document.write( "Question 247671: The perimeter of a rectangle is 106 mm. The length is one more than three times the width. What is the area? \n" ); document.write( "

Algebra.Com's Answer #180601 by oberobic(2304) $\"\"$ $\"About$

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P = 2L + 2W
\n" ); document.write( "L = 3W + 1 (length is 1 more than 3*width)
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\n" ); document.write( "substituting
\n" ); document.write( "P = 2(3W+1) + 2W
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\n" ); document.write( "P = 106 (given)
\n" ); document.write( ".
\n" ); document.write( "substituting for P
\n" ); document.write( "2(3W+1) + 2W = 106
\n" ); document.write( ".
\n" ); document.write( "multiply thru
\n" ); document.write( "6W + 2 + 2W = 106
\n" ); document.write( ".
\n" ); document.write( "collect terms and subtract 2 from both sides
\n" ); document.write( "8W = 104
\n" ); document.write( ".
\n" ); document.write( "divide both sides by 8
\n" ); document.write( "W = 13
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\n" ); document.write( "substitute back into the formula to solve for L
\n" ); document.write( "L = 3W+1 = 3(13)+1 = 40
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\n" ); document.write( "Checking to see if P = 2L + 2W
\n" ); document.write( "2L = 80
\n" ); document.write( "2W = 26
\n" ); document.write( "P = 106
\n" ); document.write( "OK
\n" ); document.write( ".
\n" ); document.write( "Check to see if L = 3W + 1
\n" ); document.write( "3W = 39
\n" ); document.write( "+ 1 = 40
\n" ); document.write( "OK \n" ); document.write( "

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