document.write( "Question 103909: suppose that the width of a rectangle is 5 inches shorter than the length and that the perimeter of the rectangle is 50.
\n" ); document.write( "set up an equation for the perimeter involving only L, the length of the rectangle.
\n" ); document.write( "sove this equation algebraically to find the length of the rectange. find the width as well.\r
\n" ); document.write( "\n" ); document.write( "thanks for the help-renee \n" ); document.write( "
Algebra.Com's Answer #75583 by stanbon(75887)\"\" \"About 
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suppose that the width of a rectangle is 5 inches shorter than the length and that the perimeter of the rectangle is 50.
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\n" ); document.write( "Let length be L ;then width = L-5
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\n" ); document.write( "\n" ); document.write( "set up an equation for the perimeter involving only L, the length of the rectangle.
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\n" ); document.write( "Perimeter = 2L + 2(L-5) = 4L-10
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\n" ); document.write( "solve this equation algebraically to find the length of the rectangle. find the width as well.
\n" ); document.write( "50 = 4L-10
\n" ); document.write( "4L = 60
\n" ); document.write( "L = 15 inches (this is the length)
\n" ); document.write( "L-5 =10 inches (this is the width)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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