document.write( "Question 76678: Find the length of each side of the rectangle so that it has the given perimeter.
\n" ); document.write( "Note: Perimeter = 2*length + 2*width
\n" ); document.write( "Length= 2x-5
\n" ); document.write( "Width= x
\n" ); document.write( "
\n" ); document.write( "p=50 in.
\n" ); document.write( "

Algebra.Com's Answer #54989 by funmath(2933)\"\" \"About 
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Find the length of each side of the rectangle so that it has the given perimeter.
\n" ); document.write( "Note: Perimeter = 2*length + 2*width
\n" ); document.write( "Length= 2x-5
\n" ); document.write( "Width= x
\n" ); document.write( "
\n" ); document.write( "p=50 in.
\n" ); document.write( "\"50=2%282x-5%29%2B2%28x%29\"
\n" ); document.write( "\"50=4x-10%2B2x\"
\n" ); document.write( "\"50=6x-10\"
\n" ); document.write( "\"50%2B10=6x-10%2B10\"
\n" ); document.write( "\"60=6x\"
\n" ); document.write( "\"60%2F6=6x%2F6\"
\n" ); document.write( "\"10=x\"
\n" ); document.write( "Length=2x-5= 2(10)-5=20-5=15 in.
\n" ); document.write( "Width=x=10 in
\n" ); document.write( "Sanity check, does a rectangle with a length of 15 in and a width of 10 in have a peimeter that is 50 in?
\n" ); document.write( "50=2(15)+2(10)
\n" ); document.write( "50=30+20
\n" ); document.write( "50=50
\n" ); document.write( "We're not off our rocker!!!!
\n" ); document.write( "Happy Calculating!!!!
\n" ); document.write( "
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