Calculate the equivalent capacity of the capacitor system between points x and y

I'm asking about this problem because I know how to calculate $V_x-V_y$, but I don't know how can I extract the equivalent capacity from there.

Using Kirchoff Laws I get the following equations

- $$V_x-V_y=q_1/C_1+q_2/C_2+q_3/C_3 $$
- $$V_x-V_y=q_4/C_4+q_3/C_3 $$
- $$V_x-V_y=q_1/C_1+q_5/C_5 $$
- $$0=-q_1/C_1-q_2/C_2+q_4/C_4 $$
- $$0=q_2/C_2+q_3/C_3-q_5/C_5 $$

($1,2,3$ express $V_x-V_y$ through all the possible paths from $x$ to $y$ I can take, while $4,5$ are loop laws for the two loops).

I can solve for $V_x-V_y$, but then here it comes the doubt.

As said I have three different ways to go from $x$ to $y$ (the first three equations are there because of that), I highlighted them in the picture (path $A$ is equation $1$, path $B$ is equation $2$, path $C$ is equation $3$)

So can I express the equilvalent capacitance of the system as the ratio of **the charge met following the particular path **and $V_x-V_y$? That would be

$$C_{equivalent}=(q_1+q_2+q_3)/V_x-V_y=(q_4+q_3)/V_x-V_y=(q_1+q_5)/V_x-V_y$$

This does not look correct at all, but it is the only reasonable way to use the potential difference to calculate capacity.

So I ask: what is the right way to link $V_x-V_y$ to $C_{equivalent}$? That is : how to calculate $C_{equivalent}$ knowing $V_x-V_y$?

I would like to use this method (using $V_x-V_y$) and not other ones (if there are any) because I would like to understand how it should work, so please help me figure out the link between $V_x-V_y$ and $C_{equivalent}$ in cases like this one.