document.write( "Question 176755: the perimeter of a rectangle is 44 inches.find the dimensions if the width is two inches longer than three times the length.then find the area of the rectangle. \n" ); document.write( "

Algebra.Com's Answer #131895 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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the perimeter of a rectangle is 44 inches.find the dimensions if the width is two inches longer than three times the length.then find the area of the rectangle
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\n" ); document.write( "2L + 2W = 44
\n" ); document.write( "simplify, divide by 2
\n" ); document.write( "L + W = 22
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\n" ); document.write( "It says; the width is two inches longer than three times the length.\"
\n" ); document.write( "The equation for this statement is:
\n" ); document.write( "W = 3L + 2
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\n" ); document.write( "substitute (3L+2) for W
\n" ); document.write( "L + (3L+2) = 22
\n" ); document.write( "4L = 22 - 2
\n" ); document.write( "4L = 20
\n" ); document.write( "L = $\"20%2F4\"$
\n" ); document.write( "L = 5 in
\n" ); document.write( ":
\n" ); document.write( "W = 3L + 2
\n" ); document.write( "W = 3(5) + 2
\n" ); document.write( "W = 17 in
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\n" ); document.write( "find the area: 17 * 5 = 85 sq/in
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\n" ); document.write( "check solutions by finding the perimeter
\n" ); document.write( "2(5) + 2(17) = 44
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