Решить ситему 2х1 + 2х2 - х3=0 3х2+4х3 = -6 х1 + х3 = 1

2х1 + 2х2 - х3=0;

3х2+4х3 = -6;

х1 + х3 = 1;        =>   x3=1-x1

2х1 + 2х2 - (1-x1)=0; => 2x2=1-3x1=>x2=(1-3x)/2

3х2+4х3 = -6;

 x3=1-x1;

x2=(1-3x)/2;

3х2+4х3 = -6;=>3(1-3x/2)+4(1-x1)=-6=>3-9x1/2+10 -4x1=0;=>-17x1+23/2=0=>x1=23/17

 x3=1-x1;

x3=1-23/17=-6/17

x2=1-(23/17)/2=-68/34=-2