Let's take a far more complex case, just to show how easy this all is using simultaneous equations:
(I picked the above problem from this site.)
Using sympy as my solver, enter the following lines:
var('ga gb gc gd ge gf gg gh gi gj n2 n3 n4 n5 n6') s2 = Eq( n2, ga*n1 + gj*n3 ) s3 = Eq( n3, gb*n2 + gh*n5 ) s4 = Eq( n4, gc*n3 + gi*n5 ) s5 = Eq( n5, gd*n4 + gg*n3 + gf*n5 ) s6 = Eq( n6, ge*n5 ) ans = solve( [s2,s3,s4,s5,s6], [n2,n3,n4,n5,n6] ) pprint( ans[n6]/n1 ) ga⋅gb⋅ge⋅(gc⋅gd + gg) ────────────────────────────────────────────────────────────────── gb⋅gd⋅gi⋅gj + gb⋅gf⋅gj - gb⋅gj - gc⋅gd⋅gh - gd⋅gi - gf - gg⋅gh + 1
Please take a moment and go above to read each of the equations I set up (\$s_2\$, etc.) You should easily be able to see how it is that I wrote those out. It's very easy. For example, node 2 (\$n_2\$) only has two terms added together: \$N_1\cdot A\$ and \$N_3\cdot J\$. You should easily see why I wrote out the equation for \$s_2\$ in the way I did.
The solver does the rest. And the answer is correct. (You can verify it by simply looking at the site I mentioned earlier. Please note that they did not use simultaneous equations to arrive at their solution. They used Mason's gain formula. But their solution is exactly the same one that I arrived at using a very simple and very well-known approach.)
Now, given the above process I've laid out, do you think you can write the appropriate equations for your case? (It's fewer equations and simpler to do.) The result, if you get it correctly handled, will have the numerator you mentioned (44) but not the same denominator value you mentioned.
(If you still need help, I'll add more details directly targeting your solution.)
In your circumstance, you need to label your nodes (you haven't done that, yet.) I believe your gains are the finite values shown on your diagram. So you can just use those values, directly, in your equation set up.
For example, I've labeled two of your nodes below:
The equation for it is: \$X_1=1\cdot R_s - 1\cdot X_2=R_s-X_2\$. You should be able to develop the equations for all of \$X_1\$ through \$X_4\$ (\$C_s=X_4\$, so that's trivial.)