document.write( "Question 432323: the length of a rectangle is five feet more than twice its width. the area is 133 sq ft. find dimensions of the rectangle. \r
\n" ); document.write( "\n" ); document.write( "SOMEBODY PLEASE HELP I KNOW I HAVE TO SET UP A COUPLE EQUATIONS BUT NOT SURE HOW. ANY HELP WOULD BE GREATLY APPRECIATED. THANKS IN ADVANCE. \n" ); document.write( "
Algebra.Com's Answer #299771 by stanbon(75887)\"\" \"About 
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the length of a rectangle is five feet more than twice its width. the area is 133 sq ft. find dimensions of the rectangle.
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\n" ); document.write( "Equations:
\n" ); document.write( "L = 2W+5
\n" ); document.write( "LW = 133
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\n" ); document.write( "Substitute for \"L\" and solve for \"W\":
\n" ); document.write( "(2W+5)W = 133
\n" ); document.write( "---
\n" ); document.write( "2W^2+5W-133 = 0
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\n" ); document.write( "Factor:
\n" ); document.write( "(W-7)(2W+19) = 0
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\n" ); document.write( "Positive solution:
\n" ); document.write( "Width = 7 ft.
\n" ); document.write( "Solve for \"Length\":
\n" ); document.write( "L = 3W+5
\n" ); document.write( "L = 3*7+5
\n" ); document.write( "Length = 26 ft
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "
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