c) x1 + 2x2 − x3 = 1, 3x1 + 4x2 − x3 = 3, 2x1 + 3x2 − x3 = 2.

You are dealing with a system of linear equations with three variables \(x_1\), \(x_2\), and \(x_3\). The system is:

\[
\begin{aligned}
1. &\quad x_1 + 2x_2 - x_3 = 1 \\
2. &\quad 3x_1 + 4x_2 - x_3 = 3 \\
3. &\quad 2x_1 + 3x_2 - x_3 = 2
\end{aligned}
\]

We will solve this system using either substitution, elimination, or matrix methods (Gaussian elimination). Let's proceed step by step with elimination.

### Step 1: Eliminate \(x_3\)

We can subtract equations to eliminate \(x_3\). Start by subtracting equation 1 from equations 2 and 3.

Subtract equation 1 from equation 2:
\[
(3x_1 + 4x_2 - x_3) - (x_1 + 2x_2 - x_3) = 3 - 1
\]
Simplifying:
\[
(3x_1 - x_1) + (4x_2 - 2x_2) = 2
\]
\[
2x_1 + 2x_2 = 2
\]
\[
x_1 + x_2 = 1 \quad \text{(Equation 4)}
\]

Subtract equation 1 from equation 3:
\[
(2x_1 + 3x_2 - x_3) - (x_1 + 2x_2 - x_3) = 2 - 1
\]
Simplifying:
\[
(2x_1 - x_1) + (3x_2 - 2x_2) = 1
\]
\[
x_1 + x_2 = 1 \quad \text{(Equation 5, same as 4)}
\]

### Step 2: Solve for \(x_1\) and \(x_2\)

From equation 4, we already know:
\[
x_1 + x_2 = 1
\]

This means \(x_2 = 1 - x_1\). Now substitute this into one of the original equations (use equation 1).

Substitute into equation 1:
\[
x_1 + 2(1 - x_1) - x_3 = 1
\]
Simplifying:
\[
x_1 + 2 - 2x_1 - x_3 = 1
\]
\[
- x_1 + 2 - x_3 = 1
\]
\[
- x_1 - x_3 = -1
\]
\[
x_1 + x_3 = 1 \quad \text{(Equation 6)}
\]

### Step 3: Solve for \(x_3\)

From equation 6, we have:
\[
x_3 = 1 - x_1
\]

### Step 4: Find the values of \(x_1\), \(x_2\), and \(x_3\)

Now substitute \(x_3 = 1 - x_1\) and \(x_2 = 1 - x_1\) into equation 4 or 5 (which are the same):
\[
x_1 + (1 - x_1) = 1
\]
Simplifying:
\[
1 = 1
\]
This is true, meaning \(x_1\) and \(x_2\) are dependent on each other, and the system has infinitely many solutions. The general solution is:
\[
x_1 = t, \quad x_2 = 1 - t, \quad x_3 = 1 - t \quad \text{for any real number } t.
\]

### Final Answer:

The solution to the system is:
\[
x_1 = t, \quad x_2 = 1 - t, \quad x_3 = 1 - t
\]
where \(t\) is any real number (the parameter).

---

Would you like more details on the elimination method or another method? 

Here are 5 follow-up questions you could consider:

1. How would you solve this system using matrices?
2. What does it mean when a system has infinitely many solutions?
3. How can you verify the solution by substitution?
4. What if we add another constraint? Will the system still have infinitely many solutions?
5. Can you interpret the geometric meaning of this solution in 3D space?

**Tip**: For systems of linear equations, try substitution when one variable is easy to isolate, or use elimination when you can quickly cancel terms.

Discover how to solve a system of linear equations with three variables using Gaussian elimination. Learn step-by-step how the system x1 + 2x2 − x3 = 1, 3x1 + 4x2 − x3 = 3, and 2x1 + 3x2 − x3 = 2 leads to a parametric solution with infinitely many solutions. Ideal for Grades 10-12 students.

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