Trouble understanding this Equality of Complex number question. I have this question here 
$$(3-2j)(x+yj)=4+j^9$$
I know that $a + bj$ and $x + yj$ are equal when $a = x$ and $b = y$.
This question is confusing me because I have had three different answers but they are all wrong. 
What feels like my closest attempt was:
$$(3-2j)(x+yj)=4+j^9.$$
$x + yj = (-3+2j)(4-j)$
$= -12 + 3j + 8j -2j^2$
$= -12 + 11j -2(-1)$
$= -12 + 11j + 2$
$= -10 + 11j \:\: \Rightarrow$
$x = -10$
$y = 11$
But the answer is supposed to be
$x = 10/13$ 
$y = 11/13$
I am just looking for some guidance on how to properly solve this and figure out where I went wrong. 
Thank you!
 A: \begin{align*}(3-2i)(x+yi)=4+i^9&\Longleftrightarrow x+yi=\frac{4+i}{3-2i}\\&\Longleftrightarrow x+yi=\frac{(4+i)(3+2i)}{(3-2i)(3+2i)}\\&\Longleftrightarrow x+yi=\frac{10+11i}{13}=\frac{10}{13}+\frac{11}{13}i\\&\Longleftrightarrow x=\frac{10}{13}\text{ and }y=\frac{11}{13}\end{align*}
A: First, let us use the mathematical symbol for the imaginary unit:  $\;i\;$ , instead of $\;j\;$ , so $\;i^2=-1\;$ . Next:
$$(3-2i)(x+iy)=3x+2y+(3y-2x)i$$
Now equate real parts and imaginary parts in both sides and you're done
A: $$\left( 3-2i \right) \left( x+yi \right) =4+{ i }^{ 9 }\\ 3x+3iy-2ix+2y=4+i\\ \\ \\ \begin{cases} 3x+2y=4 \\ 3y-2x=1\quad  \end{cases}\Rightarrow \begin{cases} 6x+4y=8 \\ 9y-6x=3 \end{cases}\Rightarrow y=\frac { 11 }{ 13 } ,x=\frac { 10 }{ 13 } $$
A: $(3-2i)(x+yi)=4+i\\
(3+2i)(3-2i)(x+yi)=(3+2i)(4+i)$
It looks like you had an operation close to this, but you only applied it to the RHS, and you had a sign flipped.
$(9 - 4i^2)(x+yi)= 12 +2i^2 +(8+3)i\\
13(x+yi)= 10  +11i\\
x+yi= \frac {10}{13}  -\frac {11}{13}i$
