x + y − 3z = −2
x − 2y + 6z = 10
−2x + y − 3z = −8

You are working with a system of three linear equations:

1. \( x + y - 3z = -2 \)
2. \( x - 2y + 6z = 10 \)
3. \( -2x + y - 3z = -8 \)

We can solve this system using substitution, elimination, or matrix methods like Gaussian elimination or Cramer's rule. Let's use elimination to solve it step-by-step.

### Step 1: Simplify the system
Label the equations for convenience:
- \( (1) \, x + y - 3z = -2 \)
- \( (2) \, x - 2y + 6z = 10 \)
- \( (3) \, -2x + y - 3z = -8 \)

### Step 2: Eliminate one variable
To eliminate \( x \), multiply equation (1) by 2 and add it to equation (3):
\[
2(x + y - 3z) = 2(-2) \implies 2x + 2y - 6z = -4
\]
Now add this to equation (3):
\[
(-2x + y - 3z) + (2x + 2y - 6z) = -8 + (-4) 
\]
\[
(0x) + 3y - 9z = -12 \implies 3y - 9z = -12 \implies y - 3z = -4 \quad (4)
\]

### Step 3: Eliminate \( x \) from another pair of equations
Subtract equation (1) from equation (2):
\[
(x - 2y + 6z) - (x + y - 3z) = 10 - (-2)
\]
\[
0x - 3y + 9z = 12 \implies -3y + 9z = 12 \implies y - 3z = -4 \quad (5)
\]

Notice that equation (4) and equation (5) are identical, so no new information is gained from this step. This means the system has dependent equations and likely has infinitely many solutions.

### Step 4: Express variables in terms of one free variable
From \( y - 3z = -4 \), solve for \( y \):
\[
y = 3z - 4
\]
Substitute \( y = 3z - 4 \) into equation (1) to solve for \( x \):
\[
x + (3z - 4) - 3z = -2
\]
Simplify:
\[
x - 4 = -2 \implies x = 2
\]

### Step 5: Final solution
Thus, the solution is:
\[
x = 2, \quad y = 3z - 4
\]
Here, \( z \) is a free variable. The solution set is dependent on \( z \).

So, the general solution is:
\[
x = 2, \quad y = 3z - 4, \quad z = z \quad (\text{free variable})
\]

Would you like more details or clarification on any part?

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Here are 5 related questions to explore this topic further:
1. How would the solution change if the system had a unique solution instead of infinitely many?
2. How can you verify that the system has infinitely many solutions?
3. Can you express the solution in vector form using parameterization?
4. What are the conditions for a system of equations to have no solution?
5. How does the determinant of the coefficient matrix relate to the number of solutions?

**Tip:** When solving a system of equations, checking for consistency between equations can help identify if the system has infinitely many or no solutions.

x + y − 3z = −2, x − 2y + 6z = 10, −2x + y − 3z = −8

Learn how to solve a system of linear equations involving three variables using the elimination method. This example illustrates how dependent equations lead to infinitely many solutions. Suitable for students in Grades 9-12.

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