5x-3y-2z=31        2x+6y+3z=4         4x + 2y - z = 30        Find x y z

We are given the following system of linear equations:

\[
5x - 3y - 2z = 31 \quad \text{(1)}
\]
\[
2x + 6y + 3z = 4 \quad \text{(2)}
\]
\[
4x + 2y - z = 30 \quad \text{(3)}
\]

We can solve this system using either substitution, elimination, or matrix methods. I will use the **elimination method** to solve for \(x\), \(y\), and \(z\).

### Step 1: Eliminate one variable

Let’s first eliminate \(z\) by manipulating equations (1) and (3).

- Multiply equation (3) by 2 to align with the \(-2z\) term in equation (1):
\[
8x + 4y - 2z = 60 \quad \text{(4)}
\]

Now subtract equation (1) from equation (4):
\[
(8x + 4y - 2z) - (5x - 3y - 2z) = 60 - 31
\]
\[
8x + 4y - 2z - 5x + 3y + 2z = 29
\]
\[
3x + 7y = 29 \quad \text{(5)}
\]

### Step 2: Eliminate \(z\) using equations (2) and (3)

- Multiply equation (2) by 1 and equation (3) by 3 to align the \(z\) terms:
\[
6x + 18y + 9z = 12 \quad \text{(6)}
\]
\[
12x + 6y - 3

5x - 3y - 2z = 31, 2x + 6y + 3z = 4, 4x + 2y - z = 30. Find x, y, z.

This page provides a detailed step-by-step solution to the system of linear equations: 5x - 3y - 2z = 31, 2x + 6y + 3z = 4, 4x + 2y - z = 30. Using the elimination method, we solve for x, y, and z, covering key algebra concepts. Suitable for high school students in Grades 9-12.

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