document.write( "Question 91865: The perimeter of a rectangle is 52 yard, and the rectangle is 168yd^2. find the dimensions of the rectangle \n" ); document.write( "
Algebra.Com's Answer #66773 by ankor@dixie-net.com(22740)\"\" \"About 
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The perimeter of a rectangle is 52 yard, and the rectangle is 168yd^2. find the dimensions of the rectangle
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\n" ); document.write( "Let x = length and y = width
\n" ); document.write( "Perimeter:
\n" ); document.write( "2x + 2y = 52
\n" ); document.write( "Simplify, divide by 2:
\n" ); document.write( "x + y = 26
\n" ); document.write( "y = (26-x): can use for substitution
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\n" ); document.write( "Area:
\n" ); document.write( "x * y = 168
\n" ); document.write( "Substitute (26-x) for y
\n" ); document.write( "x(26-x) = 168
\n" ); document.write( "26x - x^2 = 168
\n" ); document.write( "0 = x^2 - 26x + 168; arrange as a quadratic equation, solve for x
\n" ); document.write( "Factors to:
\n" ); document.write( "(x - 12)(x - 14) = 0
\n" ); document.write( ":
\n" ); document.write( "x = +12 and x = +14
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\n" ); document.write( "Length 14 and width 12, would be the answer
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check using the perimeter
\n" ); document.write( "2(14) + 2(12) = 52
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\n" ); document.write( "Did this make sense to you? Any questions? \n" ); document.write( "
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