https://file.aiquickdraw.com/mathbot/file/1727292032818b5bmclzu.jpeg

The image presents a system of three linear equations:

\[
\begin{cases}
2x - y + 2z = 6 \\
3x + 2y - z = 4 \\
4x + 3y - 3z = 1
\end{cases}
\]

I will solve this system using either substitution or elimination. Let's proceed with the elimination method to find the values of \(x\), \(y\), and \(z\).

### Step 1: Eliminate one variable
We'll start by eliminating \(z\) by adding/subtracting the equations.

#### Equation 1: \( 2x - y + 2z = 6 \)
#### Equation 2: \( 3x + 2y - z = 4 \)
#### Equation 3: \( 4x + 3y - 3z = 1 \)

First, I will multiply Equation 2 by 2 so that the coefficient of \(z\) matches that of Equation 1:

\[
2(3x + 2y - z) = 2(4) \Rightarrow 6x + 4y - 2z = 8
\]

Now, I will add this to Equation 1:

\[
(2x - y + 2z) + (6x + 4y - 2z) = 6 + 8
\]

This simplifies to:

\[
8x + 3y = 14 \quad \text{(Equation A)}
\]

Next, I will eliminate \(z\) between Equation 2 and Equation 3. I'll multiply Equation 2 by 3 and Equation 3 by 1 to align the coefficients of \(z\):

\[
3(3x + 2y - z) = 3(4) \Rightarrow 9x + 6y - 3z = 12
\]
\[
1(4x + 3y - 3z) = 1(1) \Rightarrow 4x + 3y - 3z = 1
\]

Now subtract Equation 3 from this result:

\[
(9x + 6y - 3z) - (4x + 3y - 3z) = 12 - 1
\]

Simplifies to:

\[
5x + 3y = 11 \quad \text{(Equation B)}
\]

### Step 2: Solve for \(x\) and \(y\)
Now we solve the system formed by Equation A and Equation B:

\[
\begin{cases}
8x + 3y = 14 \quad \text{(Equation A)} \\
5x + 3y = 11 \quad \text{(Equation B)}
\end{cases}
\]

Subtract Equation B from Equation A:

\[
(8x + 3y) - (5x + 3y) = 14 - 11
\]

Simplifies to:

\[
3x = 3
\]

So:

\[
x = 1
\]

### Step 3: Substitute \(x = 1\) into one of the equations
Substitute \(x = 1\) into Equation B:

\[
5(1) + 3y = 11 \Rightarrow 5 + 3y = 11
\]

Simplifies to:

\[
3y = 6 \Rightarrow y = 2
\]

### Step 4: Solve for \(z\)
Now substitute \(x = 1\) and \(y = 2\) into any original equation. Let's use Equation 1:

\[
2(1) - (2) + 2z = 6
\]

Simplifies to:

\[
2 - 2 + 2z = 6 \Rightarrow 2z = 6 \Rightarrow z = 3
\]

### Final Solution:
The solution to the system is:
\[
x = 1, \quad y = 2, \quad z = 3
\]

Let me know if you'd like more details or have any questions!

Here are 5 related questions:
1. How can you solve this system using the substitution method?
2. What would happen if you used matrices to solve this system?
3. Can you verify the solution by substituting values back into all three original equations?
4. How do you solve a system of equations with more than three variables?
5. What are some real-life applications of solving systems of equations?

**Tip:** When using the elimination method, always align variables carefully and verify each step to avoid calculation errors!

Solve the system of equations: 2x - y + 2z = 6, 3x + 2y - z = 4, 4x + 3y - 3z = 1.

Learn how to solve the system of linear equations using the elimination method. The system includes three variables: 2x - y + 2z = 6, 3x + 2y - z = 4, and 4x + 3y - 3z = 1. The solution involves eliminating variables to find the values of x, y, and z. Suitable for Grades 9-12.

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