Solve the following system of equations:

$$
\begin{aligned}
& |x-3|=5-y \\
& |y-5|+|x-3|=x+y .
\end{aligned}
$$
I. solution. (1) the left side is non-negative, so the right side is also: $5-y \geq 0$ and thus (2) the first term on the left side is equal to the left side of (1). Adding the two equations, $y$ cancels out:

$$
3 \cdot|x-3|=x+5
$$

From here, depending on whether

$$
\begin{gathered}
x<3, \quad \text { or } \quad x \geq 3, \\
3(3-x)=x+5, \quad 3(x-3)=x+5 \\
x_{1}=1, \quad x_{2}=7 .
\end{gathered}
$$

Thus, according to (1),

$$
y_{1}=3, \quad y_{2}=1,
$$

so the system of equations is satisfied by two pairs of $x, y$ values.

Attila Herczegh (Budapest, Konyves Kalman Gymnasium, 3rd grade)

II. solution. Let's plot our equations in the $x, y$ Cartesian coordinate system. (1) in the case of $x-3<0$ takes the form $y=x+2$, and in the case of $x-3 \geq 0$ it takes the form $y=8-x$. The common starting point of the two half-lines is the point $x=3, y=5$ (the thick broken line in the diagram).

![](https://cdn.mathpix.com/cropped/2024_05_02_df20e869379887cf2ecdg-1.jpg?height=445&width=637&top_left_y=1008&top_left_x=728)

(2) the left side cannot have non-negative numbers in both terms, otherwise we would end up with the contradiction $-8=0$. The form of the equation is

$$
\begin{array}{lclll}
x-3 \geq 0 & \text { and } & y-5<0 & \text { in which case } & y=1 \\
x-3<0 & \text { and } & y-5<0 & \text { in which case } & y=4-x \\
x-3<0 & \text { and } & y-5 \geq 0 & \text { in which case } & x=-1
\end{array}
$$

in the plot, two half-lines connect to the endpoints of a segment, to $(3,1)$ and $(-1,4)$ (drawn with a thick dashed line in the diagram; the thin dashed lines delimit the regions of the plane corresponding to the 4 possible sign variations of the expressions $x-3$ and $y-5$).

The two broken lines have two common points, the coordinate pairs of these points give the solutions to our system of equations:

$$
M_{1} \text { gives } \quad x_{1}=1, \quad y_{1}=3, \quad M_{2} \text { gives } \quad x_{2}=7, \quad y_{2}=1 \text {. }
$$

Oszkar Biro (Budapest, 11th district, Vali Uti Secondary School, 8th grade)