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)\"\" \"About 
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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
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\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
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\n" ); document.write( "difference: 204, confirms our solutions\r
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