We have a system of three linear equations:

$2x - y + z = 1 \quad \text{(Equation 1)}$ $3x - 3y + 4z = 5 \quad \text{(Equation 2)}$ $4x - 2y + 3z = 4 \quad \text{(Equation 3)}$

Let's solve this system step by step using substitution or elimination. First, we will use the method of elimination.

Step 1: Eliminate one variable

To eliminate $y$, let's first manipulate equations (1) and (2).

Multiply equation (1) by 3 to align the coefficients of $y$:

$6x - 3y + 3z = 3 \quad \text{(Equation 4)}$

Now subtract equation (2) from equation (4):

$(6x - 3y + 3z) - (3x - 3y + 4z) = 3 - 5$ $6x - 3y + 3z - 3x + 3y - 4z = -2$ $3x - z = -2 \quad \text{(Equation 5)}$

Step 2: Eliminate $y$ again with equations (1) and (3)

Multiply equation (1) by 2 to align the coefficients of $y$:

$4x - 2y + 2z = 2 \quad \text{(Equation 6)}$

Now subtract equation (3) from equation (6):

$(4x - 2y + 2z) - (4x - 2y + 3z) = 2 - 4$ $4x - 2y + 2z - 4x + 2y - 3z = -2$ $-z = -2$ $z = 2$

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

Now that we know $z = 2$, substitute this into equation (5):

$3x - 2 = -2$ $3x = 0$ $x = 0$

Now, substitute $x = 0$ and $z = 2$ into equation (1):

$2(0) - y + 2 = 1$ $-y + 2 = 1$ $-y = -1$ $y = 1$

Final Solution:

The solution to the system of equations is: $x = 0, \quad y = 1, \quad z = 2$

Would you like more details on the steps or have any further questions?

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Here are five related questions that expand on this topic:

1. How do you solve a system of linear equations using the substitution method?
2. What happens if there is no unique solution to the system?
3. How would you solve this system using matrices (Gauss elimination)?
4. What is the geometric interpretation of a system of three linear equations in 3D space?
5. Can this system of equations have infinite solutions? If so, how can we determine that?

Tip: If you have more than three equations, consider using matrix methods or software tools like Python, MATLAB, or a graphing calculator to solve the system more efficiently!