document.write( "Question 1031819: The length of a particular rectangle is 8 inches more than 2 times its width. A new rectangle is formed by tripling the width. The area is 30 square inches more than the area of original rectangle. Find the dimensions of the original rectangle. \n" ); document.write( "
Algebra.Com's Answer #646481 by mananth(16946)\"\" \"About 
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The length of a particular rectangle is 8 inches more than 2 times its width.\r
\n" ); document.write( "\n" ); document.write( "let width be x\r
\n" ); document.write( "\n" ); document.write( "length will be 2x+8\r
\n" ); document.write( "\n" ); document.write( "Area = x(2x+8)\r
\n" ); document.write( "\n" ); document.write( " A new rectangle is formed by tripling the width.\r
\n" ); document.write( "\n" ); document.write( "new width = 3x
\n" ); document.write( "length = 2x+8\r
\n" ); document.write( "\n" ); document.write( "Area = 3x(2x+8)\r
\n" ); document.write( "\n" ); document.write( " The area is 30 square inches more than the area of original rectangle.
\n" ); document.write( "3x(2x+8) -x(2x+8)=30\r
\n" ); document.write( "\n" ); document.write( "2x(2x+8) = 30\r
\n" ); document.write( "\n" ); document.write( "4x^2 +8x -30=0\r
\n" ); document.write( "\n" ); document.write( "4x^2+12x-10x-30=0\r
\n" ); document.write( "\n" ); document.write( "4x(x+3)-10(x+3)=0\r
\n" ); document.write( "\n" ); document.write( "(x+3)(4x-10)=0\r
\n" ); document.write( "\n" ); document.write( "x=-3 OR 5/2\r
\n" ); document.write( "\n" ); document.write( "Taking positive value width = 5/2 \r
\n" ); document.write( "\n" ); document.write( "Length = 2x+8\r
\n" ); document.write( "\n" ); document.write( "=2*5/2 + 8\r
\n" ); document.write( "\n" ); document.write( "=13
\n" ); document.write( "the dimensions of the original rectangle 2.5 in by 13 in
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