Solve the following system of equations:

$$
\frac{x}{a}=\frac{y}{b}=\frac{z}{c}, x+y+z=d
$$
First solution. From the first equation,

$$
y=\frac{x}{a} b \quad \text { and } \quad z=\frac{x}{a} c
$$

Substituting these values into the second equation,

$$
x+\frac{x}{a} b+\frac{x}{a} c=d
$$

which gives

$$
x=a \frac{d}{a+b+c}
$$

Furthermore,

$$
y=\frac{b}{a} x=b \frac{d}{a+b+c}
$$

and

$$
z=\frac{c}{a} x=c \frac{d}{a+b+c}
$$

(Tibor Pálos, Budapest.)

Second solution. If

$$
\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=r
$$

then

$$
x+y+z=r(a+b+c)=d
$$

so

$$
r=\frac{d}{a+b+c}
$$

and thus

$$
x=a \frac{d}{a+b+c}, \quad y=b \frac{d}{a+b+c} \text { and } z=c \frac{d}{a+b+c} .
$$

Number of solutions: 36.