## Task A-2.4.

Find all pairs $(x, y)$ of real numbers for which

$$
x y^{3}=-135, \quad(x+y) y=-6
$$
## Solution.

Let's introduce the substitution $a=xy, b=y^2$.

Then the equations become $ab=-135, a+b=-6$. According to Viète's formulas, $a$ and $b$ satisfy the quadratic equation $t^2 + 6t - 135 = 0$.

Solving this equation, we get the solutions $t=-15$ and $t=9$.

We have two possibilities: $y^2=9, xy=-15$ or $y^2=-15, xy=9$. We discard the second possibility because the square of a real number cannot be negative.

Therefore, $y^2=9$, so $y=3$ or $y=-3$. From $xy=-15$, we get two solutions, $(x, y)=(-5,3)$ and $(x, y)=(5,-3)$.

1 point