We have a system of three linear equations with three variables:

1. \( x + y - 2z = 5 \)
2. \( x + y + z = 5 \)
3. \( x + 3y + z = -9 \)

To solve this system, let's use the method of elimination or substitution.

### Step 1: Express one variable in terms of others

From Equation 2:

\[
x + y + z = 5 \implies z = 5 - x - y
\]

### Step 2: Substitute into other equations

Substitute \( z = 5 - x - y \) into Equation 1:

\[
x + y - 2z = 5 \implies x + y - 2(5 - x - y) = 5
\]

\[
x + y - 10 + 2x + 2y = 5 \implies 3x + 3y - 10 = 5
\]

\[
3x + 3y = 15 \implies x + y = 5
\]

### Step 3: Use the new expression in another equation

Now we know \( x + y = 5 \) from both Equations 2 and the above step. Use this in Equation 3:

\[
x + 3y + z = -9 \implies x + 3y + (5 - x - y) = -9
\]

\[
x + 3y + 5 - x - y = -9 \implies 2y + 5 = -9
\]

\[
2y = -14 \implies y = -7
\]

### Step 4: Find other variables

Now that \( y = -7 \), substitute it back into \( x + y = 5 \):

\[
x - 7 = 5 \im

Learn how to solve a system of three linear equations with three variables: x + y - 2z = 5, x + y + z = 5, and x + 3y + z = -9. This problem involves using the substitution and elimination methods to find the values of x, y, and z. Suitable for high school students.

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