the width were decreased by 1 in. the area would be increased by 204 in squared. \n" ); document.write( "2L(W-1) - LW = 204 \n" ); document.write( "2LW - 2L - LW = 204 \n" ); document.write( "2LW - LW - 2L = 204 \n" ); document.write( "LW - 2L = 204 \n" ); document.write( "Factor out L \n" ); document.write( "L(W-2) = 204 \n" ); document.write( "Replace L with (W+3) \n" ); document.write( "(W+3)(W-2) = 204 \n" ); document.write( "FOIL \n" ); document.write( "W^2 - 2W + 3W - 6 = 204 \n" ); document.write( "A quadratic equation \n" ); document.write( "W^2 + W - 210 = 0 \n" ); document.write( "Factors to \n" ); document.write( "(W-14)(W+15) = 0 \n" ); document.write( "The positive solution \n" ); document.write( "W = 14 inches is the original width \n" ); document.write( "then \n" ); document.write( "14 + 3 = 17 inches is the original length \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check this by finding the area of each rectangle: \n" ); document.write( "2(17) * (14-1) = 442 \n" ); document.write( "17 * 14 = 238 \n" ); document.write( "--------------------- \n" ); document.write( "difference: 204, confirms our solutions\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |