Dimensions of fence (ACT)

The Johnsons can afford 220 feet of fencing material. They want to fence in a rectangular area with a length 20 feet more than twice the width. What will the dimension be?
A.)  40 ft x 80 ft
B.)  45 ft x 110 ft
C.)  30 ft x 80 ft
D.)  100 ft x 25 ft
E.)  50 ft x 60 ft 

w = width
2w + 20 = length

Perimeter of a rectangle is 2(width) + 2(length)

2w + 2(2w + 20) = 220 {perimeter is 2(width) + 2(length)}
2w + 4w + 40 = 20 {used distributive property}
6w + 40 = 220 {combined like terms}
6w = 180 {subtracted 40 from both sides}
w = 30 {divided both sides by 6}
2w + 20 = 80 {substituted 30, in for w, into 2w + 20}

width = 30 ft
length = 80 ft 

C.)  30 ft x 80 ft

- Algebra House