Determine the complex number $z$ satisfying the equation $2z-3i\bar{z}=-7+3i$.  Note that $\bar{z}$ denotes the conjugate of $z$.
To solve for $z$, I need to separate the real and imaginary parts of both sides of the equation. I can write $z$ as $a+bi$, where $a$ and $b$ are real numbers, and then use the fact that $\bar{z}=a-bi$. Substituting these expressions into the equation, I get $2(a+bi)-3i(a-bi)=-7+3i$. Expanding and simplifying, I get $2a-3ai+2bi+3bi^2=-7+3i$. Using the fact that $i^2=-1$, I get $2a-3ai+2bi-3b=-7+3i$. Now I can compare the real and imaginary parts of both sides. For the real parts, I get $2a-3b=-7$. For the imaginary parts, I get $-3a+2b=3$. These are two linear equations in two variables, which I can solve by elimination or substitution. I choose to eliminate $a$ by multiplying the first equation by $3$ and the second by $2$, and then adding them. This gives me $6a-9b+(-6a+4b)=-21+6$. Simplifying, I get $-5b=-15$. Solving for $b$, I get $b=3$. To find $a$, I can plug $b=3$ into either equation. I choose the first one. This gives me $2a-3(3)=-7$. Simplifying, I get $2a-9=-7$. Adding $9$ to both sides, I get $2a=2$. Dividing by $2$, I get $a=1$. Therefore, $z=a+bi=1+3i$. 