document.write( "Question 108418: A rectangle is twice as long as it is wide. If its length and width are both decreased by 4 cm, its area is decreased by 164 cm2. Find its original dimensions. \n" ); document.write( "
Algebra.Com's Answer #79084 by ankor@dixie-net.com(22740)\"\" \"About 
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Let x = the original width
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\n" ); document.write( "\"A rectangle is twice as long as it is wide.\"
\n" ); document.write( "2x = original length
\n" ); document.write( "Then
\n" ); document.write( "2x^2 = original area
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\n" ); document.write( "If its length and width are both decreased by 4 cm, its area is decreased by 164 cm2.
\n" ); document.write( "(2x - 4) by (x - 4) = new dimensions
\n" ); document.write( "(2x^2 - 12x + 16)= new area
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\n" ); document.write( "Original area - New area = 164 sq/cm
\n" ); document.write( "2x^2 - (2x^2 - 12x + 16) = 164
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\n" ); document.write( "2x^2 - 2x^2 + 12x - 16 = 164 removing brackets, changes the signs
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\n" ); document.write( "12x = 164 + 16; conveniently, the 2x^2 are eliminated, add 16 to both sides
\n" ); document.write( "12x = 180
\n" ); document.write( "12x = 180/12
\n" ); document.write( "x = 15 cm is the original width
\n" ); document.write( "Then
\n" ); document.write( "2(15) = 30 cm is the original length
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\n" ); document.write( "Check by subtracting the new area from the original area
\n" ); document.write( "(30*15) - (26*11) =
\n" ); document.write( " 450 - 286 = 164; confirms our solution\r
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