Vi nummererer likningene:

$$\begin{cases} x + y - z = 13 & \quad \text{(I)} \\ 2x + y + z = 27 & \quad \text{(II)} \\ x - 3y - 2z = -9 & \quad \text{(III)} \end{cases}$$

Steg 1: Eliminer $x$ fra likning II og III.

$\textcolor{seagreen}{\text{(II)} - 2 \cdot \text{(I)}}$:

$$\textcolor{seagreen}{(2x + y + z) - 2(x + y - z) = 27 - 26}$$ $$\textcolor{seagreen}{-y + 3z = 1 \quad \text{(IV)}}$$

$\textcolor{steelblue}{\text{(III)} - \text{(I)}}$:

$$\textcolor{steelblue}{(x - 3y - 2z) - (x + y - z) = -9 - 13}$$ $$\textcolor{steelblue}{-4y - z = -22 \quad \Leftrightarrow \quad 4y + z = 22 \quad \text{(V)}}$$

Steg 2: Eliminer $y$ fra likning V.

Fra $\textcolor{seagreen}{\text{(IV)}}$: $y = 3z - 1$. Vi setter inn i $\textcolor{steelblue}{\text{(V)}}$:

$$4(3z - 1) + z = 22$$ $$12z - 4 + z = 22$$ $$13z = 26 \implies z = 2$$

Steg 3: Finn $y$ og $x$.

Fra (IV): $y = 3 \cdot 2 - 1 = 5$

Fra (I): $x + 5 - 2 = 13 \implies x = 10$

Kontroll i (III): $10 - 3 \cdot 5 - 2 \cdot 2 = 10 - 15 - 4 = -9$ ✓

$\underline{\underline{x = 10,\ y = 5,\ z = 2}}$