We herschrijven het onderstaande stelsel dusdanig dat we alle oplossingen kunnen aflezen.
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
2x_1 &-&x_2 &-&x_3&=&7& (i)\\
-x_1&+&3x_2&+&4x_3&=&7& (ii)\\
x_1 &+&2x_2 &+&x_3&=&8& (iii)\\
\end{array}
\right.
\end{equation}$$
We verwisselen $(i)$ en $(iii)$:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &+&2x_2 &+&x_3&=&8& (i)\\
-x_1&+&3x_2&+&4x_3&=&7& (ii)\\
2x_1 &-&x_2 &-&x_3&=&7& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(ii)$ precies $(i)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &+&2x_2 &+&x_3&=&8& (i)\\
&&5x_2&+&5x_3&=&15& (ii)\\
2x_1 &-&x_2 &-&x_3&=&7& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(iii)$ precies $-2 \times (i)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &+&2x_2 &+&x_3&=&8& (i)\\
&&5x_2&+&5x_3&=&15& (ii)\\
&-&5x_2 &-&3x_3&=&-9& (iii)\\
\end{array}
\right.
\end{equation}$$
We vermenigvuldigen $(ii)$ met $\frac{1}{5}$:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &+&2x_2 &+&x_3&=&8& (i)\\
&&x_2&+&x_3&=&3& (ii)\\
&-&5x_2 &-&3x_3&=&-9& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(i)$ precies $-2\times (ii)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &&&-&x_3&=&2& (i)\\
&&x_2&+&x_3&=&3& (ii)\\
&-&5x_2 &-&3x_3&=&-9& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(iii)$ precies $5\times (ii)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &&&-&x_3&=&2& (i)\\
&&x_2&+&x_3&=&3& (ii)\\
&&&&2x_3&=&6& (iii)\\
\end{array}
\right.
\end{equation}$$
We vermenigvuldigen $(iii)$ met $\frac{1}{2}$:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &&&-&x_3&=&2& (i)\\
&&x_2&+&x_3&=&3& (ii)\\
&&&&x_3&=&3& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(i)$ precies $(iii)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &&&&&=&5& (i)\\
&&x_2&+&x_3&=&3& (ii)\\
&&&&x_3&=&3& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(ii)$ precies $-1 \times (iii)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &&&&&=&5& (i)\\
&&x_2&&&=&2& (ii)\\
&&&&x_3&=&3& (iii)\\
\end{array}
\right.
\end{equation}$$
Uit dit laatste stelsel lezen we de oplossing $(x_1,x_2,x_3)=(5,0,3)$ direct af. En omdat we enkel elementaire operaties hebben gebruikt is dit dus ook de oplossing van het stelsel waarmee we begonnen zijn.
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
2x_1 &-&x_2 &-&x_3&=&7& (i)\\
-x_1&+&3x_2&+&4x_3&=&7& (ii)\\
x_1 &+&2x_2 &+&x_3&=&8& (iii)\\
\end{array}
\right.
\end{equation}$$
We verwisselen $(i)$ en $(iii)$:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &+&2x_2 &+&x_3&=&8& (i)\\
-x_1&+&3x_2&+&4x_3&=&7& (ii)\\
2x_1 &-&x_2 &-&x_3&=&7& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(ii)$ precies $(i)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &+&2x_2 &+&x_3&=&8& (i)\\
&&5x_2&+&5x_3&=&15& (ii)\\
2x_1 &-&x_2 &-&x_3&=&7& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(iii)$ precies $-2 \times (i)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &+&2x_2 &+&x_3&=&8& (i)\\
&&5x_2&+&5x_3&=&15& (ii)\\
&-&5x_2 &-&3x_3&=&-9& (iii)\\
\end{array}
\right.
\end{equation}$$
We vermenigvuldigen $(ii)$ met $\frac{1}{5}$:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &+&2x_2 &+&x_3&=&8& (i)\\
&&x_2&+&x_3&=&3& (ii)\\
&-&5x_2 &-&3x_3&=&-9& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(i)$ precies $-2\times (ii)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &&&-&x_3&=&2& (i)\\
&&x_2&+&x_3&=&3& (ii)\\
&-&5x_2 &-&3x_3&=&-9& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(iii)$ precies $5\times (ii)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &&&-&x_3&=&2& (i)\\
&&x_2&+&x_3&=&3& (ii)\\
&&&&2x_3&=&6& (iii)\\
\end{array}
\right.
\end{equation}$$
We vermenigvuldigen $(iii)$ met $\frac{1}{2}$:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &&&-&x_3&=&2& (i)\\
&&x_2&+&x_3&=&3& (ii)\\
&&&&x_3&=&3& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(i)$ precies $(iii)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &&&&&=&5& (i)\\
&&x_2&+&x_3&=&3& (ii)\\
&&&&x_3&=&3& (iii)\\
\end{array}
\right.
\end{equation}$$
We tellen bij $(ii)$ precies $-1 \times (iii)$ op:
$$\begin{equation}
\left \{
\begin{array}{rrrrrrrr}
x_1 &&&&&=&5& (i)\\
&&x_2&&&=&2& (ii)\\
&&&&x_3&=&3& (iii)\\
\end{array}
\right.
\end{equation}$$
Uit dit laatste stelsel lezen we de oplossing $(x_1,x_2,x_3)=(5,0,3)$ direct af. En omdat we enkel elementaire operaties hebben gebruikt is dit dus ook de oplossing van het stelsel waarmee we begonnen zijn.