What is the value of z
 in the following system of equations?
 
	x+2y+3z=0
2x+y+3z=0
3x+2z=1

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To solve for \( z \) in the system of equations:

\[
\begin{cases}
x + 2y + 3z = 0 \\
2x + y + 3z = 0 \\
3x + 2z = 1
\end{cases}
\]

we'll follow these steps:

1. **Eliminate \( y \)** from the first two equations to simplify.
2. **Use substitution or elimination** to isolate \( z \) in the reduced system.

### Step 1: Eliminate \( y \)

1. Multiply the first equation by \( 2 \) and subtract the second equation from it:
   \[
   2(x + 2y + 3z) - (2x + y + 3z) = 0
   \]
   This simplifies to:
   \[
   2x + 4y + 6z - 2x - y - 3z = 0
   \]
   \[
   3y + 3z = 0
   \]
   \[
   y + z = 0 \quad \Rightarrow \quad y = -z
   \]

### Step 2: Substitute \( y = -z \) into the third equation

3. Substitute \( y = -z \) in the third equation:
   \[
   3x + 2z = 1
   \]

Now we have:
\[
\begin{cases}
y = -z \\
3x + 2z = 1
\end{cases}
\]

### Step 3: Substitute \( y = -z \) into one of the original equations to find \( x \)

4. Substitute into

What is the value of z in the following system of equations?

x + 2y + 3z = 0
2x + y + 3z = 0
3x + 2z = 1

Find the value of z in a system of linear equations with three variables: x + 2y + 3z = 0, 2x + y + 3z = 0, and 3x + 2z = 1. Learn the step-by-step elimination and substitution techniques used to isolate z in this algebraic problem. Suitable for students in Grades 9-12.

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