document.write( "Question 484352: How do you solve this? \r
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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" );
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Algebra.Com's Answer #331401 by htmentor(1343)![]() ![]() 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( " |