Solve the system of equations

$$
\begin{aligned}
& x^{2}+y^{2}=13 \\
& x^{3}+y^{3}=35
\end{aligned}
$$
Solution. Let's introduce new unknowns. Let $a=x+y$ and $b=xy$. Then

$$
x^{2}+y^{2}=(x+y)^{2}-2xy=a^{2}-2b=13
$$

and

$$
x^{3}+y^{3}=(x+y)^{3}-3xy(x+y)=a^{3}-3ab=a\left(a^{2}-3b\right)=35
$$

Thus, the new system of equations is:

$$
\begin{aligned}
a^{2}-2b & =13 \\
a\left(a^{2}-3b\right) & =35
\end{aligned}
$$

From three times the (1) equation: $\quad-6b=3(13-a^{2})$.

From two times the (2) equation: $\quad a(2a^{2}-6b)=70$.

Substituting the previous result into this: $\quad a(2a^{2}+3(13-a^{2}))=70$.

The resulting equation is cubic:

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

Notice that the original system of equations has the solution $x=2$ and $y=3$, so $a=5$ is a solution to the cubic equation.

Divide the left side by $(a-5)$:

$$
\frac{a^{3}-39a+70}{a-5}=a^{2}+5a-14
$$

The solutions to the quadratic equation $a^{2}+5a-14=0$ are $a=2$ and $a=-7$. Therefore, $a^{3}-39a+70=(a-5)(a-2)(a+7)=0$. From (1) we have $b=\frac{a^{2}-13}{2}$, so the solutions are:

$$
\begin{array}{ll}
a=5, & b=6 \\
a=2, & b=-\frac{9}{2} \\
a=-7, & b=18
\end{array}
$$

From these, the appropriate values of $x$ can be found using Vieta's formulas: the solutions to the equation $z^{2}-az+b=0$ are $x$ and $y$ (in any order). Thus,

1. For $a=5, b=6$:

$$
z^{2}-5z+6=0, \quad \text{which gives } x_{1}=2, y_{1}=3 \text{ and } x_{2}=3, y_{2}=2
$$

2. For $a=2, b=-\frac{9}{2}$:

$$
\begin{gathered}
z^{2}-2z-\frac{9}{2}=0, \quad \text{which gives: } \\
x_{3}=1-\frac{\sqrt{22}}{2}, \quad y_{3}=1+\frac{\sqrt{22}}{2} \text{ and } x_{4}=1+\frac{\sqrt{22}}{2}, \quad y_{4}=1-\frac{\sqrt{22}}{2}.
\end{gathered}
$$

3. For $a=-7, b=18$:

$$
z^{2}+7z+18=0
$$

In this case, we do not get real roots.

The obtained solutions satisfy the original system of equations.