a) Find all solutions of the system of equations (1)-(3). - Furthermore, provide a system of equations that differs from the given one by exactly one coefficient and

b) has infinitely many solutions,

c) which is not satisfied by any triple of real numbers $x, y, z$.

$$
\begin{gathered}
2 x+3 y+z=1 \\
4 x-y+2 z=2 \\
8 x+5 y+3 z=4
\end{gathered}
$$
a) We notice that (1) and (2) form a system of equations with two unknowns, $2x + z = u$ and $y$. Subtracting (2) from twice (1), we get $7y = 0$, hence $y = 0$, and thus from (1),

$$
u = 2x + z = 1
$$

Based on this, from (3),

$$
(8x + 4z) - z = 4u - z = 4, \quad z = 0
$$

Finally, from (4), $x = 1/2$.

Accordingly, the only solution can be $x = 1/2, y = z = 0$, and it is easily verified that this solution satisfies the system.

b) When transforming (2) and (1) as above, $y$ also cancels out if, for example, we replace $-y$ in (2) with $6y$, i.e.,

$$
4x + 6y + 2z = 2
$$

Thus, $\left(2^{*}\right)$ merely repeats the requirement of (1) and can be ignored. By arbitrarily choosing the value of $y$, from (1) we get $u = 1 - 3y$, and thus from (3), and again from (1),

$$
\begin{aligned}
-z = 4 - 5y - 4u = 7y, & \quad x = \frac{u - z}{2} = 2y + \frac{1}{2}, \quad \text{that is} \\
x = 2y + 1/2, & \quad y = y, \quad z = -7y
\end{aligned}
$$

satisfies the system $(1), \left(2^{*}\right), (3)$ for any $y$.

c) As we have seen, the system (1), (2) uniquely determines $y = 0, u = 1$. This does not satisfy the equation

$$
3u + 5y = 4, \quad \text{that is} \quad 6x + 5y + 3z = 4
$$

Therefore, no triplet of numbers $x, y, z$ satisfies the system (1), (2), $\left(3^{**}\right)$.

Lempert László (Budapest, Radnóti M. Gymnasium II. Class)

Note. We can also obtain infinitely many solutions to the b) question in a different form, for example, if we replace (3) with $8x + 5y + 4z = 4$, which is the sum of twice (1) and (2). By arbitrarily choosing $x$,

$$
x = x, \quad y = 0, \quad z = 1 - 2x
$$

satisfies the modified system.