Answer
$x=1; y=3; z=-2$
Work Step by Step
We need to solve the given system of equations:
$x+y-z=6 ~~(1)\\ 3x-2y+z=-5~~(2) \\ x+3y-2z=14~~(3)$
First, we will add equation (1) and equation (2) to eliminate $z$. So, we get:
$4x-y=1~~(4)$
Now, multiply equations (2) by $2$ and add it to equation (3) to eliminate $z$. So, we get:
$6x+x-4y+3y=-10+14 \implies 7x-y=4~~(5)$
Now, solve equations (4) and (5) to solve $x$ and $y$.
$-4x+7x=-1+4 \implies x=1$
Back substitute the value of $x$ into Equation (4) to solve for $y$:
$(4)(1)-y=1 \implies y=3$
Finally, back substitute the values of $x, y$ into Equation (1) to solve for $z$:
$1+3-z=6 \\ z=-2$
So, $x=1; y=3; z=-2$
