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

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.
