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