document.write( "Question 305672: what are original dimensions- The length of a rectangle is three times its width. If the width were increased by 4 and the length remained the same, the resulting rectangle would have an area of 231 square inches. Find the demensions of the ORIGINAL rectangle.
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Algebra.Com's Answer #218845 by mananth(16946)\"\" \"About 
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what are original dimensions- The length of a rectangle is three times its width. If the width were increased by 4 and the length remained the same, the resulting rectangle would have an area of 231 square inches. Find the demensions of the ORIGINAL rectangle.\r
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\n" ); document.write( "\n" ); document.write( "let the width be x\r
\n" ); document.write( "\n" ); document.write( "length will be 3x\r
\n" ); document.write( "\n" ); document.write( "width increased by 4 = x+4 \r
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\n" ); document.write( "\n" ); document.write( "Length = 3x\r
\n" ); document.write( "\n" ); document.write( "3x(x+4)= 231\r
\n" ); document.write( "\n" ); document.write( "3x^2 + 12x =231\r
\n" ); document.write( "\n" ); document.write( "3x^2 +12x-231=0\r
\n" ); document.write( "\n" ); document.write( "3(x^2+4x-77) =0\r
\n" ); document.write( "\n" ); document.write( "3(x+11)(x-7)=0\r
\n" ); document.write( "\n" ); document.write( "x=-11 OR x =7\r
\n" ); document.write( "\n" ); document.write( "the width cannot be negative \r
\n" ); document.write( "\n" ); document.write( "so width = 7\r
\n" ); document.write( "\n" ); document.write( "length = 3x = 3*7 =21\r
\n" ); document.write( "\n" ); document.write( "7 by 21 are the dimensions of the original rectangle
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