We are given the following system of equations:

1. $x - 2y + 4z = 3$
2. $x + 3y - 2z = 6$
3. $x - 4y + 3z = -5$

To solve this system of equations, we will use the substitution method or elimination method. Let's use the elimination method for this case.

Step 1: Eliminate one variable

First, we'll eliminate $x$ by subtracting the first equation from the second equation and subtracting the first equation from the third equation.

Subtract equation 1 from equation 2:

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

Subtract equation 1 from equation 3:

$(x - 4y + 3z) - (x - 2y + 4z) = -5 - 3$ Simplifying: $x - x - 4y + 2y + 3z - 4z = -8$ $-2y - z = -8 \quad \text{(Equation 5)}$

Step 2: Solve the new system of two equations

Now we have the following system of two equations:

1. $5y - 6z = 3$
2. $-2y - z = -8$

Solve for $y$ in terms of $z$ from equation 5:

$-2y - z = -8 \quad \Rightarrow \quad -2y = z - 8 \quad \Rightarrow \quad y = \frac{8 - z}{2}$

Substitute $y = \frac{8 - z}{2}$ into equation 4:

Substitute into $5y - 6z = 3$: $5 \left( \frac{8 - z}{2} \right) - 6z = 3$ Simplifying: $\frac{5(8 - z)}{2} - 6z = 3$ $\frac{40 - 5z}{2} - 6z = 3$ Multiply everything by 2 to clear the fraction: $40 - 5z - 12z = 6$ $40 - 17z = 6$ $-17z = 6 - 40$ $-17z = -34$ $z = 2$

Step 3: Solve for $y$

Substitute $z = 2$ back into $y = \frac{8 - z}{2}$: $y = \frac{8 - 2}{2} = \frac{6}{2} = 3$

Step 4: Solve for $x$

Now that we know $y = 3$ and $z = 2$, substitute these values into one of the original equations. Let's use the first equation $x - 2y + 4z = 3$: $x - 2(3) + 4(2) = 3$ $x - 6 + 8 = 3$ $x + 2 = 3$ $x = 1$

Final Solution:

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

Would you like a detailed explanation of any step, or do you have other questions?

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Here are some related questions that might help deepen your understanding:

1. How do you apply the substitution method to solve a system of linear equations?
2. What is the difference between the elimination method and substitution method in solving systems of equations?
3. How do you check the solution to a system of linear equations?
4. What happens if the system of equations has no solution or infinite solutions?
5. How can matrices be used to solve a system of linear equations?

Tip: When solving systems of equations, always check if the variables in the equations are independent and if the system has a unique solution, infinitely many solutions, or no solution at all!