HINT Your question can be equivalently written as: $ \begin{align*} 3ab=2cd \tag{1} \\ bc+ad = ab \tag{2} \end{align*} $
You can eliminate $d$ from the system by taking $a \times (1) + 2c \times (2)$, when you will get the equation: $ \ldots (\text{do the algebra and find the equation}) $ Collect together all terms on one side and factor the equations. You will find that either $b = 0$ or $\ldots$. (Fill in the blank.)
Case 1: Suppose $b = 0$. Can you handle this case?
Case 2: Suppose $\ldots$ holds. Then in this case, $a = c = 0$. (Can you prove this? It is slightly nontrivial to show this.) In this case, what can you say about $b$ and $d$?
So, to conclude, what are all the solutions of the equation?