Find the tension in each segment of rope suspending two hanging masses 
I have the situation in the image where two masses are suspended along a piece of rope (resulting in three segments of rope). I need to find the tension in each of the three segments of rope. 
I know how to solve the similar problem with only one mass (and two segments of rope) by finding the horizontal and vertical components for each segment of rope, noting that the system is in equilibrium and then finding two equations in two unknowns and solving.
However, I am a little confused now that there is a second mass involved. If I follow the same method. I have three unknown tensions, but only 2 equations (one for the horizontal and one for the vertical component), and then I can't solve.
What am I missing here? Is this solvable (I suspect so)? 
My only other idea is that I need to look at each mass separately somehow, so that I sort of follow the same method as for one mass but do it twice (once for each mass).
 A: Somebody correct me if I'm wrong but I do believe you would have five tensions. There would be the tension in the one rope joining the two ropes holding the masses, the tension in the two ropes from the intersection to the mass, then the tension in the two rope from the top of the diagram to the intersection of all the ropes.
The tension in the rope from intersection to mass would be independent from each other, whereas the other three would be dependent on those masses and each other.
A: Edit
I was wrong.  Look at Floris's answer.
If you split it into sections it should be easier than combining it into one.
If you have $F_1$ , $F_2$, $F_3$ as the tensions of the strings
$$
F_{1x} = F_{2x}
$$
$$
F_{2x} = F_{3x}
$$
$$
F_{1y} + F_{2y} = 12g
$$
$$
F_{2y} + 7g = F_{3y}
$$
Then you can use trigonometry and then eliminate.
Edit:
Eliminate $F_{2y}$
$$
F_{1y} + F_{3y} = 19g
$$
Then Trig
$$
F_1sin(45) + F_3sin(30) = 19g
$$
Along with
$$
F_1cos(45) = F_3cos(30) 
$$
A: You can indeed balance the horizontal forces at each point, and the sum of vertical components should equal the weight.
That does seem to leave you with an over constrained problem (four equations with three unknowns) which will only have a solution when the angles are chosen "just so".
If one of the angles was not given you could solve. Pick one and prove that the value for the angle you calculate is the once given... Or prove that there is no solution! Although it is easier to assume the angles as given, and make one of the weights "unknown". Then solve for the value of the weight that gets you the angles.
