Solve the system of linear equations. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, express x1, x2, and x3 in terms of the parameter t.)
2x1	 + 	x2	 − 	5x3	 = 	5
4x1			 + 	2x3	 = 	12
−2x1	 + 	3x2	 − 	19x3	 = 	−9

Let's solve the given system of linear equations step by step.

The system is:
\[ 
\begin{cases}
2x_1 + x_2 - 5x_3 = 5 \\
4x_1 + 2x_3 = 12 \\
-2x_1 + 3x_2 - 19x_3 = -9 
\end{cases}
\]

First, we will write the system in matrix form as \( A \mathbf{x} = \mathbf{b} \).

\[ 
A = \begin{pmatrix}
2 & 1 & -5 \\
4 & 0 & 2 \\
-2 & 3 & -19
\end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix}
x_1 \\
x_2 \\
x_3
\end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix}
5 \\
12 \\
-9
\end{pmatrix}
\]

Next, we will use Gaussian elimination to solve the system.

Step 1: Normalize the first row by dividing it by 2.
\[ 
\begin{pmatrix}
1 & \frac{1}{2} & -2.5 & \mid & 2.5 \\
4 & 0 & 2 & \mid & 12 \\
-2 & 3 & -19 & \mid & -9
\end{pmatrix}
\]

Step 2: Eliminate the first element of the second and third rows.
\[ 
R2 \leftarrow R2 - 4R1
\]
\[ 
R3 \leftarrow R3 + 2R1
\]

\[ 
\begin{pmatrix}
1 & \frac{1}{2} & -2.5 & \mid & 2.5 \\
0 & -2 & 12 & \mid & 2 \\
0 & 4 & -24 & \mid & -4
\end{pmatrix}
\]

Step 3: Normalize the second row by dividing it by -2.
\[ 
\begin{pmatrix}
1 & \frac{1}{2} & -2.5 & \mid & 2.5 \\
0 & 1 & -6 & \mid & -1 \\
0 & 4 & -24 & \mid & -4
\end{pmatrix}
\]

Step 4: Eliminate the second element of the third row.
\[ 
R3 \leftarrow R3 - 4R2
\]

\[ 
\begin{pmatrix}
1 & \frac{1}{2} & -2.5 & \mid & 2.5 \\
0 & 1 & -6 & \mid & -1 \\
0 & 0 & 0 & \mid & 0
\end{pmatrix}
\]

Step 5: Express the variables in terms of a parameter \( t \).
Let \( x_3 = t \).

From the second row:
\[ 
x_2 - 6x_3 = -1 \implies x_2 - 6t = -1 \implies x_2 = 6t - 1 
\]

From the first row:
\[ 
x_1 + \frac{1}{2}x_2 - 2.5x_3 = 2.5 \implies x_1 + \frac{1}{2}(6t - 1) - 2.5t = 2.5 
\]
\[ 
x_1 + 3t - 0.5 - 2.5t = 2.5 \implies x_1 + 0.5t - 0.5 = 2.5 
\]
\[ 
x_1 = 2.5 - 0.5t + 0.5 \implies x_1 = 3 - 0.5t 
\]

Thus, the system has infinitely many solutions:
\[ 
(x_1, x_2, x_3) = (3 - 0.5t, 6t - 1, t) 
\]

If you have any questions or need further details, please let me know.

Here are some more questions you might want to ask:
1. How to solve a system of linear equations using the inverse matrix method?
2. How to determine if a system of equations has no solution?
3. What is the difference between consistent and inconsistent systems?
4. Can you explain the row reduction method in linear algebra?
5. What are the applications of systems of linear equations in real life?

**Tip:** Always check the determinant of the coefficient matrix to quickly determine if a system of linear equations is invertible.

Learn how to solve the system of linear equations using Gaussian elimination. This detailed solution covers key linear algebra concepts, including matrix form, Gaussian elimination steps, and how to express solutions in terms of a parameter. Suitable for students in Grades 11-12.

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