срочно!!!! очень надо!!! 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
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=1-23/17=-6/17
x2=1-(23/17)/2=-68/34=-2
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Answers & Comments
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