Thanks a lot topsquark, you are encouraging.
I'll check out the details tomorrow, but I think I understood what you did.
At first I was trying to solve the problem with a single Cartesian graph with a unique origin. But I read in the book "Fundamentals of Physics" that I must separate the problem into 2 "systems", so I forgot about my first idea. For me it looked much more simple putting all the equations with respect to the same reference system. Especially because when I have vectors, say for example $\displaystyle P_1=3 i + 7j$ and $\displaystyle P_2=\sin \zeta 3i$ and that I want to sum up these vectors, if they are expressed into two different reference systems I can't add them that easily while if they are expressed with respect to the same referential system it's just a common addition. I'm not sure you can understand what I'm trying to say...
And for about
I'm curious how you got the acceleration without recognizing that the tensions were equal. There is a way to do this without considering tensions, but it is a much more complicated problem to visualize and I'm assuming you didn't do the problem in this way.

Okay, checking what I did : "The acceleration is only occurring in the xaxis. So I get that the force that matters to the movement of the bodies is only acting on the xaxis. $\displaystyle \sum F_x=a_xm \Leftrightarrow a(m_1+m_2)=m_2g\sin \beta m_1g\sin \alpha \Leftrightarrow a=\frac{m_2g\sin \beta m_1g\sin \alpha}{m_1+m_2}$." Here note that the xaxis from the 1st body is not the same that the xaxis from the 2nd body. In other words I followed what all students do : separate the problem into 2 separated Free Body Diagram, as you did. Is it clear?
When I assume that the acceleration is only occurring along the xaxis it is because I know that the bodies won't leave out the triangle, otherwise they would be flying and the normal force would then disappear and we would fall into a physically impossibility.
EDIT: Ah topsquark, I know that "the tensions were equal". In fact I wrote that $\displaystyle T_1+T_2=T$ where $\displaystyle T$ is the total tension in the cord and $\displaystyle T_1$ is the tension created by the first body while $\displaystyle T_2$ is the tension created by the second one. T would be the sum of the 2 tensions which is also the total tension in the cord at any point in it. But I know I didn't do well on that part, so that's why I'll check out what you answered to me.
EDIT2: Oh... you are saying that in my example $\displaystyle T_1$ would be equal to $\displaystyle T_2$? If so... then I don't really understand why, but I will investigate.
But now my question is : Do my calculus of the acceleration is wrong because I neglected the tensions? It's like I assumed that they would cancel out... without knowing it. Ah yes... I can't do what I did. I must show that $\displaystyle T_1=T_2$ before doing what I did.