document.write( "Question 164311This question is from textbook
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document.write( ": The width of a rectangle is 22 centimeters shorter that its length. The perimiter of the rectangular is 160. find its length. \n" );
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Algebra.Com's Answer #121053 by algebrapro18(249)![]() ![]() ![]() You can put this solution on YOUR website! The width of a rectangle is 22 centimeters shorter that its length. The perimiter of the rectangular is 160. find its length.\r \n" ); document.write( "\n" ); document.write( "let w be the width and L be the length(both in centemeters).\r \n" ); document.write( "\n" ); document.write( "The width of a rectangle is 22 centimeters shorter that its length. \n" ); document.write( "converting this into an equation you get \n" ); document.write( "w = L - 22\r \n" ); document.write( "\n" ); document.write( "The perimiter of the rectangular is 160. find its length. \n" ); document.write( "converting this into an equation you get \n" ); document.write( "2L + 2w(equation of the perimeter of a rectangle) = 160\r \n" ); document.write( "\n" ); document.write( "now we substitute the first equation into the second one and solve for the length.\r \n" ); document.write( "\n" ); document.write( "2L + 2w = 160 \n" ); document.write( "2L + 2(L - 22) = 160 \n" ); document.write( "2L + 2L -44 = 160 \n" ); document.write( "4L -44 = 160 \n" ); document.write( "4L = 204 \n" ); document.write( "L = 51 \n" ); document.write( " \n" ); document.write( " |