Solve the following system of equations:

$$
\begin{aligned}
& x^{2}+y^{2}=13 \\
& x^{3}+y^{3}=35
\end{aligned}
$$
$$
\begin{aligned}
& x^{2}+y^{2}=13, \\
& x^{3}+y^{3}=35 .
\end{aligned}
$$

1. Our system can be easily transformed so that the sum $x+y=u$ and the product $x y=v$ are considered as new unknowns:

$$
\begin{gathered}
x^{2}+y^{2}=(x+y)^{2}-2 x y=u^{2}-2 v=13 \\
x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)=u^{3}-3 u v=35
\end{gathered}
$$

From here, eliminating $v$:

$$
u^{3}-39 u+70=0
$$

Notice that $u_{1}=2$ satisfies the equation, and thus the left side can be factored as $(u-2)$ and a quadratic polynomial:

$$
(u-2)\left(u^{2}+2 u-35\right)=0 .
$$

This yields two more roots from the equation:

$$
u^{2}+2 u-35=0
$$

which are $u_{2}=5, u_{3}=-7$. Therefore, the left side of (3) can be written as:

$$
(u-2)(u-5)(u+7)
$$

This cannot be zero if $u$ differs from 2, 5, and -7, so (3) has no more roots.

The corresponding values of $v$ are $v_{1}=\left(u_{1}^{2}-13\right) / 2=-4.5 ; v_{2}=6, v_{3}=18$.

2. Now, from the system

$$
\begin{gathered}
x+y=u, \\
x y=v
\end{gathered}
$$

eliminating $y$:

$$
x^{2}-u x+v=0, \quad x_{1,2}=\frac{1}{2}\left(u \pm \sqrt{u^{2}-4 v}\right) .
$$

Substituting the corresponding pairs of $u$ and $v$, we get two pairs of solutions for the original system based on the first two pairs:

$$
\begin{array}{lll}
u=2, v=-4.5 \text { gives: } & x_{1}=3.345, & y_{1}=-1.345 \\
& x_{2}=-1.345, & y_{2}=3.345 \\
u=5, v=6 \text { gives: } & x_{3}=3, & y_{3}=2 \\
& x_{4}=2, & y_{4}=3
\end{array}
$$

The value pair $u=-7, v=18$ does not yield real values for $x, y$. - This completes the solution.

Remark. It is easy to recognize that the integer pairs $(2,3)$ and $(3,2)$ satisfy the system. Based on this, we can continue the solution graphically. (1) is satisfied by the points on the circle centered at the origin with radius $r=\sqrt{13}$, and (2) by the points $(\sqrt[3]{35}=3.271 ; 0)$ and $(0 ; 3.271)$ on the axes, which are inside the circle, since $\sqrt{13}=3.605$ > 3.271.

![](https://cdn.mathpix.com/cropped/2024_05_02_4c7057ea8c89b2375de7g-2.jpg?height=995&width=1152&top_left_y=126&top_left_x=473)

However, we expect that the curve representing (2) will exit the circle, as for every real $x$ there is exactly one point on the curve according to $y=\sqrt[3]{35-x^{3}}$. We indeed find an exit point between $x=3.3$ and $x=3.4$ - and by the symmetry of the equations, the fourth solution also emerges by swapping $x$ and $y$.

Thus, we can conclude that there are no further solutions, as the curve representing (2) is monotonically decreasing.