Решить двумя способами: (1-2)x+(1-2)y-(2*(1)-4)z=0 (1+1)x-(1+1)y+(3*(1)+3)z=4 (3*(-1)+3)x+(1+1)y+(1+1)2=4*1+4

  [latex]\left\{\begin{matrix} (1-2)x+(1-2)y-(2*(1)-4)z=0\\ (1+1)x-(1+1)y+(3*(1)+3)z=4\\ (3*(-1)+3)x+(1+1)y+(1+1)z=4*1+4 \end{matrix}\right.\Rightarrow\\\Rightarrow\left\{\begin{matrix} -x-y+2z=0\\ 2x-2y+6z=4\\ 2y+2z=8 \end{matrix}\right.\\ 1\quad cn.:\\ \left\{\begin{matrix} -x-y+2z=0\\ 2x-8+z2+6z=4\\ y=4-z \end{matrix}\right.\Rightarrow\left\{\begin{matrix} -6+4z-4+z+2z=0\\ x=6-4z\\ y=4-z \end{matrix}\right.\Rightarrow\\\Rightarrow\left\{\begin{matrix} z=\frac{10}7\\ x=\frac27\\ y=\frac{18}7 \end{matrix}\right.[/latex] [latex]\left\{\begin{matrix} -x-y+2z=0\\ 2x-2y+6z=4\\ 2y+2z=8 \end{matrix}\right.\Rightarrow\left\{\begin{matrix} -2x-2y+4z=0\\ 2x-2y+6z=4\\ 2y+2z=8 \end{matrix}\right.\Rightarrow\\\Rightarrow\left\{\begin{matrix} -2x-2y+4z=0\\ -4y+10z=4\\ 4y+4z=16 \end{matrix}\right.\Rightarrow\left\{\begin{matrix} -2x-2y+4z=0\\ -4y+10z=4\\ 14z=20 \end{matrix}\right.\Rightarrow\left\{\begin{matrix} x=\frac27\\ y=\frac{18}7\\ z=\frac{10}7 \end{matrix}\right.[/latex]