Answer
$x=-3; y=\dfrac{1}{2}; z=1$
Work Step by Step
We need to solve the given system of equations:
$x+2y-z=-3 ~(1)\\ 2x-4y+z=-7~(2) \\ -2x+2y-3z=4~(3)$
First, we will add equation (2) and equation (3) to eliminate $x$. So, we get:
$-2y-2z=-3(4)$
Now, multiply equation (2) by $2$ and add it to equation (3) to eliminate $x$. So, we get:
$4y+2y-2z-3z=-6+4 \implies 6y-5z=-2~~(5)$
Now, simplify equations (4) and (5) to solve $y$ and $z$.
$-6z-5z=-9-2 \implies z=1$
Back substitute the value of $z$ into Equation (4) to solve for $y$:
$2y=3-2z \\ 2y=3-(2)(1) \implies y=\dfrac{1}{2}$
Finally, back substitute the values of $y, z$ into Equation (1) to solve for $y$:
$x+(2)(\dfrac{1}{2})-1=-3 \implies x=-3$
So, $x=-3; y=\dfrac{1}{2}; z=1$
