Answer
$x=1; y=3; z=-2$
Work Step by Step
We need to solve the given system of equations:
$x-y+z=-4 ~~(1)\\ 2x-3y+4z=-15~~(2) \\ 5x+y-2z=12~~(3)$
First, we will add equation (1) and equation (3) to eliminate $z$. So, we get:
$6x-z=8~~(4)$
Now, multiply equation (3) by $3$ and add it to equation (2) to eliminate $y$. So, we get:
$2x+15x+4z-6z=-15+36 \implies 17x-2z=21~~(5)$
Now, simplify equations (4) and (5) to solve $x$ and $z$.
$-12x+17x=-16+21 \implies x=1$
Back substitute the value of $x$ into Equation (4) to solve for $y$:
$(6)(1)-z=8 \implies z=-2$
Finally, back substitute the values of $x, z$ into Equation (1) to solve for $y$:
$1-y+(-2)=-4 \implies y=3$
So, $x=1; y=3; z=-2$
