<s>[INST] What method can be used to solve a numeric finding puzzle involving getting the right sum? I've been trying to solve this puzzle at no avail. Can somebody help me?
The problem is as follows:

In each circle from the figure below write down an integer number
  between $1$ to $9$. Each one of them must be different from each
  other. The resulting sum from those numbers joined by three circles in
  a straight line and those joined by arrows in the direction indicated be equal to $18$. Which
  number must be written in the circle painted by an orange shade?


$\textrm{The alternatives given were:}$


*

*7

*4

*5

*3

*6


The only thing I could come up with was to make a sequence of dividing $18$ into different choices for summing that number between $1$ to $9$.
Therefore:
$1+8+9=18$
$2+7+9=18$
$3+6+9=18$
$4+5+9=18$
and the rest would be repeating those numbers in different order like
$5+4+9=18$
$6+3+9=18$
$7+2+9=18$
$8+1+9=18$
So the preceding can't be:
But this other combination can also reach to $18$:
$8+7+3=18$
$8+6+4=18$
However at this point I got tangled up with many choices therefore end up with clear clue to take to get $18$.
How can I solve this problem without just guessing?. Is there an algorithm, formula or method that can be followed orderly to avoid wasting much time?. 
I would like someone could re drawn the question and explain me with much details possible.
Please do not just fill out the drawing right away. I know maybe you have the ability to do this, but this is not what I'm looking for but rather a more step-by-step solution which has proven that it works in many similar scenarios.
Edit: I currently replaced the previous picture to this version which is the right one.
 [/INST] Fellow the image Below:
$A + B + C + D +E +F + G+ H+ X = 1 + 2+3+4+5+6+7+8+9 = 45$
so
$X+ C +F = (A+B+C+D+E+F+G+H+X) - (B+E+H) - (A+D+G) = 45 - 18 -18 = 9$.
so
$2X + A+E +C +F = (X+C+F)+(A+E+X)= 9 + 18 = 27$
so 
$A+B+C+D+E+F + 3X = (2X+A+E+C+F) + (B+E+D) = 27 + 18 = 45$
so
$A+B+C + 3X = (A+B+C+D+E+F) - (D+E+F) = 45-18 = 27$.
so 
$3X = (A+B+C + 3X) - (A+B+C) = 27-18 = 9$.
So $X = \frac {3X}3 = \frac 93 = 3$.

</s>