Pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and $33x+56y=\frac{x}{x^2+y^2}$. Let (x,y) be a pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and  $33x+56y=\frac{x}{x^2+y^2}$. If $|x|+|y|=\frac{p}{q}$ (where p and q are relatively prime), then find the value (6p – q).
I used the concept $\frac{x}{y}=t$ while solving i get $56t+33=-\frac{1}{t^2+1}$ and $33t+56=\frac{t}{t^2+1}$ on dividing the two equation i end  up getting quadratic equation but that is not helping me
 A: Both of the existing answers posted here are incorrect, so here I will start with @Aditya Dwivedi's solution and go from there. 
Let $ z = x - yi $ 
Then, subtract the first equation times $i$ from the second equation.
\begin{align*}
&33x - 56y - 56xi - 33yi =  \frac{x + yi}{x^2 + y^2} \\
\implies &33(x-yi) - 56(y + xi) = \frac{1}{x - yi} \\
&\text{you can see that $y + xi = i \cdot (x - yi) $, so} \\
\implies & z(33 - 56i) = \frac{1}{z} \implies z^2 = \frac{1}{33-56i}
\end{align*}
$33 - 56i = (7 - 4i)^2$, so $ z = \pm \frac{1}{7 - 4i} = \pm(\frac{7}{65} + \frac{4}{65} i) \implies (x, y) = (\pm \frac{7}{65}, \mp \frac{4}{65}) $. Thus, $|x| + |y| = \boxed{\frac{11}{65}} \hspace{.5em} \blacksquare $
A: Let 
$z =x +iy
\\ \bar{z}=x-iy
\\ now \ subtract \ 2 \ equations \ by multiplying \ first \ with \ i $
$$\bar{z}(33-56i) = \frac{1}{z}$$
Now you can proceed
A: Equation (1) $\times x$ + Equation (2) $\times y$ when added together gives
\begin{align*}
56x^2+66xy+56y^2&=0\\
28x^2+33xy+28y^2&=0\\
28\left(x+\frac{33}{56}y\right)^2+\frac{2047}{112}y^2&=0.
\end{align*}
Since left side is always non-negative (for $x,y \in \mathbb{R}$), this can only happen when both
\begin{align*}
x+\frac{33}{56}y&=0\\
y^2&=0
\end{align*}
This gives $x=y=0$ as the only solution. But this does not satisfy the original equation (denominator will become $0$). Thus no such real $x$ and $y$ can exist.
