29. [15] Let $(x, y)$ be a pair of real numbers satisfying
$$
56 x+33 y=\frac{-y}{x^{2}+y^{2}}, \quad \text { and } \quad 33 x-56 y=\frac{x}{x^{2}+y^{2}} \text {. }
$$

Determine the value of $|x|+|y|$.
Answer:
$\frac{11}{65}$ Observe that
$$
\frac{1}{x+y i}=\frac{x-y i}{x^{2}+y^{2}}=33 x-56 y+(56 x+33 y) i=(33+56 i)(x+y i) \text {. }
$$

So
$$
(x+y i)^{2}=\frac{1}{33+56 i}=\frac{1}{(7+4 i)^{2}}=\left(\frac{7-4 i}{65}\right)^{2} .
$$

It follows that $(x, y)= \pm\left(\frac{7}{65},-\frac{4}{65}\right)$.