document.write( "Question 834441: The length of a rectangle is 4yd longer than its width. If the perineter of the rectangle is 48 yd, find its area. \n" ); document.write( "

Algebra.Com's Answer #502996 by JulietG(1812) $\"\"$ $\"About$

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P = 2L + 2W (the perimeter fence takes 2 pieces of length and 2 pieces of width)
\n" ); document.write( "P = 48
\n" ); document.write( "L = W + 4
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\n" ); document.write( "Now substitute the value of L into the original equation
\n" ); document.write( "48 = 2(W+4) + 2W
\n" ); document.write( "Distribute the 2
\n" ); document.write( "48 = 2W + 8 + 2W
\n" ); document.write( "Subtract 8 from each side
\n" ); document.write( "40 = 2W + 2W
\n" ); document.write( "Add the Ws
\n" ); document.write( "40 = 4W
\n" ); document.write( "Divide each side by 4
\n" ); document.write( "10 = W
\n" ); document.write( ".
\n" ); document.write( "If the width is 10, then the length is 4yd longer, or 14.
\n" ); document.write( "Let's add them up.
\n" ); document.write( "2W (20) + 2L (28) does indeed = 48.
\n" ); document.write( ".
\n" ); document.write( "Now that you have the W & L, multiply them to get the area.
\n" ); document.write( "10 * 14 = ?? \n" ); document.write( "

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