Solve the following system of equations:

$$
\begin{aligned}
a x+b y-c z & =a b \ldots \\
3 a x-b y+2 c z & =a(5 c-b) \ldots \\
3 y+2 z & =5 a \ldots
\end{aligned}
$$

What case arises if $a=0$ or if $8 b+15 c=0$?
To eliminate $z$, we multiply the terms of 3) by -c and add them to the terms of 2); then, multiplying the terms of 1) by 2, we also subtract them from the terms of 2). This results in:

$$
3 a x-(3 c+b) y=-a b \ldots
$$

and

$$
5 a x+b y=5 a c+a b \ldots
$$

or

$$
3(a x-c y)=b(y-a) \ldots
$$

or

$$
5 a(x-c)+b(y-a)=0 \ldots
$$

Clearly, 5a) is satisfied by: $x=c, y=a$; but this pair of values also satisfies 4a). Now from 3): $z=a$. According to this, the solution of the system of equations is:

$$
x=c, \quad y=a, \quad z=a
$$

This system of equations has only this one solution and it is determined, except when (based on 4) and 5))

$$
3 a b+5 a(3 c+b)=a(8 b+15 c)=0
$$

If $a=0$, then our system of equations becomes:

$$
b y-c z=0, \quad-b y+2 c z=0, \quad 3 y+2 z=0
$$

The solution to this is: $y=0, z=0$ for any values of $b$ and $c$. The value of $x$ is arbitrary. The system of equations has infinitely many solutions; thus, we are dealing with an indeterminate case.

(If $6=c=0$, then $y$ and $z$ can take any value.)

If, however, $8 b+15 c=0$, that is, $c=-\frac{8 b}{15}$, then from 4) and 5) we get:

$$
5 a x+b y=-\frac{5 a b}{3} \quad \text { and } \quad 5 a x+b y=-\frac{5 a b}{3}
$$

thus we are again faced with an indeterminate case.