Answer another hard math problem

Beth is two years older than Julio. Gerald is twice as old as Beth. Debra is twice as old as Gerald. The sum of their ages is 38. How old is Beth?
How do I set this up to find the answer? Thanks!

Let the ages be b, j, g, and d
then
"Beth is two years older than Julio" becomes b = j + 2
Gerald is twice as old as Beth becomes g = 2b
Debra is twice as old as Gerald becomes d = 2g
"The sum of their ages is 38." becomes b + j + g + d = 38
b = j + 2
g = 2b = 2(j + 2)
d = 2g = 2(2b) = 4b = 4(j+2)
so:
j + j +2 + 2j +4 + 4j + 8 = 38
==>8j + 14 = 38
==> 8j = 24
==> j = 3
So, beth is 5 years old

or
j = b-2
g = 2b
d = 4b
so: b-2 + b + 2b + 4b = 38
==> 8b = 40
==> b = 5
or bet is 5 years old.

Beth =B Julio=J Gerald=G Debra=D

Beth is 2 yrs older than Julio hence B= J+2
Gerald is twice as old as Beth so G= (J+2) x 2
Debra is twice as old as G therefore she is 4x older than Beth so D= (J+2) x 4

I'd use guess and check
because there are four of them the 38/4=9.5 so Debra must be older than that and Julio must be much younger i'd use guess and check
5,7,14,28 sum=54 to high
1,3,6,12 sum=22 to low
Julio's age is between 5 and 1 so 3,5,10,20 sum= 38

Start with the least common denominator and build from there.
Julio is X
Beth = X+2 Julio + 2
Gerald = 2(X+2) Twice as old as Beth
Debra = 2(2(X+2) = 4(X+2) Twice as old as Gerald
Now add everything up: The sum of their ages is 38.
X + X+2 + 2(X+2) + 4(X+2) = 38
X + 7(X+2) = 38 [combine like things]
8x +14 = 38 [subtract 14 from each side]
8x = 24 [divide 24 by 8]
24/8=3
Julio = 3
Beth = 5 Beth is 2 years older than Julio. 3+2 = 5
Gerald = 10 Gerald is twice as old as Beth. 5*2 = 10
Debra= 20 Debra is twice as old as Gerald. 10*2 = 20
20+10+5+3=38
Best wishes

Let Beth be "x" years old
Name: Beth Julio Gerald Debra
age: x x-2 2x 4x
We know
x+x-2+2x+4x = 38
8x = 40 or x = 5
So their ages are as follows:
Name: Beth Julio Gerald Debra
age: 5 3 10 20

set it up as x+(x+2)+2x+4x=38