Система уравнений 2ху-xz+yz=2xyz 1/x-1/z=3 1/y+1/z=2

$\left \{ {{2xy-xz+yz=2xyz} \atop { \frac{1}{x} - \frac{1}{z} =3} ||*xyz (x \neq 0, y \neq 0, z \neq 0} \atop { \frac{1}{y}+\frac{1}{z}=2 ||*xyz \right. \\ \left \{ {{2xy-xz+yz=2xyz} \atop { yz - xy =3xyz} ||*(-1)} \atop { xz+xy=2xyz \right. \\\left \{ {{2xy-xz+yz=2xyz} \atop {xy - yz =-3xyz}} \atop { xy+xz=2xyz \right\\\left \{ {{2xy+xy+xy-xz+xz+yz-yz=2xyz-3xyz+2xyz} \atop {xy - yz =-3xyz}} \atop { xy+xz=2xyz \right \\ \left \{ {{4xy=xyz ||/xy} \atop {xy - yz =-3xyz}} \atop { xy+xz=2xyz \right$
$\left \{ {{z=4} \atop {xy - yz =-3xyz} ||/y} \atop { xy+xz=2xyz ||/x}\right\\ \left \{ {{z=4} \atop {x - z =-3xz}} \atop { y+z=2yz\right \\ \left \{ {{z=4} \atop {x - 4 =-3x*4}} \atop { y+4=2y*4\right\\ \left \{ {{4=z} \atop {x+12x=4}} \atop { y-8y=-4\right\\ \left \{ {{z=4} \atop {x= \frac{4}{13} }} \atop { y= \frac{4}{7} \right$
Проверка.
32/7*13-16/13+16/7=128/13*7; 32-16*7+16*13=128; 32+16*6=128; 128=128.
13/4-1/4=3; 3=3.
7/4+1/4=2; 2=2.
Ответ: $x= \frac{4}{13}; y= \frac{4}{7}; z=4.$