document.write( "Question 163842: The length of a rectangle is 7 units more than its width. If the width is doubled and the length is increased by 2, the area is increased by 42 square units. Use an algebraic solution to find the dimensions of the original rectangle. \n" ); document.write( "
Algebra.Com's Answer #120706 by nerdybill(7384)\"\" \"About 
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Let x = width of original rectangle
\n" ); document.write( "then
\n" ); document.write( "x+7 = length of original rectangle
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\n" ); document.write( "Then this:
\n" ); document.write( "\"If the width is doubled and the length is increased by 2, the area is increased by 42 square units.\" is translated to:
\n" ); document.write( "(2x)(x+7+2) = x(x+7) + 42
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\n" ); document.write( "Now we can solve for x:
\n" ); document.write( "(2x)(x+7+2) = x(x+7) + 42
\n" ); document.write( "(2x)(x+9) = x(x+7) + 42
\n" ); document.write( "2x^2 + 18x = x^2 +7x + 42
\n" ); document.write( "x^2 + 11x - 42 = 0
\n" ); document.write( "Factoring the left:
\n" ); document.write( "(x+14)(x-3) = 0
\n" ); document.write( "So,
\n" ); document.write( "x = {3, -14}
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\n" ); document.write( "We can toss the negative solution leaving us with:
\n" ); document.write( "x = 3 units (width of original rectangle)
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\n" ); document.write( "length:
\n" ); document.write( "x+7 = 3+7 = 10 units (length of original rectangle)
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\n" ); document.write( "ANS: dimension of original rectangle: 3 units by 10 units
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