document.write( "Question 811419: The area of a rectangular conference room is represented by the equation A = (x2 + 17x + 72) feet2.\r
\n" ); document.write( "\n" ); document.write( "If x = 6 feet and the width = (x + 8) feet, what is the perimeter, in feet, of the room?
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Algebra.Com's Answer #488735 by Stitch(470)\"\" \"About 
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The equation for the area of a rectancle is A = L * W.
\n" ); document.write( "The equation for the perimeter of a rectangle is P = 2L + 2W.
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\n" ); document.write( "Given: X = 6
\n" ); document.write( "Given: \"A+=+X%5E2+%2B+17X+%2B+72\"
\n" ); document.write( "Given: \"W+=+X+%2B+8\"
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\n" ); document.write( "Since we know that X = 6, we can solve for W.
\n" ); document.write( "\"W+=+X+%2B+8\"
\n" ); document.write( "\"W+=+%286%29+%2B+8\"
\n" ); document.write( "\"highlight%28W+=+14%29\"
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\n" ); document.write( "Now we can solve for the area by plugging 6 into the equation for X.
\n" ); document.write( "Given: \"A+=+X%5E2+%2B+17X+%2B+72\"
\n" ); document.write( "\"A+=+%286%29%5E2+%2B+17%2A%286%29+%2B+72\"
\n" ); document.write( "Simplify
\n" ); document.write( "\"A+=+36+%2B+102+%2B+72\"
\n" ); document.write( "Combine like terms.
\n" ); document.write( "\"highlight_green%28A+=+210%29\"
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\n" ); document.write( "Now we can solve for the length.
\n" ); document.write( "\"A+=+L%2AW\"
\n" ); document.write( "\"210+=+L+%2A+%2814%29\"
\n" ); document.write( "Divide both sides by 14.
\n" ); document.write( "\"highlight%2815+=+L%29\"
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\n" ); document.write( "Now we can use the perimeter equation.
\n" ); document.write( "\"P+=+2L+%2B+2W\"
\n" ); document.write( "\"P+=+2%2A%2815%29+%2B+2%2A%2814%29\"
\n" ); document.write( "Simplify
\n" ); document.write( "\"P+=+30+%2B+28\"
\n" ); document.write( "\"highlight_green%28P+=+58%29\"
\n" ); document.write( "The perimeter of the room is 58 feet.
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