document.write( "Question 495819: The length of a rectangle is 5 feet more than twice its width. Find the dimensions of the rectangle if the perimeter of the rectangle is 163 feet. \n" ); document.write( "
Algebra.Com's Answer #336203 by algebrahouse.com(1659)\"\" \"About 
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\"The length of a rectangle is 5 feet more than twice its width. Find the dimensions of the rectangle if the perimeter of the rectangle is 163 feet. \"\r
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\n" ); document.write( "\n" ); document.write( "x = width
\n" ); document.write( "2x + 5 = length {length is 5 more than twice width}\r
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\n" ); document.write( "\n" ); document.write( "Perimeter of a rectangle is 2(width) + 2(length)\r
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\n" ); document.write( "\n" ); document.write( "2x + 2(2x + 5) = 163 {perimeter is 2(width) + 2(length)}
\n" ); document.write( "2x + 4x + 10 = 163 {used distributive property}
\n" ); document.write( "6x + 10 = 163 {subtracted 10 from both sides}
\n" ); document.write( "6x = 153 {subtracted 10 from both sides}
\n" ); document.write( "x = 25.5 {divided both sides by 6}
\n" ); document.write( "2x + 5 = 56 {substituted 25.5, in for x, into 2x + 5}\r
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\n" ); document.write( "\n" ); document.write( "width = 25.5 ft
\n" ); document.write( "length = 56 ft
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