Solve the following system of equations:

$$
\begin{aligned}
& x+y=a \ldots \\
& \frac{1}{x}+\frac{1}{y}=\frac{1}{b} \ldots
\end{aligned}
$$

Determine for what values of $a$ and $b$ the roots are real. What relationship must hold between $a$ and $b$ for one of the roots to be three times the other?
$1^{0}$. From equation 2): $\frac{x+y}{x y}=\frac{1}{b}$, so $x y=a b$.

Accordingly, $x$ and $y$ are the roots of the equation

$$
u^{2}-a u+a b=0 \ldots
$$

These are real if

$$
a^{2}-4 a b=a(a-4 b) \geqq 0
$$

Therefore, if $a>0$, it is necessary that $b \leqq \frac{a}{4}$.

If $a<0$, then it is necessary that $b \geqq \frac{a}{4}$.

(In summary, we can say that $b$ and 0 must be on the same side of $\frac{a}{4}$.)

The value pairs that satisfy the system of equations are:

$$
x=u_{1,2}=\frac{1}{2}[a \pm \sqrt{a(a-4 b)}], \quad y=u_{2,1}=\frac{1}{2}[a \mp \sqrt{a(a-4 b)}]
$$

$2^{0}$. If $x=3 y$, then on the one hand $x+y=4 y=a$, and on the other hand $x y=3 y^{2}=a b$ or $3 y^{2}-a b=0$.

Since

$$
y=\frac{a}{4}, \quad 3 \cdot \frac{a^{2}}{16}-a b=0
$$

thus

$$
a(3 a-16 b)=0
$$

We can exclude the case $a=0$, because then $x=y=0$ and thus $3 a-16 b=0$ is the desired relationship between $a$ and $b$. Indeed, if

$$
b=\frac{3 a}{16}, \quad \text { then } \quad a-4 b=a-\frac{3}{4} a=\frac{a}{4}
$$

and thus

$$
x=\frac{1}{2}\left[a+\sqrt{\frac{a^{2}}{4}}\right]=\frac{3 a}{4}, \quad y=\frac{1}{2}\left[a-\sqrt{\frac{a^{2}}{4}}\right]=\frac{a}{4}
$$

Joó Endre (Bencés g. VI. o. Kőszeg.)