document.write( "Question 50301This question is from textbook Beginning Algebra
\n" ); document.write( ": The length of a rectangle is 2 inches more than twice its width. If the perimeter of the rectangle is 34 inches , find the dimensions of the rectangle.\r
\n" ); document.write( "\n" ); document.write( "Please help solve.\r
\n" ); document.write( "\n" ); document.write( "This is what i got. I do not think I did it correctly.\r
\n" ); document.write( "\n" ); document.write( "2X+2(2X+2)=34 OR 2X+4X+4=34 OR 6X=30 OR X=30/6 OR X=5
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Algebra.Com's Answer #33566 by tutorcecilia(2152)\"\" \"About 
You can put this solution on YOUR website!
You are correct!! I came up with the same answer:\r
\n" ); document.write( "\n" ); document.write( "P=2(length+width)
\n" ); document.write( "34=2[(2w+2)+w]
\n" ); document.write( "34=2[(2w+2)+w]
\n" ); document.write( "34=2[2w+2+w]
\n" ); document.write( "34=2[3w+2]
\n" ); document.write( "17=3w+2
\n" ); document.write( "17-2=3w+2-2
\n" ); document.write( "15/3=3w/3
\n" ); document.write( "5=w
\n" ); document.write( ".
\n" ); document.write( "Check by plugging (w=5) back into the original equation]
\n" ); document.write( "34=2[(2(5)+2)+(5)]
\n" ); document.write( "34=2[(2(5)+2)+(5)]
\n" ); document.write( "34=34 [Checks out]
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