Question

Solve the following system of linear equations in three variables
$$3x-2y+z=2,2x+3y-z=5,x+y+z=6\text{.}$$

Answer 1

$3x-2y+z=2\dots \left(1\right)2x+3y-z=5\dots \left(2\right)x+y+z=6\dots \left(3\right)$
Substituting $x=1$ in $\left(4\right),5+y=7\Rightarrow y=2$
Substituting $x=1,y=2$ in (3), $1+2+z=6$ we get, $z=3$ Therefore, $x=1,y=2,z=3$