document.write( "Question 756044: The length of a rectangle is 6 more inches than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.\r
\n" ); document.write( "\n" ); document.write( "Suppose width is x inches than length is x+6
\n" ); document.write( "Area of a rectangle is= l*b
\n" ); document.write( " 91=(x)*(x+6)
\n" ); document.write( " 91= x^2+6x
\n" ); document.write( "0=x^2+6x-91
\n" ); document.write( "0=x^2+13x-7x-91
\n" ); document.write( "0=x(x+13)-7(x+13)
\n" ); document.write( "0=(x-7)(x+13)
\n" ); document.write( "0=(x-7) OR 0=(x+13)
\n" ); document.write( "7=x OR -13=x (negative width not possible)
\n" ); document.write( "x=7 inches width and length is six more so 7+6=13
\n" ); document.write( "Answer is 7 and 13 are the length and width of the rectangle
\n" ); document.write( "I hope u find my solution useful \n" ); document.write( "

Algebra.Com's Answer #459964 by nihar*2013(34) $\"\"$ $\"About$

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The length of a rectangle is 6 more inches than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle.\r
\n" ); document.write( "\n" ); document.write( "Suppose width is x inches than length is x+6
\n" ); document.write( "Area of a rectangle is= l*b
\n" ); document.write( " 91=(x)*(x+6)
\n" ); document.write( " 91= x^2+6x
\n" ); document.write( "0=x^2+6x-91
\n" ); document.write( "0=x^2+13x-7x-91
\n" ); document.write( "0=x(x+13)-7(x+13)
\n" ); document.write( "0=(x-7)(x+13)
\n" ); document.write( "0=(x-7) OR 0=(x+13)
\n" ); document.write( "7=x OR -13=x (negative width not possible)
\n" ); document.write( "x=7 inches width and length is six more so 7+6=13
\n" ); document.write( "Answer is 7 and 13 are the length and width of the rectangle
\n" ); document.write( "I hope u find my solution useful \n" ); document.write( "

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