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The system of equations given is:

\[
\begin{aligned}
x + y + z &= 3 \\
2x - y + 2z &= 5 \\
3x + y + z &= 7
\end{aligned}
\]

To solve this, we can use either substitution, elimination, or matrix methods. Let's solve it step by step.

### Step 1: Add equations (1) and (3)
From equations (1) and (3):

\[
( x + y + z ) + ( 3x + y + z ) = 3 + 7 \implies 4x + 2y + 2z = 10
\]

This simplifies to:

\[
2x + y + z = 5 \quad \text{(Equation A)}
\]

### Step 2: Subtract equation (1) from equation (3)
From equations (3) and (1):

\[
( 3x + y + z ) - ( x + y + z ) = 7 - 3 \implies 2x = 4
\]

This gives:

\[
x = 2
\]

### Step 3: Substitute \( x = 2 \) into equation (1)
Substitute \( x = 2 \) into the first equation:

\[
2 + y + z = 3 \implies y + z = 1 \quad \text{(Equation B)}
\]

### Step 4: Substitute \( x = 2 \) into equation (2)
Substitute \( x = 2 \) into the second equation:

\[
2(2) - y + 2z = 5 \implies 4 - y + 2z = 5 \implies -y + 2z = 1 \quad \text{(Equation C)}
\]

### Step 5: Solve the system of equations (B) and (C)
Now we have the system:
1. \( y + z = 1 \)
2. \( -y + 2z = 1 \)

Add both equations:

\[
(y + z) + (-y + 2z) = 1 + 1 \implies 3z = 2 \implies z = \frac{2}{3}
\]

### Step 6: Substitute \( z = \frac{2}{3} \) into equation (B)
Substitute \( z = \frac{2}{3} \) into equation \( y + z = 1 \):

\[
y + \frac{2}{3} = 1 \implies y = 1 - \frac{2}{3} = \frac{1}{3}
\]

### Final solution:
\[
x = 2, \quad y = \frac{1}{3}, \quad z = \frac{2}{3}
\]

This corresponds to option (b) in the provided choices.

Would you like further details on this solution?

Here are 5 related questions:
1. How can you verify solutions of a system of equations?
2. What other methods can solve systems of linear equations?
3. How does matrix algebra relate to solving linear equations?
4. What are the advantages of using substitution versus elimination?
5. How can graphical methods solve systems of linear equations?

**Tip**: Always substitute your final solution back into the original equations to check for consistency!

Consider the following system of equations:

x + y + z = 3
2x - y + 2z = 5
3x + y + z = 7
Find x, y, and z from the given choices:
a) x = 3/2, y = 2/3, z = 5/3
b) x = 2, y = 1/3, z = 2/3
c) x = 2, y = 1, z = 0
d) x = 2, y = -1/3, z = 4/3

Learn how to solve the given system of linear equations: x + y + z = 3, 2x - y + 2z = 5, and 3x + y + z = 7. Understand the step-by-step process to find the values of x, y, and z using substitution and elimination methods. This problem is ideal for students in Grades 9-12.

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