Solve the following equation:

Question

\begin{align*} 7yz+3zx&=4xy,\\ 21yz-3zx&=4xy,\\ x+2y+3z&=19. \end{align*}

Answer: $(x,y,z)=(7,3,2)$

I'm not able to solve this. Steps?

Answer 1

Hint:

Adding the first two equations we have $$28yz=8xy\qquad\iff\qquad y(7z-2x)=0$$

Now, try two cases:

1) when $y=0$,

2) when $7z-2x=0$,