So based on the diagram here's what I've attempted:
Note: $\bar{g}$ is the Earth's gravitational constant and $T$ is the tension force acting on the Blocks.
$a_1,a_2,a_3$ will be the acceleration for Block A, B, and C respectively.
Block A:
$T-3\bar{g}=3a_1$
Block B:
$T-5\bar{g}=5a_2$
Block C:
$2T-6\bar{g}=6a_3$
Then from equation Block A and C we find that $a_1=a_3$ (acceleration of Block A is equal to Block C).
And from the constant length from the strings and differentiating twice we we get: $a_1+a_2+2a_3=0$ implies $3a_1=-a_2$.
Upon subtracting equation Block B from A,
$2\bar{g}=3a_1-5a_2$, if we solve for $a_1$ we find that it's equal to $\frac{\bar{g}}{9}$ which doesn't seem to be the answer.
Can someone point out my error and explain it to me?
(This question was already asked on Physics stackexchange but was closed down before anyone could answer it. Here is the original post https://physics.stackexchange.com/questions/802718/pulley-mechanics-problem)
Your answer is correct. Your system is
$$\begin{cases} T - m_A g = m_A a_A \\ T - m_B g = m_B a_B \\ 2 T - m_C g = m_C a_C \\ a_A + a_B + 2 a_C = 0 \\ \end{cases}$$
You can solve it by rewriting all the accelerations in the kinematic constraint in terms of the tension:
$$\left( \frac{T}{m_A} - g \right) + \left( \frac{T}{m_B} - g \right) + 2 \left( \frac{2 T}{m_C} - g \right) = 0$$
$$\implies \left( \frac{1}{m_A} + \frac{1}{m_B} + \frac{4}{m_C} \right) T = 4 g$$
$$\implies T = \left[ \frac{4}{1 + \beta^{-1} + 4 \gamma^{-1}} \right] (m_A g)$$
where we define the dimensionless mass ratios $\beta = \frac{m_B}{m_A}$ and $\gamma = \frac{m_C}{m_A}$. The accelerations can then by substituting the solution for $T$:
$$\begin{cases} a_A = \left[ \frac{3 - \beta^{-1} - 4 \gamma^{-1}}{1 + \beta^{-1} + 4 \gamma^{-1}} \right] (g) \\ a_B = -\left[ \frac{1 - 3 \beta^{-1} + 4 \gamma^{-1}}{1 + \beta^{-1} + 4 \gamma^{-1}} \right] (g) \\ a_C = \left[ \frac{-1 - \beta^{-1} + 4 \gamma^{-1}}{1 + \beta^{-1} + 4 \gamma^{-1}} \right] (g) \\ \end{cases}$$
For your problem, $\beta = \frac{5}{3}$ and $\gamma = 2$. Substituting, you indeed find that $a_A = \frac{1}{9} g$, along with $a_B = -\frac{1}{3} g$ and $a_C = \frac{1}{9} g$.