You have a system of three equations:

1. $x = 4$
2. $x + y = -6$
3. $4x - 3y + 2z = 26$

Step 1: Solve for $y$

From equation (1), we know $x = 4$. Substitute this into equation (2):

$4 + y = -6$

Solve for $y$:

$y = -6 - 4 = -10$

Step 2: Solve for $z$

Now substitute $x = 4$ and $y = -10$ into equation (3):

$4(4) - 3(-10) + 2z = 26$

Simplify the equation:

$16 + 30 + 2z = 26$

$46 + 2z = 26$

Solve for $z$:

$2z = 26 - 46 = -20$

$z = \frac{-20}{2} = -10$

Final Solution:

• $x = 4$
• $y = -10$
• $z = -10$

Would you like more details on any step? Here are some follow-up questions:

1. How do we substitute values in systems of equations?
2. What are the different methods for solving systems of equations?
3. Can substitution always be used to solve such systems?
4. How can I verify the solution to a system of equations?
5. What happens if a system has no solution or infinitely many solutions?

Tip: Always check your solutions by plugging them back into the original equations.