Question: If $z=x+jy$, where $x$ and $y$ are real, find the values of $x$ and $y$ when

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> $\frac{3z}{1-j}+\frac{3z}{j}= \frac{4}{3-j}.$
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My attempt:

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> $\frac{3(x+jy)}{1-j}+\frac{3(x+jy)}{j}= \frac{4}{3-j}$
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> $\frac{3x+3jy}{1-j}+\frac{3x+3jy}{j}= \frac{4}{3-j}.$
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I'm not sure how to go on after this step above. Please help; I have an exam in a few hours.

Answer: 
Since\begin{align}\frac{3z}{1-i}+\frac{3z}i&=3z\left(\frac1{1-i}+\frac1i\right)\\&=3z\left(\frac12-\frac i2\right)\\&=z\left(\frac32-\frac32i\right),\end{align}you have\begin{align}z&=\frac{\frac4{3-i}}{\frac32-\frac32i}\\&=\frac{\frac{6}{5}+\frac{2 i}{5}}{\frac32-\frac32i}\\&=\frac4{15}+\frac8{15}i.\end{align}
The complex equation is
$$\frac{3x+3jy)}{1-j}+\frac{3x+3jy}{j}=\frac{4}{3-j}.$$ Here $j^2=-1$.
Multiply first, second and third term term by $1+j,-j$, and $3+j$ up and down and separate real and imaginary parts as
$$(-6/5+3x/2+3y/2)+j(-2/5-3x/2+3y/2)=0,\implies x+y=4/5,-x+y=4/15$$ these two simultaneous Eqs. then give $x=4/15,y=8/15$.