## Task A-1.2.

Determine all real solutions of the system of equations

$$
\begin{aligned}
& \frac{x y}{4 y-3 x}=20 \\
& \frac{x z}{2 x-3 z}=15 \\
& \frac{z y}{4 y-5 z}=12
\end{aligned}
$$
## First Solution.

First, note that none of the numbers $x, y, z$ can be equal to 0, because then the left side of one of the equations would be equal to zero.

Let the numbers $a, b$, and $c$ be such that

$$
a=\frac{1}{x}, \quad b=\frac{1}{y}, \quad c=\frac{1}{z}
$$

By taking the reciprocal of both sides of all the equations in the system and substituting the new unknowns, we get

$$
\begin{aligned}
& 4 a-3 b=\frac{1}{20} \\
& 2 c-3 a=\frac{1}{15} \\
& 4 c-5 b=\frac{1}{12}
\end{aligned}
$$

Multiply the first equation by 3 and the second by 4, and then add them. We get

$$
8 c-9 b=\frac{5}{12}
$$

Now subtract from the obtained equation the third equation of the system multiplied by two, we get $b=\frac{1}{4}$.

Substituting the value of $b$ into the system of equations for $a, b$, and $c$, we get: $a=\frac{1}{5}$ and $c=\frac{1}{3}$.

Since $a, b$, and $c$ are the reciprocals of the unknowns $x, y$, and $z$, we get

$$
x=5, \quad y=4, \quad z=3
$$

By checking, we see that none of the denominators on the left sides of the original system are equal to zero, and that the obtained values for $x, y$, and $z$ are indeed (the only) solution to the system.