document.write( "Question 484352: How do you solve this? \r
\n" ); document.write( "\n" ); document.write( "The dimensions of a rectangle are such that its length is 3 inches more than its width. If length were doubled and width decreased by 1, area would be increased by 176 inches squared. What are the length and width?
\n" ); document.write( "

Algebra.Com's Answer #331401 by htmentor(1343)\"\" \"About 
You can put this solution on YOUR website!
The dimensions of a rectangle are such that its length is 3 inches more than its width. If length were doubled and width decreased by 1, area would be increased by 176 inches squared. What are the length and width?
\n" ); document.write( "==========================================================================
\n" ); document.write( "Given: l = w + 3 [1]
\n" ); document.write( "The original area = A = lw
\n" ); document.write( "For the new area we can write
\n" ); document.write( "2l*(w-1) = A + 176
\n" ); document.write( "2(w+3)*(w-1) = A + 176
\n" ); document.write( "Substitute the value for l in [1] into the expressions for the two areas:
\n" ); document.write( "A = w(w+3) [2]
\n" ); document.write( "A + 176 = 2(w+3)(w-1) -> A = 2(w+3)(w-1) - 176
\n" ); document.write( "Since the LHS's are equal, we can equate the RHS's:
\n" ); document.write( "w(w+3) = 2(w+3)(w-1) - 176
\n" ); document.write( "Simplify and solve for w:
\n" ); document.write( "w^2 + 3w = 2w^2 + 4w - 6 - 176
\n" ); document.write( "w^2 + w - 182 = 0
\n" ); document.write( "w = (-1 +/- sqrt(1 + 4*182))/2
\n" ); document.write( "w = (-1 +/- 27)/2
\n" ); document.write( "Taking the positive solution, we have w = 13
\n" ); document.write( "Therefore l = 13 + 3 = 16
\n" ); document.write( "Ans: w = 13, l = 16
\n" ); document.write( "
\n" );