SOLUTION: how do you solve: the length of a rectangle is 2 less than 4 times the width and the perimeter is 56?

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Question 161427: how do you solve:
the length of a rectangle is 2 less than 4 times the width and the perimeter is 56?
Found 2 solutions by checkley77, nerdybill:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
2L+2W=PERIMETER
L=4W-2
2(4W-2)+2W=56
8W-4+2W=56
10W=56+4
10W=60
W=60/10
W=6 ANSWER FOR THE WIDTH.
L=4*6-2
L=24-2
L=22 ANSWER FOR THE LENGTH.
PROOF:
2*22+2*6=56
44+12=56
56=56

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!

the length of a rectangle is 2 less than 4 times the width and the perimeter is 56?
.
Let w = width of rectangle
then from "the length of a rectangle is 2 less than 4 times the width"
4w-2 = length of rectangle
.
Perimeter is 2(L + W)
2[(4w-2) + w] = 56
2[5w-2] = 56
10w-4 = 56
10w = 60
w = 6 (width)
.
length:
4w-2 = 4(6)-2 = 24-2 = 22 (length)