Answer
$x=8; y=2; z=0$ or, $(8,2,0)$
Work Step by Step
We need to solve the given system of equations:
$$x-y=6 ~~~~ ( 1)\\ 2x-3z=16 ~~~(2) \\ 2y+z=4 ~~~~(3)$$
First, we will multiply equation $(3)$ by $3$ to get:
$$ 6y+3z=12\quad \quad(4)$$
Now, add equations $(2)$ and $(4)$ to obtain:
$$2x-3z+6y+3z=16+12\\ 2x+6y=28 ~~~~(5)$$
In order to solve for $x$, we will multiply equation $(1)$ by $6$ and then add the new equation and equation $(5)$ to get:
$$6x-6y+2x+6y=36+28 \\ 8x =64 \\ x=8$$
Now, back substitute $x$ into Equation $(1)$ to solve for $y$:
$$8-y=6 \\ y=8-6 \\ y=2$$
Finally, back substitute $y$ into Equation $(3)$ to solve for $z$:
$$(2)(2)+z=4 \\ 4+z=4 \\ z=4-4 \\ z=0$$
So, the solution is $x=8; y=2; z=0$ or, $(8,2,0)$
