The system of equations from the image is:

$-4x + 6y + 3z = -10$ $4x - 10y - z = -10$ $-4x + y - z = -9$

Let's solve this system step-by-step using either substitution or elimination. I'll proceed with the elimination method.

Step 1: Add the first and third equations

By adding these two equations, we aim to eliminate $x$.

$(-4x + 6y + 3z) + (-4x + y - z) = -10 + (-9)$ Simplifying:

$-8x + 7y + 2z = -19$

This is our new equation (Equation 4): $-8x + 7y + 2z = -19 \quad \text{(Equation 4)}$

Step 2: Add the second and third equations

We also eliminate $x$ from the second and third equations.

$(4x - 10y - z) + (-4x + y - z) = -10 + (-9)$ Simplifying:

$-9y - 2z = -19$ This is our new equation (Equation 5): $-9y - 2z = -19 \quad \text{(Equation 5)}$

Step 3: Solve for $y$ and $z$

Now we solve the system formed by Equations 4 and 5:

From Equation 5: $-9y - 2z = -19 \quad \text{(Equation 5)}$ Solve for $z$: $z = \frac{-19 + 9y}{-2} = \frac{19 - 9y}{2}$

Step 4: Substitute $z$ back into Equation 4 and solve for $y$

I will calculate the solution next.It seems I made a mistake in interpreting the structure of the system in terms of substitution.

I will recalculate the solution