SOLUTION: the perimeter of a rectangle is twice the sum of its length and its width, the perimeter is 40 meters and te length is 2 meters more than twice its width what is the length

Question 285902: the perimeter of a rectangle is twice the sum of its length and its width, the perimeter is 40 meters and te length is 2 meters more than twice its width what is the length
Found 2 solutions by richwmiller, mgmoeab:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
2L+2W=40
L=2+2W

Answer by mgmoeab(37) (Show Source): You can put this solution on YOUR website!
1).- P = 2(W + L)-------------OR-------------- P= 2W +2L (distributing)
where:
p= perimeter
w= width
l= length
They tell you that "the length is 2 meters more than twice it width"
2).- L = 2 + 2W
If we substitute (2) in (1), then:
P = 2W + 2 ( 2 + 2W)
P = 2W + 4 + 4W
P = 6W + 4
We also know that the perimeter is 40 meters, so:
P = 40 = 6W + 4
Solving for W, you get:
40 = 6W + 4
36 = 6W
W = 6.
But we want to know what L equals. We go back to equation 2 and subtitue W =6.
L = 2 + 2(6)
L= 14
WHICH IS TRUE..
P = 2(6) + 2 (14)
P = 12 + 28
P = 40

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