Solve the following system of equations:

$$
\begin{gathered}
8 x^{2}-26 x y+15 y^{2}+116 x-150 y+360=0 \\
8 x^{2}+18 x y-18 y^{2}+60 x+45 y+108=0
\end{gathered}
$$
Both given equations can easily be factored on the left side. Indeed, considering $y$ as a parameter, the equations can be written as:

$$
\begin{aligned}
& 8 x^{2}-2(13 y-58) x=-\left(15 y^{2}-150 y+360\right) \\
& 8 x^{2}+6(3 y+10) x=9\left(2 y^{2}-5 y-12\right)
\end{aligned}
$$

and by taking half of each side, both sides can be completed to perfect squares. From $\left(1^{\prime}\right)$:

$$
\left(2 x-\frac{13 y-58}{4}\right)^{2}=\left(\frac{13 y-58}{4}\right)^{2}-\frac{1}{2}\left(15 y^{2}-150 y+360\right)=\left(\frac{7 y-22}{4}\right)^{2}
$$

thus

$$
\left(\frac{8 x-13 y+58}{4}\right)^{2}-\left(\frac{7 y-22}{4}\right)^{2}=\frac{1}{2}(2 x-5 y+20)(4 x-3 y+18)=0
$$

and similarly $\left(2^{\prime}\right)$ becomes:

$$
\frac{1}{2}(4 x-3 y+12)(2 x+6 y+9)=0
$$

Both $\left(1^{\prime \prime}\right)$ and $\left(2^{\prime \prime}\right)$ can be satisfied in two ways, depending on whether the first or second factor on the left side is 0. By pairing these conditions, we get 4 systems of linear equations with two unknowns:
I. $\left\{\begin{array}{l}2 x-5 y+20=0, \\ 4 x-3 y+12=0\end{array}\right.$
II. $\left\{\begin{array}{l}2 x-5 y+20=0 \\ 2 x+6 y+9=0\end{array}\right.$
III. $\left\{\begin{array}{l}4 x-3 y+18=0, \\ 4 x-3 y+12=0 ;\end{array}\right.$
IV. $\left\{\begin{array}{l}4 x-3 y+18=0, \\ 2 x+6 y+9=0\end{array}\right.$

System III is contradictory, and from the other three systems, we get one solution pair each:

$\begin{array}{ll}\text { I. } x=0, y=4 ; & \text { II. } x=-7.5, y=1 ; \quad \text { IV. } x=-4.5, y=0 .\end{array}$

András Krasznai (Gyöngyös, Vak Bottyán Gymn., III. grade)

János Pető (Budapest, Kölcsey F. Gymn., III. grade)

Remarks. 1. The images of equations (1) and (2) in the Cartesian coordinate system are two intersecting pairs of lines.

![](https://cdn.mathpix.com/cropped/2024_05_02_0a686251b32f9c65f952g-1.jpg?height=724&width=1049&top_left_y=1622&top_left_x=533)

2. Of course, we perform the above preparatory calculations even if we want to solve $\left(1^{\prime}\right)$ and $\left(2^{\prime}\right)$ for $x$ by forming the discriminant and finding it to be a perfect square in both cases; after all, the quadratic formula is also derived by completing the square.