document.write( "Question 941930: The perimeter of a rectangle is 22 inches. The length of the rectangle is 10 inches more than the width of the rectangle. Find the length and width. \r
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document.write( "I know that the perimeter of a rectangle is P=2l+2w, so 22=2l+2w. I think you have to add the 10 multiplied by the 2l, but I'm not sure. \n" );
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Algebra.Com's Answer #574184 by macston(5194)![]() ![]() You can put this solution on YOUR website! width=W length=L=W+10 \n" ); document.write( "P=2L+2W=2(L+W) \n" ); document.write( "22=2(L+W) divide both sides by 2 \n" ); document.write( "11=L+W Substitute for length \n" ); document.write( "11=W+10+W \n" ); document.write( "11=2W+10 Subtract 10 from each side \n" ); document.write( "11-10=2W+10-10 \n" ); document.write( "1=2W divide both sides by 2 \n" ); document.write( "1/2=W \n" ); document.write( "L=W+10=10 1/2 \n" ); document.write( "CHECK: \n" ); document.write( "P=2(L+W) \n" ); document.write( "22=2(10 1/2 + 1/2) \n" ); document.write( "22=2(11) \n" ); document.write( "22=22 \n" ); document.write( " |