376. Solve the systems of equations:
a) $\left\{\begin{array}{l}\frac{1}{x}+\frac{2}{y}-\frac{3}{z}=3 \\ \frac{4}{x}-\frac{1}{y}-\frac{2}{z}=5 \\ \frac{3}{x}+\frac{4}{y}+\frac{1}{z}=23\end{array}\right.$
b) $\left\{\begin{array}{l}\frac{1}{x-y}+x^{2}=1, \\ \frac{x^{2}}{x-y}=-2 .\end{array}\right.$
$\triangle$ a) Let's introduce the substitutions:

$$
\frac{1}{x}=t, \quad \frac{1}{y}=u, \quad \frac{1}{z}=v
$$

We will have:

$$
t+2 u-3 v=3, \quad 4 t-u-2 v=5, \quad 3 t+4 u+v=23
$$

This system of linear equations can be solved in various ways. For example, add the first and third equations and subtract the second: $7 u=21, u=3$.

Substitute the value $u=3$ into the first and third equations. After simplifications, we get:

$$
\mathrm{t}-3 v=-3, \quad 3 t+v=11
$$

From the last system, $t=3, v=2$. Now it is easy to compute $x, y$ and $z$.

b) Let

$$
\frac{1}{x-y}=z, \quad x^{2}=t
$$

Then

$$
z+t=1, \quad z t=-2
$$

From here, $z=-1, t=2$ (the case $z=2, t=-1$ is impossible since $t \geq 0$). We will have:

$$
\frac{1}{x-y}=-1, \quad x^{2}=2
$$

The last system is easily solved.

Answer: a) $(1 / 3 ; 1 / 3 ; 1 / 2) ; \quad$ b) $(\sqrt{2} ; 1+\sqrt{2}),(-\sqrt{2} ; 1-\sqrt{2})$.