document.write( "Question 1004773: A rectangle is twice as long as it is wide. If the length is increased by 4 cm and its width is decreased by 3 cm, the new rectangle formed has an area of 100 sq cm. Find the dimensions of the original rectangle. \n" ); document.write( "
Algebra.Com's Answer #621159 by addingup(3677)\"\" \"About 
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L= 2W
\n" ); document.write( "(2W+4)(W-3)= 100
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\n" ); document.write( "Solve for W:
\n" ); document.write( "(W-3) (2 W+4) = 100
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\n" ); document.write( "Expand out terms of the left hand side:
\n" ); document.write( "2W^2-6W+4W-12 = 100 Add subtract on left:
\n" ); document.write( "2W^2-2W-12= 100
\n" ); document.write( "Divide both sides by 2:
\n" ); document.write( "W^2-W-6 = 50 Fa
\n" ); document.write( "Add 6 to both sides:
\n" ); document.write( "W^2-W = 56
\n" ); document.write( "Subtract 56 on both sides:
\n" ); document.write( "W^2-W-56= 0 Factor the equation: 7*8= 56 and 8-7= 1
\n" ); document.write( "(W-8)(W+7)= 0
\n" ); document.write( "Split into 2 equations:
\n" ); document.write( "W-8= 0 or W+7= 0
\n" ); document.write( "W= 8 or W= -7 Toss the -7, we are not looking for a negative number.
\n" ); document.write( "Our width is 8. Proof:
\n" ); document.write( "(2W+4)(W-3)= 100
\n" ); document.write( "((2*8)+4)(8-3)= 100
\n" ); document.write( "20*5= 100
\n" ); document.write( "100=100 We've got the right answer \n" ); document.write( "
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