https://file.aiquickdraw.com/mathbot/file/17267231263672a4eewt5.jpg

The system of equations shown in the image is:

\[
\begin{aligned}
1. \ 2x + y - 3z &= 9 \\
2. \ 4x - 2y + 2z &= 6 \\
3. \ 2x - 2y - 3z &= 0 \\
\end{aligned}
\]

To solve this system algebraically, let's proceed step by step:

### Step 1: Eliminate one variable (e.g., \(z\))
We'll use equations (1) and (3) to eliminate \(z\).

- Multiply equation (1) by 2 to match the coefficient of \(z\) in equation (3):
  \[
  2(2x + y - 3z) = 2(9) \implies 4x + 2y - 6z = 18
  \]
- Now subtract equation (3) from this:
  \[
  (4x + 2y - 6z) - (2x - 2y - 3z) = 18 - 0
  \]
  Simplifying:
  \[
  2x + 4y - 3z = 18 \implies 2x + 4y = 18 \implies x + 2y = 9 \quad (Equation 4)
  \]

### Step 2: Use another pair of equations
Now let's use equation (2) and (3) to eliminate \(z\).

- Add equation (2) and (3) to eliminate \(z\):
  \[
  (4x - 2y + 2z) + (2x - 2y - 3z) = 6 + 0
  \]
  Simplifying:
  \[
  6x - 4y - z = 6 \quad (Equation 5)
  \]

### Step 3: Solve for \(x\) and \(y\)
We now have two simpler equations to solve:
1. \(x + 2y = 9\) (from step 1)
2. \(6x - 4y - z = 6\) (from step 2)

At this point, substitute values back into original system

Solve the system algebraically: 2x + y - 3z = 9, 4x - 2y + 2z = 6, 2x - 2y - 3z = 0.

Learn how to solve the system of linear equations 2x + y - 3z = 9, 4x - 2y + 2z = 6, and 2x - 2y - 3z = 0 algebraically using elimination and substitution. This problem covers key algebraic methods, ideal for high school students in Grades 10-12.

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