Using the techniques discussed in this section, solve the following system of linear equations. Enter the solution as an ordered triple. For parametric solutions use

*z* = *t*

as the parameter. If a solution does not exist, enter DNE for all coordinates.

−3*x* + 7*y* − 8*z*

=

9

*x* − *y* + 2*z*

=

−9

−4*y* + *z*

=

7

To solve the given system of linear equations, we'll use a combination of elimination and substitution methods. We want to find values for \(x\), \(y\), and \(z\) that satisfy all three equations. Here are the equations provided:

\[
\begin{cases}
-3x + 7y - 8z = 9 \\
x - y + 2z = -9 \\
-4y + z = 7
\end{cases}
\]

### Step 1: Write the System in Matrix Form

We can express the system as an augmented matrix:

\[
\left[
\begin{array}{ccc|c}
-3 & 7 & -8 & 9 \\
1 & -1 & 2 & -9 \\
0 & -4 & 1 & 7
\end{array}
\right]
\]

### Step 2: Use Gaussian Elimination to Find the Solution

We'll perform row operations to bring the matrix to reduced row echelon form (RREF). Let's proceed with the steps:

1. **Swap Row 1 and Row 2** to have a leading 1 in the first row:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & -1 & 2 & -9 \\
   -3 & 7 & -8 & 9 \\
   0 & -4 & 1 & 7
   \end{array}
   \right]
   \]

2. **Add 3 times Row 1 to Row 2**:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & -1 & 2 & -9 \\
   0 & 4 & -2 & -18 \\
   0 & -4 & 1 & 7
   \end{array}
   \right]
   \]

3. **Add Row 2 to Row 3**:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & -1 & 2 & -9 \\
   0 & 4 & -2 & -18 \\
   0 & 0 & -1 & -11
   \end{array}
   \right]
   \]

4. **Multiply Row 3 by -1** to get a leading 1:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & -1 & 2 & -9 \\
   0 & 4 & -2 & -18 \\
   0 & 0 & 1 & 11
   \end{array}
   \right]
   \]

5. **Add 2 times Row 3 to Row 2**:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & -1 & 2 & -9 \\
   0 & 4 & 0 & 4 \\
   0 & 0 & 1 & 11
   \end{array}
   \right]
   \]

6. **Divide Row 2 by 4**:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & -1 & 2 & -9 \\
   0 & 1 & 0 & 1 \\
   0 & 0 & 1 & 11
   \end{array}
   \right]
   \]

7. **Add Row 2 to Row 1**:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & 0 & 2 & -8 \\
   0 & 1 & 0 & 1 \\
   0 & 0 & 1 & 11
   \end{array}
   \right]
   \]

8. **Subtract 2 times Row 3 from Row 1**:
   \[
   \left[
   \begin{array}{ccc|c}
   1 & 0 & 0 & -30 \\
   0 & 1 & 0 & 1 \\
   0 & 0 & 1 & 11
   \end{array}
   \right]
   \]

### Step 3: Interpret the Results

The matrix is now in reduced row echelon form, corresponding to the following system of equations:

\[
\begin{cases}
x = -30 \\
y = 1 \\
z = 11
\end{cases}
\]

### Solution

The solution as an ordered triple is:

\[
(x, y, z) = (-30, 1, 11)
\]

### Conclusion

The solution to the system of equations is \((-30, 1, 11)\).

If you have any questions or want further details about any step, feel free to ask!

#### Related Questions:

1. How can we verify the solution of a system of linear equations?
2. What are other methods to solve a system of linear equations apart from Gaussian elimination?
3. How does the augmented matrix method help in finding solutions to linear systems?
4. What is the difference between Row Echelon Form (REF) and Reduced Row Echelon Form (RREF)?
5. What are the possible outcomes when solving a system of linear equations?

#### Tip:
Always double-check each row operation when using Gaussian elimination to ensure no arithmetic mistakes are made.

Using the techniques discussed in this section, solve the following system of linear equations. Enter the solution as an ordered triple. For parametric solutions use z = t as the parameter. If a solution does not exist, enter DNE for all coordinates.
−3x + 7y − 8z = 9
x − y + 2z = −9
−4y + z = 7

Learn how to solve a system of linear equations using Gaussian elimination. This example walks through the step-by-step solution of the system: −3x + 7y − 8z = 9, x − y + 2z = −9, and −4y + z = 7, using matrix representation and row operations. Suitable for students in Grades 10-12.

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