We are given the following system of equations:

$1) \quad y + z - 2w = 0$ $2) \quad 2x - 3y - 3z + 6w = 2$ $3) \quad 4x + y + z - 2w = 4$

Step 1: Solve the first equation for $y$

From the first equation $y + z - 2w = 0$, we can solve for $y$:

$y = 2w - z$

Step 2: Substitute $y = 2w - z$ into the second and third equations

Substituting into the second equation:

$2x - 3(2w - z) - 3z + 6w = 2$ Simplify the expression: $2x - 6w + 3z - 3z + 6w = 2$ $2x = 2$ $x = 1$

Substituting into the third equation:

$4x + (2w - z) + z - 2w = 4$ Substitute $x = 1$: $4(1) + 2w - z + z - 2w = 4$ $4 = 4$

This is an identity, meaning the third equation does not provide any additional information, and the solution depends on $w$ and $z$.

Step 3: Express the general solution

Since $y = 2w - z$ and $x = 1$, we have a solution in terms of the free variables $w$ and $z$. The general solution is:

$x = 1, \quad y = 2w - z, \quad z = z, \quad w = w$

Final Answer:

The solution to the system is:

$x = 1, \quad y = 2w - z, \quad z = z, \quad w = w$

This represents a family of solutions parameterized by $w$ and $z$.

Would you like a more detailed explanation or have any questions?

Here are 5 related questions for further exploration:

1. How would the solution change if we add another equation to this system?
2. Can this system of equations be represented in matrix form?
3. How can we interpret free variables geometrically in a system of linear equations?
4. How do you determine if a system of linear equations has no solution, one solution, or infinitely many solutions?
5. What methods can be used to solve systems of equations with more than four variables?

Tip: When dealing with underdetermined systems (more unknowns than equations), expect a family of solutions dependent on free variables.