document.write( "Question 1109207: the length of a rectangle is 2 cm less than twice the width. If the area is 84 square cm, find the dimensions of the rectangle \n" ); document.write( "
Algebra.Com's Answer #724255 by addingup(3677)\"\" \"About 
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L = 2W-2
\n" ); document.write( "L x W = 84
\n" ); document.write( "substitute for L:
\n" ); document.write( "(2W - 2) x W = 84
\n" ); document.write( "2W^2 - 2W = 84
\n" ); document.write( "divide all sides by 2:
\n" ); document.write( "W^2 - W = 42
\n" ); document.write( "W^2 - W - 42 = 0
\n" ); document.write( "42 is divisible by 6 and 7 (6*7 = 42) and 6 - 7 = -1:
\n" ); document.write( "(W^2 + 6W) + (-7W - 42)
\n" ); document.write( "Factor out W from W^2 + 6W and factor out -7 from -7W - 42:
\n" ); document.write( "W(W + 6) - 7(W+6)
\n" ); document.write( "Now we have a common term on both sides, W + 6. Let's factor it out and we get:
\n" ); document.write( "(W+6)(W-7)
\n" ); document.write( "so our answer is:
\n" ); document.write( "W+6 or W-7
\n" ); document.write( "W = -6 or W = 7
\n" ); document.write( "------------------------
\n" ); document.write( "Since we are not looking for a negative number, let's try the 7:
\n" ); document.write( "L = 2(7) - 2 = 12
\n" ); document.write( "So, 7 is our width and 12 our length.
\n" ); document.write( "Area = L x W = 12 x 7 = 84 Correct \n" ); document.write( "
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