Solve the system. If there is no solution or if there are infinitely many solutions and the​ system's equations are​ dependent, so state.

4 x4x

plus+

yy

equals=

1212

xx

plus+

yy

minus−

zz

equals=

99

2 x2x

plus+

3 y3y

plus+

zz

equals=

00

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Part 1

Select the correct choice below and fill in any answer boxes within your choice.

A.

There is one solution. The solution set is 

​{left parenthesis nothing comma nothing comma nothing right parenthesisenter your response here,enter your response here,enter your response here​}.

​(Simplify your​ answers.)

B.

There are infinitely many solutions.

C.

There is no solution.

To solve the given system of equations, we will use the methods of linear algebra. Here are the equations:

1. \( 4x + y = 12 \)
2. \( x + y - z = 9 \)
3. \( 2x + 3y + z = 0 \)

We need to determine if there is a unique solution, no solution, or infinitely many solutions. We'll use the augmented matrix and row reduction (Gaussian elimination) to find the solution.

First, let's write the augmented matrix for the system:
\[
\begin{pmatrix}
4 & 1 & 0 & | & 12 \\
1 & 1 & -1 & | & 9 \\
2 & 3 & 1 & | & 0
\end{pmatrix}
\]

We'll perform row operations to transform this matrix into row-echelon form.

### Step 1: Make the leading coefficient of the first row 1
\[
R1 = \frac{1}{4}R1 \rightarrow \begin{pmatrix}
1 & \frac{1}{4} & 0 & | & 3 \\
1 & 1 & -1 & | & 9 \\
2 & 3 & 1 & | & 0
\end{pmatrix}
\]

### Step 2: Eliminate the leading coefficient in the second and third rows
\[
R2 = R2 - R1 \rightarrow \begin{pmatrix}
1 & \frac{1}{4} & 0 & | & 3 \\
0 & \frac{3}{4} & -1 & | & 6 \\
2 & 3 & 1 & | & 0
\end{pmatrix}
\]

\[
R3 = R3 - 2R1 \rightarrow \begin{pmatrix}
1 & \frac{1}{4} & 0 & | & 3 \\
0 & \frac{3}{4} & -1 & | & 6 \\
0 & \frac{11}{2} & 1 & | & -6
\end{pmatrix}
\]

### Step 3: Make the leading coefficient of the second row 1
\[
R2 = \frac{4}{3}R2 \rightarrow \begin{pmatrix}
1 & \frac{1}{4} & 0 & | & 3 \\
0 & 1 & -\frac{4}{3} & | & 8 \\
0 & \frac{11}{2} & 1 & | & -6
\end{pmatrix}
\]

### Step 4: Eliminate the leading coefficient in the first and third rows
\[
R1 = R1 - \frac{1}{4}R2 \rightarrow \begin{pmatrix}
1 & 0 & \frac{1}{3} & | & 1 \\
0 & 1 & -\frac{4}{3} & | & 8 \\
0 & \frac{11}{2} & 1 & | & -6
\end{pmatrix}
\]

\[
R3 = R3 - \frac{11}{2}R2 \rightarrow \begin{pmatrix}
1 & 0 & \frac{1}{3} & | & 1 \\
0 & 1 & -\frac{4}{3} & | & 8 \\
0 & 0 & \frac{17}{3} & | & -50
\end{pmatrix}
\]

### Step 5: Make the leading coefficient of the third row 1
\[
R3 = \frac{3}{17}R3 \rightarrow \begin{pmatrix}
1 & 0 & \frac{1}{3} & | & 1 \\
0 & 1 & -\frac{4}{3} & | & 8 \\
0 & 0 & 1 & | & -\frac{150}{17}
\end{pmatrix}
\]

### Step 6: Eliminate the leading coefficient in the first and second rows
\[
R1 = R1 - \frac{1}{3}R3 \rightarrow \begin{pmatrix}
1 & 0 & 0 & | & \frac{17}{17} + \frac{50}{17} = \frac{17}{17} = 1 \\
0 & 1 & -\frac{4}{3} & | & 8 + \frac{600}{51} = 8 - \frac{200}{17} = \frac{136-200}{17} = \frac{264}{17} = 0 \\
0 & 0 & 1 & | & -\frac{150}{17}
\end{pmatrix}
\]

\[
R2 = R2 + \frac{4}{3}R3 \rightarrow \begin{pmatrix}
1 & 0 & 0 & | & 1 \\
0 & 1 & 0 & | & \frac{136}{17} + \frac{200}{17} = 0 \\
0 & 0 & 1 & | & -\frac{150}{17}
\end{pmatrix}
\]

This gives us the solution:
\[
\begin{cases}
x = 1 \\
y = 0 \\
z = -\frac{150}{17}
\end{cases}
\]

So, the solution is \((1, 0, -\frac{150}{17})\).

### Answer:
A. There is one solution. The solution set is \((1, 0, -\frac{150}{17})\).

Would you like further details or have any questions?

Here are some additional questions you might consider:

1. How do you perform row reduction on a matrix?
2. Can you explain the method of Gaussian elimination in detail?
3. What are the different types of solutions for a system of linear equations?
4. How do you determine if a system of equations has no solution?
5. What is the difference between dependent and independent systems of equations?

**Tip:** Practice solving different types of linear systems to become more familiar with the various methods and their applications.

Learn how to solve a system of linear equations with three variables: 4x + y = 12, x + y - z = 9, 2x + 3y + z = 0. Detailed step-by-step solution using Gaussian elimination and matrix operations. Suitable for Grades 10-12 students studying advanced algebra.

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