Atwood Machine: Find M in Terms of m1 & m2

[Ryan95]

Homework Statement

In the Atwood’s machine, what should M be, in terms of m1 and m2, so that it doesn’t move?

Homework Equations

F=ma

The Attempt at a Solution

I've set T1=Mg as T1 is the tension of the rope attached to M. m1 and m2 are both connected together by the same rope so I assumed T1=2T2 .
I set up the equations
T2-m1g=m1a
T2-m2g=-m2a (negative due to m1 and m2 moving in opposite directions).
I then added the two together to get
T2=[m1(g+a)+m2(g-a)]/2.
Which, since T1=2T2=Mg, it can be expressed as
M=[m1(g+a)+m2(g-a)]/g (I think).
Which is where I'm stuck now. I'm unsure of how to get rid of g and a in order to have my answer in terms of just m1 and m2. It's probably pretty simple, but I can't quite get it.. It seems as g can be canceled out from the current equation, but I'm not sure about a.
I've tried canceling out the g and then solving for a which gives me
a=M/m₁-m₂
But upon plugging that in as a, I am unsure of how to proceed.

 

[haruspex]

Please either describe the set-up or post a diagram. Not all Atwood's machines are the same.

 

[Ryan95]

Right, apologies. I've added a photo.

 

[Chestermiller]

You have 2 linear algebraic equations in two unknowns, T2 and a. Solve for them separately.