Solve the following system of equations:

$$
3\left(x+\frac{1}{x}\right)=4\left(y+\frac{1}{y}\right)=5\left(z+\frac{1}{z}\right)
$$

$$
xy + yx + zx = 1
$$
Let us introduce the symbols $A, B, C$ for the quantities $x y, y z$, and $z x$, respectively, which appear in (2). With these, (2) becomes:

$$
A+B+C=1
$$

From (1), we know that $x, y, z$ are non-zero; with this condition, we can multiply (1) by $(x y z)$ without introducing new roots:

$$
3\left(x^{2} y z+y z\right)=4\left(y^{2} x z+x z\right)=5\left(z^{2} x y+x y\right)
$$

All terms appearing here can be easily expressed in terms of $A, B, C$, for example, $x^{2} y z=(x y)(x z)=A C$, so from (1) we get

$$
3(A C+B)=4(A B+C)=5(B C+A)
$$

The factors in parentheses can be factored into products using (2'). For example, from (2') we have $B=-A-C+1$, so

$$
A C+B=A C-A-C+1=(A-1)(C-1)
$$

and similarly for the others. Thus,

$$
3(A-1)(C-1)=4(A-1)(B-1)=5(B-1)(C-1)
$$

and from (2') we have

$$
(A-1)+(B-1)+(C-1)=-2
$$

This gives us a significantly simpler system of equations for the quantities $u=A-1, v=B-1, w=C-1$:

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

The double equality clearly determines the ratio of $u, v$, and $w$ (assuming none of them is zero), and the other equality then determines their values.

Notice that up to (3), we tried to maintain the "symmetry" of the equations - for example, we introduced new unknowns for $x y, y z$, and $z x$; we did not split (1) into two equations; or in (1'), we did not substitute any of $A, B, C$ with the value derived from (2'), but all three. Furthermore, we significantly exploited the fact that 1 stands on the right side of (2), but the method we used (at least up to this point) works even if the numbers $3,4,5$ in (1) are replaced by arbitrary numbers.

Let us return to (3). We saw that there are two cases:
a) one of $u, v, w$ is zero;
b) none of $u, v, w$ is zero.

In the latter case, as stated earlier, we can determine the ratio of $u, v$, and $w$:

$$
u: v: w=5: 3: 4
$$

and thus from the second equation in (3) we get $u=-5 / 6, v=-3 / 6, w=-4 / 6$; or $A=x y=u+1=1 / 6, B=y z=v+1=1 / 2, C=x z=w+1=1 / 3$. These values naturally satisfy the system of equations (1')-(2'); furthermore, $A B C=x^{2} y^{2} z^{2}=1 / 36$, so $x y z=1 / 6$ or $-1 / 6$. From this, in the first case,

$$
x=\frac{x y z}{y z}=\frac{x y z}{B}=\frac{1}{3}, \quad y=\frac{x y z}{C}=\frac{1}{2}, \quad z=\frac{x y z}{A}=1
$$

and in the second case,

$$
x=-\frac{1}{3}, \quad y=-\frac{1}{2}, \quad z=-1
$$

and both naturally satisfy the original system of equations.

There is still the case a) to consider, that is, if one of $u, v, w$ is zero. In this case, according to (3), the value of each of the pairwise products is zero, so at least two of $u, v$, and $w$ are zero - and then the third value is -2. This means that two of $A, B, C$ are 1, and one is -1. Thus, $A B C=-1$, which, compared to $A B C=x^{2} y^{2} z^{2}$, shows that no new solution arises from this.

In summary, the system of equations has two solutions: $(1 / 3,1 / 2,1)$ and $(-1 / 3,-1 / 2,-1)$.