document.write( "Question 1065152: A new community center is being built in Safe Harbor. The perimeter of the rectangular playing field is 444 yards. The length of the field is 2 yards less than triple the width. What are the dimensions of the playing field? \n" ); document.write( "
Algebra.Com's Answer #680277 by JulietG(1812)\"\" \"About 
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Remember that a perimeter is like a fence around. To fence an area, you need twice the width and twice the length.
\n" ); document.write( "P = 2W + 2L
\n" ); document.write( "What do we know from the question?
\n" ); document.write( "P = 444
\n" ); document.write( "L = 3W - 2
\n" ); document.write( "Plug those values into the first equation.
\n" ); document.write( "444 = 2W + 2(3W-2)
\n" ); document.write( "444 = 2W + 6W - 4
\n" ); document.write( "448 = 8W
\n" ); document.write( "Divide
\n" ); document.write( "56 = W
\n" ); document.write( "L = 3W - 2
\n" ); document.write( "L = 3(56) - 2
\n" ); document.write( "L = 168-2
\n" ); document.write( "L = 166
\n" ); document.write( "Let make sure that's correct by putting the values back into the original equation.
\n" ); document.write( "P = 2W + 2L
\n" ); document.write( "444 = 2(56) + 2(166)
\n" ); document.write( "444 = 112 + 332
\n" ); document.write( "444 = 444
\n" ); document.write( "Success!
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