## Task 2 - 211212

Determine all ordered pairs $(x ; y)$ of non-zero real numbers $x, y$ that satisfy the following system of equations:

$$
\begin{aligned}
(x+y)^{2}+3(x+y) & =4 \\
\frac{1}{x}+\frac{1}{y} & =-\frac{1}{6}
\end{aligned}
$$
I. If a pair $(x ; y)$ of non-zero real numbers satisfies the system of equations (1), (2), then: The number

$$
z=x+y
$$

satisfies, according to (1), the equation

$$
z^{2}+3 z-4=0
$$

From this, it follows that

$$
z_{1,2}=-\frac{3}{2} \pm \sqrt{\frac{9}{4}+4}=-\frac{3}{2} \pm \frac{5}{2}
$$

thus either $z=1$ or $z=-4$.

From (2), it further follows by multiplying with $6xy$ that

$$
6(x+y)=-x y \quad \text { so } \quad x y=-6 z
$$

In the case $z=1$, (3) and (4) state

$$
x+y=1 \quad ; \quad x y=-6
$$

In the case $z=-4$, (3) and (4) state

$$
x+y=-4 \quad ; \quad x y=24
$$

Substituting $y$ from (5) or (5') into (6) or (6') respectively, we get

$$
x(1-x)=-6 \quad \text { or } \quad x(-4-x)=24
$$

From (7), it follows that

$$
x^{2}-x-6=0 \quad ; \quad x_{1,2}=\frac{1}{2} \pm \sqrt{\frac{1}{4}+6}=\frac{1}{2} \pm \frac{5}{2}
$$

thus either $x=3$ and then, according to (5), $y=-2$ or $x=-2$ and then, according to (5), $y=3$.

From ( $7^{\prime}$ ), however, it follows that $x^{2}+4 x+24=0$, and this equation has no real solutions $x$ because $2^{2}-24<0$. Therefore, only the pairs $(3 ;-2)$ and $(-2 ; 3)$ can satisfy the system of equations.

II. They satisfy this system of equations, as can be shown by a simple check for both pairs due to the commutative law. Therefore, the pairs $(3 ;-2)$ and $(-2 ; 3)$ are the only solutions.