document.write( "Question 82261: The length of a rectangle is 3in more than twice its width. If the perimeter of the rectangle is 36in, find the width of the rectangle.
\n" ); document.write( "(a)7in
\n" ); document.write( "(b)6in
\n" ); document.write( "(c)4in
\n" ); document.write( "(d)5in \n" ); document.write( "
Algebra.Com's Answer #58928 by tutorcecilia(2152)\"\" \"About 
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Use the formula for the perimeter of a rectangle:
\n" ); document.write( "P=2(length+width)
\n" ); document.write( "P=36
\n" ); document.write( "Length = 3in more than twice its width = 2 x Width +3 = 2w+3
\n" ); document.write( "Width=w
\n" ); document.write( ".
\n" ); document.write( "P=2(l+w) [plug-in the values and solve for the width(w)]
\n" ); document.write( "36=2[(2w+3)+w]
\n" ); document.write( "36/2=2/2(2x+3+w)
\n" ); document.write( "18=2w+3+w
\n" ); document.write( "18=3w+3
\n" ); document.write( "18-3=3w
\n" ); document.write( "15=3w
\n" ); document.write( "15/3=3w/3
\n" ); document.write( "5=w
\n" ); document.write( ".
\n" ); document.write( "checking:
\n" ); document.write( "Length = 2w+3=2(5)+3=13
\n" ); document.write( "Width=5
\n" ); document.write( ".
\n" ); document.write( "36=2(13+5)
\n" ); document.write( "36=2(18)
\n" ); document.write( "36=36 [checks out]\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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