We are given a system of three equations with three variables $x$, $y$, and $z$: $3xy = 2(x+y)$ $5yz = 6(y+z)$ $4zx = 3(z+x)$ We need to find the values of $x$, $y$, and $z$.

1. Problem Description

We are given a system of three equations with three variables $x$, $y$, and $z$:

$3xy = 2(x+y)$

$5yz = 6(y+z)$

$4zx = 3(z+x)$

We need to find the values of $x$, $y$, and $z$.

2. Solution Steps

First, we rewrite the equations:

$3xy = 2x + 2y$

$5yz = 6y + 6z$

$4zx = 3z + 3x$

Now, divide each equation by $xy$, $yz$, and $zx$ respectively:

$\frac{3xy}{xy} = \frac{2x}{xy} + \frac{2y}{xy} \Rightarrow 3 = \frac{2}{y} + \frac{2}{x}$

$\frac{5yz}{yz} = \frac{6y}{yz} + \frac{6z}{yz} \Rightarrow 5 = \frac{6}{z} + \frac{6}{y}$

$\frac{4zx}{zx} = \frac{3z}{zx} + \frac{3x}{zx} \Rightarrow 4 = \frac{3}{x} + \frac{3}{z}$

Let $a = \frac{1}{x}$, $b = \frac{1}{y}$, and $c = \frac{1}{z}$. Then the equations become:

$3 = 2b + 2a$

$5 = 6c + 6b$

$4 = 3a + 3c$

Dividing the equations by 2, 6, and 3 respectively, we get:

$1.5 = a + b$

$\frac{5}{6} = b + c$

$\frac{4}{3} = a + c$

We have the system:

$a + b = \frac{3}{2}$ (1)

$b + c = \frac{5}{6}$ (2)

$a + c = \frac{4}{3}$ (3)

Subtract equation (2) from equation (1):

$a - c = \frac{3}{2} - \frac{5}{6} = \frac{9-5}{6} = \frac{4}{6} = \frac{2}{3}$ (4)

Add equation (3) and equation (4):

$2a = \frac{4}{3} + \frac{2}{3} = \frac{6}{3} = 2$

$a = 1$

From equation (1):

$b = \frac{3}{2} - a = \frac{3}{2} - 1 = \frac{1}{2}$

From equation (3):

$c = \frac{4}{3} - a = \frac{4}{3} - 1 = \frac{1}{3}$

Since $a = \frac{1}{x}$, $b = \frac{1}{y}$, and $c = \frac{1}{z}$, we have:

$x = \frac{1}{a} = \frac{1}{1} = 1$

$y = \frac{1}{b} = \frac{1}{\frac{1}{2}} = 2$

$z = \frac{1}{c} = \frac{1}{\frac{1}{3}} = 3$

3. Final Answer

$x = 1$, $y = 2$, $z = 3$