Có 3.(x-1)=2.(y-2)
=> $\frac{y-2}{3}$ = $\frac{x-1}{2}$
Có 4.(y-2)=3.(z-3)
=> $\frac{z-3}{4}$ = $\frac{y-2}{3}$
=> $\frac{y-2}{3}$ = $\frac{x-1}{2}$= $\frac{z-3}{4}$
Có 2x + 3y -z = 50 => 2x+ 3y= 50+z
Áp dụng TC dãy tỉ số bằng nhau
$\frac{y-2}{3}$ = $\frac{x-1}{2}$= $\frac{3y-6}{9}$ = $\frac{2x-2}{4}$= $\frac{2x-2+3y-6}{9+4}$= $\frac{50+z-8}{13}$= $\frac{42+z}{13}$
=> $\frac{z-3}{4}$= $\frac{42+z}{13}$
=> 13(z-3)= 4(42+z)
<=> 13z- 39= 168+ 4z
<=> 9z= 207
<=> z= 23
=> y= 17, x= 11
Vậy z=23, y= 17, x=11
