document.write( "Question 812230: Nick drew a rectangle that is 4 times as long as it is wide. Stacey increased each dimension of Nick's rectangle by 5 inches. If the perimeter of Stacey's rectangle is 60 inches, what are the dimensions of Nick's rectangle?
\n" ); document.write( " I tried 4(w+5)+ 4(4w+5)=60 \n" ); document.write( "

Algebra.Com's Answer #489220 by erica65404(394) $\"\"$ $\"About$

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you have the dimensions right
\n" ); document.write( "L=4w+5
\n" ); document.write( "w=w+5
\n" ); document.write( "but the equation is set up wrong
\n" ); document.write( "you put p=4L+4w
\n" ); document.write( "The real Perimeter equation is
\n" ); document.write( "P=2L+2w
\n" ); document.write( "60=2(4w+5)+2(w+5)
\n" ); document.write( "60=8w+10+2w+10
\n" ); document.write( "60=10w+20
\n" ); document.write( "40=10w
\n" ); document.write( "w=4 The width for nick's rectangle is 4 inches and since the length is 4 times as long as the width (4w); the length is 16. Since Stacey increased each length and width by 5 inches, Nicks perimeter should be 20inches smaller than Staceys.
\n" ); document.write( "The perimeter is 40 inches.
\n" ); document.write( "check your answer
\n" ); document.write( "4+16+4+16=40
\n" ); document.write( "Final answer: Width is 4, length is 16, and the perimeter is 40. \n" ); document.write( "

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