Solve the systems:

a) $x+y+z=6, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{11}{6}, x y+y z+x z=11$

b) $x(y+z)=2, y(z+x)=2, z(x+y)=3$

c) $x^{2}+y^{2}+x+y=32, 12(x+y)=7 x y$

d) $\frac{x^{2}}{y}+\frac{y^{2}}{x}=\frac{7}{2}, \frac{1}{y}+\frac{1}{x}=\frac{1}{2}$

e) $x+y+z=1, x y+x z+y z=-4, x^{3}+y^{3}+z^{3}=1$;

f) $x^{2}+y^{2}=12, x+y+x y=9$
a) From the last two equations, it follows that $x y z=6$. Thus, $x, y, z$ are uniquely determined up to permutation. But one solution - $(1,2,3)$ - is obvious.

b) Subtracting the first two equations, we get $x z=y z$. Since $z \neq 0$ (this is evident from the last equation), then $x=y$. Then from the last equation, $x z=3 / 2$. Substituting into the first equation, we get $x^{2}=1 / 2$.

c) Let $u=x+y, v=x y$. Then the equations can be written as $u^{2}+u-2 v=32,12 u=7 v$. Substituting $v$ from the second equation into the first, we get $7 u^{2}-17 u-224=0$, from which $u_{1}=7, u_{2}=-32 / 7$. Accordingly, $v_{1}=12, v_{2}=-384 / 49$. In the first case, $x, y$ are the roots of the quadratic equation $t^{2}-7 t+12=0$, that is, $\{x, y\}=\{3,4\}$; in the second case, we get the quadratic equation $49 t^{2}+224 t-384=0$. Its roots are $\frac{-16 \pm 8 \sqrt{10}}{7}$.

d) The equations can be written as $2\left(x^{3}+y^{3}\right)=7 x y, 2(x+y)=x y$. Dividing the first equation by the second, we get $x^{2}-x y+y^{2}=7$. Let $u=x+y, v=x y$. Then $u^{2}-3 v=7,2 u=v$. Substituting, we get $u^{2}-6 u-7=0$, from which $u_{1}=7, u_{2}=-1$. Accordingly, $v_{1}=14, v_{2}=-2$. The equation $t^{2}-7 t+14=0$, corresponding to the first case, has no roots; in the second case, we get the quadratic equation $t^{2}+t-2=0$. Its roots are 1 and -2.

e) One solution - $\{1,2,-2\}$ - is obvious. There are no other solutions - see the solution of problem $\underline{61039}$.
f) Let $u=x+y, v=x y$. Then the equations can be written as $u^{2}-2 v=12, u+v=9$. Substituting, we get $u^{2}+2 u-30=0$, from which

$u=-1 \pm \sqrt{31} . \quad x, y$ are the roots of a quadratic equation of the form $t^{2}-u t+(9-u)=0$, that is, equal to

$\frac{u \pm \sqrt{u^{2}+4 u-36}}{\Sigma^{2}}=\frac{u \pm \sqrt{2 u-6}}{2}$. The expression under the square root is positive only for the first value of $u=\sqrt{31}^{2}-1$

## Answer

a) $\{1,2,3\}$;
b) $\left( \pm \frac{\sqrt{2}}{2}, \pm \frac{\sqrt{2}}{2}, \pm \frac{3 \sqrt{2}}{2}\right)$
c) $\{3,4\}$,
$\left\{\frac{-16+8 \sqrt{10}}{7}, \frac{-16-8 \sqrt{10}}{7}\right\}$
d) $\{1,-2\}$;
e) $\{1,2,-2\}$;
f) $\left\{\frac{\sqrt{31}-1+\sqrt{2 \sqrt{31}-8}}{2}\right.$
$\left.\frac{\sqrt{31}-1-\sqrt{2 \sqrt{31}-8}}{2}\right\}$