Solve the following system of equations - graphically as well:

$$
x+|y|=3, \quad 2|x|-y=3
$$
I. solution. Let's examine the 4 possible cases based on the signs of $x$ and $y$ separately. We know that

$$
\begin{array}{lll}
\text { if } u \geq 0, & \text { then } & |u|=u \text { and } \\
\text { if } u<0, & \text { then } & |u|=-u
\end{array}
$$

1) If $x \geq 0$, then $y \geq 0$, the system is

$$
x+y=3, \quad 2 x-y=3
$$

from which, in the usual way, $x=2, y=1$, both positive, we have a solution. Similarly,
2) $\left.\begin{array}{l}x \geq 0 \\ y<0\end{array}\right\}$ in the case of $\left.\quad \begin{array}{r}x-y=3 \\ 2 x-y=3\end{array}\right\}$, we get $\quad x=0, \quad y=-3$, a solution;
3) $\left.\begin{array}{l}x<0 \\ y \geq 0\end{array}\right\}$ in the case of $\left.\quad \begin{array}{r}x+y=3 \\ -2 x-y=3\end{array}\right\}$, we get $\quad x=-6, \quad y=9$, a solution;
4) $\left.\begin{array}{l}x<0 \\ y<0\end{array}\right\}$ in the case of $\left.\quad \begin{array}{r}x-y=3 \\ -2 x-y=3\end{array}\right\}$, we get $\quad x=0, \quad y=-3$,

here $x$ does not meet the assumption, so there is no solution that satisfies the assumption $x<0, y<0$.

According to these, the system has 3 solutions:

$$
\begin{array}{ll}
x_{1}=2, & y_{1}=1 \\
x_{2}=-6, & y_{2}=9 \\
x_{3}=0, & y_{3}=-3
\end{array}
$$

The graphical solution can be performed on a diagram, as the assumptions 1)-4) use only one quadrant of the Cartesian coordinate system, and these quadrants do not overlap each other.

![](https://cdn.mathpix.com/cropped/2024_05_02_1c3afefc6519d99d3391g-1.jpg?height=856&width=639&top_left_y=1236&top_left_x=732)

The equation $x+|y|=3$ is represented by the continuously drawn graph on the diagram, and the equation $2|x|-y=3$ by the dashed one. The common points $M_{1}, M_{2}, M_{3}$ correspond to the above 3 solutions.

Valkó Ágnes (Budapest, Szilágyi E. lg. II. o. t.)

Remark. According to the first equation, $|y|=3-x \geq 0$, which shows that the graph has points only at $x \leq 3$ abscissas, in other words: the function $|y|=3-x$ is only defined at $x \leq 3$. The graph is naturally symmetric with respect to the $X$-axis. - Similarly, from the second equation, $|x|=(3+y) / 2$, so the function $y=2|x|-3$ has a value set of $y \geq-3$, and its graph is symmetric with respect to the $Y$-axis.

Márki László (Budapest, Fazekas M. gyak. g. II. o. t.)

II. solution. We can remove the absolute value sign by squaring and taking the square root, as $|x|=\sqrt{x^{2}}$.

Substituting $y=2|x|-3$ from the second equation into the first one, we get

$$
\begin{gathered}
|2| x|-3|=3-x \\
\sqrt{\left(2 \sqrt{x^{2}}-3\right)^{2}}=3-x \\
4 x^{2}-12 \sqrt{x^{2}}+9=9-6 x+x^{2} \\
x^{2}+2 x=4 \sqrt{x^{2}} \\
x^{4}+4 x^{3}-12 x^{2}=x^{2}\left(x^{2}+4 x-12\right)=x^{2}(x+6)(x-2)=0
\end{gathered}
$$

Therefore, if the system of equations has a solution, the value of $x$ can only be $0, -6$, or 2. We calculate the corresponding $y$ values from the second equation: $-3, 9$, and 1. The pairs $(0,-3), (-6,9), (2,1)$ also satisfy the first equation, so they are solutions of the system.

Óhegyi Ernö (Budapest, Rákóczi F. g. I. o. t.)

Remark. The solution of the quartic equation was facilitated by the fact that we could factor out the $x^{2}$ term. If we eliminate $x$, we similarly arrive at the equation

$$
y^{4}-4 y^{3}-42 y^{2}-36 y+81=0
$$

By searching for integer roots, this can also be solved.