document.write( "Question 64023This question is from textbook Algebra Structure and Method Book1
\n" ); document.write( ": A rectangle is three times as long as it is wide.If its length and width are both decreased by 2 cm,its area is decreased by 36 cm squared.Find its original dimensions.Make a sketch as in Oral Exercise 2. \n" ); document.write( "
Algebra.Com's Answer #44768 by Earlsdon(6294)\"\" \"About 
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The area of a rectangle is given by:
\n" ); document.write( "\"A+=+L%2AW\"
\n" ); document.write( "In your rectangle: \"L+=+3W\"
\n" ); document.write( "If you decrease the length, L, and the width, W, each by 2 cm, the area, A, is decreased by 36 sq.cm. So, puting this into an equation, you'll get:
\n" ); document.write( "\"%28L-2%29%28W-2%29+=+A-36\" Substituting 3W for L, and L*W for A, you now have:
\n" ); document.write( "\"%283W-2%29%28W-2%29+=+3W%5E2-36\" Simplifying this, you'll get:
\n" ); document.write( "\"3W%5E2-8W%2B4+=+3W%5E2+-+36\" Further simplification yields:
\n" ); document.write( "\"-8W%2B4+=+-36\" and:
\n" ); document.write( "\"-8W+=+-40\" Dividing by -8:
\n" ); document.write( "\"W+=+5\" and
\n" ); document.write( "\"L+=+3W\" so:
\n" ); document.write( "\"L+=+15\"\r
\n" ); document.write( "\n" ); document.write( "The original length, L = 15 cm
\n" ); document.write( "The original width, W = 5 cm\r
\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "Original area:
\n" ); document.write( "\"A1+=+%2815%29%285%29\"
\n" ); document.write( "\"A1+=+75+cm%5E2\"\r
\n" ); document.write( "\n" ); document.write( "After decreasing the lenth and width by 2 cm:
\n" ); document.write( "\"A2+=+%2813%29%283%29\"
\n" ); document.write( "\"A2+=+39\" sq.cm.\r
\n" ); document.write( "\n" ); document.write( "The decrease in area is:
\n" ); document.write( "\"75-39+=+36\"sq.cm.
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