Question

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

Answer 1

 

$left{begin{matrix} (1-2)x+(1-2)y-(2*(1)-4)z=0\ (1+1)x-(1+1)y+(3*(1)+3)z=4\ (3*(-1)+3)x+(1+1)y+(1+1)z=4*1+4 end{matrix}right.Rightarrow\Rightarrowleft{begin{matrix} -x-y+2z=0\ 2x-2y+6z=4\ 2y+2z=8 end{matrix}right.\ 1quad cn.:\ left{begin{matrix} -x-y+2z=0\ 2x-8+z2+6z=4\ y=4-z end{matrix}right.Rightarrowleft{begin{matrix} -6+4z-4+z+2z=0\ x=6-4z\ y=4-z end{matrix}right.Rightarrow\Rightarrowleft{begin{matrix} z=frac{10}7\ x=frac27\ y=frac{18}7 end{matrix}right.$

$left{begin{matrix} -x-y+2z=0\ 2x-2y+6z=4\ 2y+2z=8 end{matrix}right.Rightarrowleft{begin{matrix} -2x-2y+4z=0\ 2x-2y+6z=4\ 2y+2z=8 end{matrix}right.Rightarrow\Rightarrowleft{begin{matrix} -2x-2y+4z=0\ -4y+10z=4\ 4y+4z=16 end{matrix}right.Rightarrowleft{begin{matrix} -2x-2y+4z=0\ -4y+10z=4\ 14z=20 end{matrix}right.Rightarrowleft{begin{matrix} x=frac27\ y=frac{18}7\ z=frac{10}7 end{matrix}right.$