document.write( "Question 964145: Suppose the length of a certain rectangle is 7 meters more than twice its width and the perimeter is 50 meters.
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document.write( "The width must be ??? meters. \n" );
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Algebra.Com's Answer #589040 by addingup(3677)![]() ![]() You can put this solution on YOUR website! 2L+2W= 50 and the problem says that L= 2W+7, so we substitute for L: \n" ); document.write( "2(2W+7)+2W= 50 Multiply on left: \n" ); document.write( "4W+14+2W= 50 Subtract 14 on both sides and add W on left: \n" ); document.write( "6W= 36 Divide both sides by 6: \n" ); document.write( "W= 6 This is the width \n" ); document.write( "And the length? The problem doesn't ask for it, but it doesn't hurt for you to have it. It's 6*2+7= 19 \n" ); document.write( " |