document.write( "Question 91974: Solve the following problems.show the equation used for the solution.\r
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document.write( "length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 34 in., find the dimensions of the rectangle.\r
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Algebra.Com's Answer #67002 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! Let w=width of the rectangle\r \n" ); document.write( "\n" ); document.write( "Then length(l)=2w+2\r \n" ); document.write( "\n" ); document.write( "Perimeter=2l+2w so our equation to solve is:\r \n" ); document.write( "\n" ); document.write( "34=2(2w+2)+2w get rid of parens\r \n" ); document.write( "\n" ); document.write( "34=4w+4+2w subtract 4 from both sides\r \n" ); document.write( "\n" ); document.write( "34-4=4w+4-4+2w collect like terms\r \n" ); document.write( "\n" ); document.write( "30=6w divide both sides by 6\r \n" ); document.write( "\n" ); document.write( "w=5in-------------------------------width of rectangle\r \n" ); document.write( "\n" ); document.write( "2w+2=2*5+2=12in--------------------length of rectangle\r \n" ); document.write( "\n" ); document.write( "CK\r \n" ); document.write( "\n" ); document.write( "34=2*5+2*12 \n" ); document.write( "34=10+24 \n" ); document.write( "34=34 \n" ); document.write( "also \n" ); document.write( "12=2*5+2 \n" ); document.write( "12=12\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( " |