Answer
$x=-3; y=2; z=1$ or, $(-3,2,1)$
Work Step by Step
We need to solve the given system of equations:
$$2x+y=-4 ~~~(1)\\ -2y+4z=0 ~~~(2) \\ 3x-2z=-11 ~~~(3)$$
First, we will divide equation $(2)$ by $2$.
$$-y+2z=0 ~~~(4)$$
Add equations $(3)$ and $(4)$ to eliminate $z$:
$$3x-2z-y+2z=-11+0 \\ 3x-y=-11 ~~~(5)$$
Now, add equations $(1)$ and $(5) $to eliminate $x$:
$$2x+y+3x-y=-4-11 \\ 5x= -15 \\ x=\dfrac{-15}{5}\\ x=-3$$
Now, back substitute the value of $x$ into Equation $(1)$ to solve for $y$:
$$(2)(-3)+y=-4 \\ -6+y=-4 \\ y=-4+6 \\ y=2$$
Finally, back substitute the value of $y$ into Equation $(2)$ to solve for $z$:
$$-2(2)+4z=0 \\-4+4z=0 \\ 4z=4 \\ z=1$$
So, the solution is: $x=-3; y=2; z=1$ or, $(-3,2,1)$
