$\Large\frac{x}{4}$ = $\Large\frac{y}{3}$
<=> $x$ = $\Large\frac{4y}{3}$
=> $x+y+z =-6$
<=> $\Large\frac{4y}{3}$$+y+z =-6$
<=> $\Large\frac{7y}{3}$$+z =-6$ (1)
.
+) $\Large\frac{y}{2}$ = $\Large\frac{z}{3}$
<=> $y$ = $\Large\frac{2z}{3}$ (2)
Thay (2) vào (1) ta có:
$\Large\frac{\Large\frac{14z}{3}}{3}$$+z =-6$
$\Large\frac{14z}{9}$$+z =-6$
=> z= $\Large\frac{-27}{7}$
=> y= $\Large\frac{-18}{7}$
=> x= $\Large\frac{-24}{7}$
