document.write( "Question 149942This question is from textbook
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document.write( ": Suppose that the width of a rectangle is 2 inches shorter than the length and that the perimeter of the rectangle is 80.
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document.write( "a) Set up an equation for the perimeter involving only L, the length of the rectangle.
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document.write( "b) Solve this equation algebraically to find the length of the rectangle. Find the width as well. \n" );
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Algebra.Com's Answer #110046 by nerdybill(7384)![]() ![]() You can put this solution on YOUR website! Suppose that the width of a rectangle is 2 inches shorter than the length and that the perimeter of the rectangle is 80. \n" ); document.write( "a) Set up an equation for the perimeter involving only L, the length of the rectangle. \n" ); document.write( "b) Solve this equation algebraically to find the length of the rectangle. Find the width as well. \n" ); document.write( ". \n" ); document.write( "Let L = length of rectangle \n" ); document.write( "L-2 = width of rectangle \n" ); document.write( ". \n" ); document.write( "Since the perimeter is twice the \"length + width\" we have: \n" ); document.write( "2(L + L - 2) = 80 (solution for a) \n" ); document.write( ". \n" ); document.write( "solving for L: \n" ); document.write( "2(L + L - 2) = 80 \n" ); document.write( "(2L - 2) = 40 \n" ); document.write( "2(L - 1) = 40 \n" ); document.write( "L - 1 = 20 \n" ); document.write( "L = 21 inches (solution for b -- length of rectangle) \n" ); document.write( "L-2 = 19 inches (solution for b -- width of rectangle) \n" ); document.write( " |