11.6. Solve the system of equations

\[
\left\{\begin{aligned}
x+2 y+3 z & =2 \\
\frac{1}{x}+\frac{1}{2 y}+\frac{1}{3 z} & =\frac{5}{6} \\
x y z & =-1
\end{aligned}\right.
\]
# Solution:

Method 1. Substituting $2 y=p, 3 z=s$ leads (after transformations) to the equivalent system $\left\{\begin{aligned} x+p+s & =2 \\ x p s & =-6 . \text { By Vieta's theorem, the numbers } x, p, s-\text { are the roots } \\ x p+p s+s x & =-5\end{aligned}\right.$ of the cubic equation $t^{3}-2 t^{2}-5 t+6=0$, which are the numbers $1,-2,3$. Depending on which variable corresponds to which root, we get 6 different solutions. Method 2. The second equation of the system, after multiplying by $6 x y z$ (which equals -6 from the third equation), transforms into the equivalent equation $6 y z+3 x z+2 x y=-5$. Multiplying this term by term by $x$, we get $6 x y z+x^{2}(3 z+2 y)=-5 x$. Substituting from the first equation $2 y+3 z=2-x$ and from the third $x y z=-1$, we obtain the cubic equation in $x$: $-6+x^{2}(2-x)=-5 x$. Solving this, we find $x=1$, or $x=-2$ or $x=3$. Each of these three cases reduces to a system of two equations with two unknowns $y$ and $z$, which can be solved by substitution.

## Answer:

$$
(1,-1,1),\left(1, \frac{3}{2},-\frac{2}{3}\right),\left(-2, \frac{1}{2}, 1\right),\left(-2, \frac{3}{2}, \frac{1}{3}\right),\left(3,-1, \frac{1}{3}\right),\left(3, \frac{1}{2},-\frac{2}{3}\right)
$$

Recommendations for checking:

| present in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| Correct solution method but incorrect answer due to  arithmetic errors | 6 points |
| Correct solution method, but one or two solutions  of the system are lost due to errors in transformations | 5 points |
| Correctly derived cubic equation in one  of the variables, but this equation is not solved | 3 points |
| Substitution is made that reduces the system to symmetric  in three unknowns | 2 points |
| All 6 solutions are provided, but the absence of others is not proven | 1 point |
| Some (not all) solutions are provided and/or any  calculations, from which the solution method is not clear |  |