SOLUTION: 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, w

Algebra ->  Equations -> SOLUTION: 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, w      Log On


   

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?
I tried 4(w+5)+ 4(4w+5)=60
Answer by erica65404(394) About Me  (Show Source):
You can put this solution on YOUR website!
you have the dimensions right
L=4w+5
w=w+5
but the equation is set up wrong
you put p=4L+4w
The real Perimeter equation is
P=2L+2w
60=2(4w+5)+2(w+5)
60=8w+10+2w+10
60=10w+20
40=10w
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.
The perimeter is 40 inches.
check your answer
4+16+4+16=40
Final answer: Width is 4, length is 16, and the perimeter is 40.