document.write( "Question 858169: A rectangle has a length that is 50 feet more than twice it's width. Find the length of the rectangle if the perimeter is 1,3000 ft. \n" ); document.write( "

Algebra.Com's Answer #517078 by ben720(159) $\"\"$ $\"About$

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Let w = width and 2w+50 = length
\n" ); document.write( "Perimeter = 2W+2Length
\n" ); document.write( "Substitute length for 2w+50
\n" ); document.write( " $\"P=2w%2B2%282w%2B50%29\"$
\n" ); document.write( "Then, distribute the 2
\n" ); document.write( " $\"P=2w%2B4w%2B100\"$
\n" ); document.write( "Combine Like Terms
\n" ); document.write( " $\"P=6w%2B100\"$
\n" ); document.write( "Substitute the perimeter for 1300
\n" ); document.write( " $\"1300=6w%2B100\"$
\n" ); document.write( "Subtract 100 from both sides
\n" ); document.write( " $\"1200=6w\"$
\n" ); document.write( "Divide both sides by 6
\n" ); document.write( " $\"200=w\"$
\n" ); document.write( "The width is 200. The length is
\n" ); document.write( " $\"2w%2B50\"$
\n" ); document.write( " $\"2%2A200%2B50\"$
\n" ); document.write( " $\"450\"$
\n" ); document.write( "The length is 450 feet. \n" ); document.write( "

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