Question

the length of a rectangle is 15 inches longer than twice its width. if the perimeter is 78 inches, find the length and width of the rectangle.

Answer 1

P = 2(l + w)
l = 2w + 15

P = 2(2w + 15 + w)
P = 4w + 30 + w
P = 5w + 30

w = (P - 30) / 5
w = 48 / 5

l = 2w + 15
l = 2(48 / 5) + 15
l = 171/5

Answer 2

eh? I got something different.
Here's what I got:

Let the length be x and the width be y

x = 2y + 15 ...(1)
2x + 2y = 78 ...(2)

From (2):
x + y = 39
x = 39 - y

--> (1):
39 - y = 2y + 15
24 = 3y
y = 8

--> (1):
x = 2(8) + 15
x = 16 + 15
x = 31

So the length is 31 inches and the width is 8 inches.

Checking:
the length is 15 inches longer than twice its width
31 = 15 + 2(8)
31 = 15 + 16
Correct.

the perimeter is 78 inches
2(31) + 2(8) = 62 + 16 = 78
Also correct.

Answer 3

Oh I found where you went wrong, Fruitless.

P = 2(2w + 15 + w)
P = 4w + 30 + w

You expanded the bracket wrong. Shoudl've been 4w + 30 + 2w, which is 6w + 30

And then 6w + 30 = 78
6w = 48
w = 8