document.write( "Question 953981: The length of a rectangle is 3 less than twice its width. The area of the rectangle is 35 cm^2. Find the dimensions of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #582632 by addingup(3677)\"\" \"About 
You can put this solution on YOUR website!
Let's call the length L and the width W.
\n" ); document.write( "The area is:
\n" ); document.write( "W*L = 35, and the problem says:
\n" ); document.write( "L = 2W-3 Let's plug this value for L into the first equation:
\n" ); document.write( "W(2W-3)= 35 Multiply on left:
\n" ); document.write( "2W^2-3W= 35 Subtract 35 from both sides:
\n" ); document.write( "2W^2-3W-35= 0 Now we have quadratic equation. Let's factor it:
\n" ); document.write( "We have a -3W in the middle. And we know we need two numbers, let's call them a and b, such that a+b = -3W And those same two numbers must: a*b = -35:
\n" ); document.write( "(W-5)(2W+7) = 0 Solve separately:
\n" ); document.write( "W-5 = 0 or 2W+7 = 0 On the first equation, add 5 to both sides. On the 2nd subtract 7 to both sides:
\n" ); document.write( "W= 5 or 2W= -7 On the second equation divide both sides by 2:
\n" ); document.write( "W= 5 or W= -7/2= -3.5
\n" ); document.write( "Let's discard the 2nd equation because we need a positive number. Let's take 5 and see. We said the area is:
\n" ); document.write( "W(2W-3)= 35
\n" ); document.write( "5(2*5-3)= 35 Solve to remove parenthesis:
\n" ); document.write( "5*7 = 35 We have the correct answer, the W=5 and the L= 7
\n" ); document.write( "
\n" );