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: 
$$\frac {3z}{1-j}+\frac{3z}j=\frac4{3-j}$$

$$\frac {3zj+3z(1-j)}{(1-j)j}=\frac4{3-j}$$

$$\frac{3z}{1+j}=\frac4{3-j}$$

$$3z=\frac{4(1+j)}{3-j}=\frac{4+4j}{3-j}=\frac{(4+4j)(3+j)}{10}=\frac{(2+2j)(3+j)}5=\frac{4+8j}5$$

$$z=\dfrac{4+8j}{15}$$
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 $x=4/15,y=8/15$.