Solve the following system of equations, if there is a single solution, answer it as an ordered pair (x, y, z), if there are infinitely many solutions, parameterize your solutions in such a way that z 
 is your free variable.

2x - 3y - 6z = 35
-9x + 5y - z = 8
-5x + 8y - 7z = 21

To solve this system of equations, we can use either the elimination method or matrix methods (such as Gaussian elimination). Let's go step-by-step with Gaussian elimination.

Given:
\[
2x - 3y - 6z = 35
\]
\[
-9x + 5y - z = 8
\]
\[
-5x + 8y - 7z = 21
\]

### Step 1: Write the system in augmented matrix form.
\[
\begin{bmatrix}
2 & -3 & -6 & | & 35 \\
-9 & 5 & -1 & | & 8 \\
-5 & 8 & -7 & | & 21 \\
\end{bmatrix}
\]

### Step 2: Use row operations to get a row-echelon form.

1. **Make the leading coefficient of the first row equal to 1** by dividing the first row by 2:
   \[
   R_1: \quad \left[1, -\frac{3}{2}, -3 \; | \; 17.5 \right]
   \]

2. **Eliminate the \(x\)-term in rows 2 and 3**:
   - For \(R_2\): \( R_2 + 9 \times R_1 \rightarrow R_2 \)
   - For \(R_3\): \( R_3 + 5 \times R_1 \rightarrow R_3 \)

   After performing these operations, we get:
   \[
   \begin{bmatrix}
   1 & -\frac{3}{2} & -3 & | & 17.5 \\
   0 & \frac{1}{2} & 26 & | & 166 \\
   0 & \frac{5}{2} & -22 & | & 108 \\
   \end{bmatrix}
   \]

3. **Make the leading coefficient of the second row equal to 1** by multiplying \( R_2 \) by 2:
   \[
   R_2: \quad \left[0, 1, 52 \; | \; 332 \right]
   \]

4. **Eliminate the \(y\)-term in row 3** by performing \( R_3 - \frac{5}{2} \times R_2 \rightarrow R_3 \):
   \[
   \begin{bmatrix}
   1 & -\frac{3}{2} & -3 & | & 17.5 \\
   0 & 1 & 52 & | & 332 \\
   0 & 0 & -282 & | & -722 \\
   \end{bmatrix}
   \]

5. **Solve for \( z \)** using the third row:
   \[
   -282z = -722 \Rightarrow z = \frac{722}{282} = \frac{361}{141} = 2.5
   \]

6. **Back-substitute \( z = 2.5 \) into \( R_2 \) and \( R_1 \) to find \( y \) and \( x \)**:
   - From \( R_2 \): \( y + 52 \times 2.5 = 332 \)
   - Solve for \( y \):
     \[
     y = 332 - 130 = 202
     \]

   - Substitute \( y = 202 \) and \( z = 2.5 \) into \( R_1 \):
     \[
     x - \frac{3}{2} \times 202 - 3 \times 2.5 = 17.5
     \]
     Solving this, we get \( x \).

After calculating

Solve the following system of equations, if there is a single solution, answer it as an ordered pair (x, y, z), if there are infinitely many solutions, parameterize your solutions in such a way that z is your free variable.

2x - 3y - 6z = 35

-9x + 5y - z = 8

-5x + 8y - 7z = 21

Explore how to solve a 3-variable system of linear equations using Gaussian elimination. This step-by-step solution demonstrates matrix row operations and techniques to determine if a system has a unique or infinite number of solutions, focusing on parameterizing solutions with free variables. Suitable for high school algebra students.

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