You have a system of three linear equations:

1. $2x - y + 3z = 14$
2. $x + y - 2z = -5$
3. $3x + y - z = 2$

Let's solve this system using the substitution or elimination method. We will start with the elimination method for simplicity.

Step 1: Eliminate one variable

From Equation (2), solve for $y$:

$x + y - 2z = -5 \implies y = -x + 2z - 5$

Step 2: Substitute $y$ into the other two equations

Substitute $y = -x + 2z - 5$ into Equation (1) and Equation (3):

For Equation (1): $2x - (-x + 2z - 5) + 3z = 14$ Simplify: $2x + x - 2z + 5 + 3z = 14$ $3x + z + 5 = 14$ $3x + z = 9 \quad \text{(Equation 4)}$

For Equation (3): $3x + (-x + 2z - 5) - z = 2$ Simplify: $3x - x + 2z - 5 - z = 2$ $2x + z - 5 = 2$ $2x + z = 7 \quad \text{(Equation 5)}$

Step 3: Solve the system of two equations (Equation 4 and Equation 5)

Now solve the system:

1. $3x + z = 9$
2. $2x + z = 7$

Subtract Equation (5) from Equation (4): $(3x + z) - (2x + z) = 9 - 7$ $x = 2$

Step 4: Substitute $x = 2$ into one of the simplified equations

Substitute $x = 2$ into Equation (5): $2(2) + z = 7$ $4 + z = 7$ $z = 3$

Step 5: Substitute $x = 2$ and $z = 3$ into the equation for $y$

Substitute $x = 2$ and $z = 3$ into $y = -x + 2z - 5$: $y = -(2) + 2(3) - 5$ $y = -2 + 6 - 5$ $y = -1$

Final Solution:

The solution to the system is: $x = 2, \quad y = -1, \quad z = 3$

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

Here are five related questions:

1. How can I check the solution to a system of linear equations?
2. What is the general method for solving systems with three variables?
3. How do you handle systems of equations with infinitely many solutions?
4. What happens when the determinant of a system of equations is zero?
5. Can you explain the matrix method for solving systems of equations?

Tip: Always check your solution by plugging the values back into the original equations to verify they satisfy all the equations.