A system of linear equations and a reduced matrix for the system are given.
 

x	 − 	y	 + 	z	 = 	3
3x			 + 	2z	 = 	7
x	 − 	4y	 + 	2z	 = 	5
    

1	0	
2
3
7
3

0	1	− 
1
3
− 
2
3
0	0	0		0
(a)
Use the reduced matrix to find the general solution of the system, if one exists. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, express your answers in terms of z as in Example 3.)
(x, y, z) = 

 
  

(b)
If multiple solutions exist, find two specific solutions. (Enter your answers as a comma-separated list of ordered triples. If there is no solution, enter NO SOLUTION.)
(x, y, z) = 


### Given:

The system of linear equations:
\[
\begin{aligned}
x - y + z &= 3 \\
3x + 2z &= 7 \\
x - 4y + 2z &= 5
\end{aligned}
\]

And the reduced row-echelon form of the augmented matrix:

\[
\begin{pmatrix}
1 & 0 & \frac{2}{3} & \frac{7}{3} \\
0 & 1 & -\frac{1}{3} & -\frac{2}{3} \\
0 & 0 & 0 & 0
\end{pmatrix}
\]

### (a) Find the general solution:

We need to interpret the reduced matrix. The last row \( 0 = 0 \) does not provide any new information, meaning the system is consistent and there are infinitely many solutions.

From the first two rows of the matrix, we can write the following equations:

1. \( x + \frac{2}{3}z = \frac{7}{3} \)
2. \( y - \frac{1}{3}z = -\frac{2}{3} \)

Solve both equations for \( x \) and \( y \) in terms of \( z \):

1. \( x = \frac{7}{3} - \frac{2}{3}z \)
2. \( y = -\frac{2}{3} + \frac{1}{3}z \)

Thus, the general solution is:

\[
(x, y, z) = \left( \frac{7}{3} - \frac{2}{3}z, -\frac{2}{3} + \frac{1}{3}z, z \right)
\]

### (b) Find two specific solutions:

We can find two specific solutions by assigning different values to \( z \).

1. Let \( z = 0 \):
\[
x = \frac{7}{3} - \frac{2}{3}(0) = \frac{7}{3}, \quad y = -\frac{2}{3} + \frac{1}{3}(0) = -\frac{2}{3}
\]
So, one specific solution is:
\[
\left( \frac{7}{3}, -\frac{2}{3}, 0 \right)
\]

2. Let \( z = 3 \):
\[
x = \frac{7}{3} - \frac{2}{3}(3) = \frac{7}{3} - 2 = \frac{1}{3}, \quad y = -\frac{2}{3} + \frac{1}{3}(3) = -\frac{2}{3} + 1 = \frac{1}{3}
\]
So, another specific solution is:
\[
\left( \frac{1}{3}, \frac{1}{3}, 3 \right)
\]

### Final Answer:

#### (a) General solution:
\[
(x, y, z) = \left( \frac{7}{3} - \frac{2}{3}z, -\frac{2}{3} + \frac{1}{3}z, z \right)
\]

#### (b) Two specific solutions:
\[
\left( \frac{7}{3}, -\frac{2}{3}, 0 \right), \left( \frac{1}{3}, \frac{1}{3}, 3 \right)
\]

Would you like any more details or have any questions?

Here are 5 related questions to deepen understanding:

1. How do you find the reduced row-echelon form of a matrix manually?
2. What do the rows in the reduced matrix signify in terms of the original system?
3. How can we determine if a system has infinitely many solutions based on its augmented matrix?
4. What would the matrix look like if there were no solutions?
5. How does parametric form help in representing general solutions for underdetermined systems?

**Tip**: Always check for consistency in the system by looking at the last row in the reduced matrix. If it corresponds to a false statement like \( 0 = 1 \), then the system has no solution!

A system of linear equations and a reduced matrix for the system are given.

x - y + z = 3
3x + 2z = 7
x - 4y + 2z = 5

Reduced matrix:
1 0 2/3 7/3
0 1 -1/3 -2/3
0 0 0 0

(a) Use the reduced matrix to find the general solution of the system, if one exists. (b) If multiple solutions exist, find two specific solutions.

Find the general and specific solutions of a system of linear equations using the reduced row-echelon form (RREF). The problem involves key concepts from linear algebra, such as parametric solutions and consistency of systems, and is suitable for students in Grades 10-12.

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