Question

consider the linear system
x+2y+3z=a
x+3y+8z=b
x+2y2z=c
where a,b and c are arbitrary constant.find all solutions of the system

Answer 1

$\begin{cases} x+2y+3z=a \\ x+3y+8z=b\\ x+2y+2z=c \end{cases}$

$\begin{cases} x+2y+3z=a \\ x+3y+8z=b\\ -z=c-a \end{cases}$

$\begin{cases} x+2y+3z=a\\ y+5z=b-a\\ z=a-c \end{cases}$

$\begin{cases} x+2y+3z=a\\ y+5(a-c)=b-a\\ z=a-c \end{cases}$

$\begin{cases} x+2y+3z=a\\ y=b-6a+5c\\ z=a-c \end{cases}$

$\begin{cases} x+2(b-6a+5c)+3(a-c)=a\\ y=b-6a+5c\\ z=a-c \end{cases}$

$\begin{cases} x=10a-2b-7c\\ y=b-6a+5c\\ z=a-c \end{cases}$

Answer:

$x=10a-2b-7c$

$y=b-6a+5c$

$z=a-c$