Answer
$x=3; y=-\dfrac{8}{3}; z=\dfrac{1}{9}$
Work Step by Step
We need to solve the given system of equations:
$x+4y-3z=-8 ~~(1)\\ 3x-y+3z=12~~(2) \\ x+y+6z=1~~(3)$
First, we will add equation (2) and equation (3) to eliminate $y$. So, we get: $4x+9z=13~~(4)$
Now, multiply equation (2) by $4$ and add it to equation (1) to eliminate $y$. So, we get: $13x+9z=40~~(5)$
Now, simplify equations (4) and (5) to solve $x$ and $z$.
$-4x+13x=13+40 \implies x=3$
Back substitute the value of $z$ into Equation (4) to solve for $y$:
$(4)(3)+9z=13 \implies z=\dfrac{1}{9}$
Finally, back substitute the values of $x, z$ into Equation (3) to solve for $y$:
$3+y+(6)(\dfrac{1}{9})=1 \implies y=-\dfrac{8}{3}$
So, $x=3; y=-\dfrac{8}{3}; z=\dfrac{1}{9}$
