Step 1: Convert all decimals to fractions

$$\begin{array}{ cccc }6~ x&-~~~5~ y&+~~3~ z&~=~13\\4~ x&+~~10~ y&+~~2~ z&~=~62\\2~ x&+~~5~ y&+~~~~ z&~=~31\end{array}$$

Step 2: Divide the second equation by $ 2 $. The result is:

$$\begin{array}{ cccc }6~ x&-~~~5~ y&+~~3~ z&~=~13\\2~ x&+~~5~ y&+~~~~ z&~=~31\\2~ x&+~~5~ y&+~~~~ z&~=~31\end{array}$$

Step 3: Swap Row 1 and Row 2. The result is:

$$\begin{array}{ cccc }2~ x&+~~5~ y&+~~~~ z&~=~31\\6~ x&-~~~5~ y&+~~3~ z&~=~13\\2~ x&+~~5~ y&+~~~~ z&~=~31\end{array}$$

Step 4: Multiply the first equation by $ -3 $ and add the result to the second equation. The result is:

$$\begin{array}{ cccc }2~ x&+~~5~ y&+~~~~ z&~=~31\\&-~~~20~ y&&~=~-80\\2~ x&+~~5~ y&+~~~~ z&~=~31\end{array}$$

Step 5: Multiply the first equation by $ -1 $ and add the result to the third equation. The result is:

$$\begin{array}{ cccc }2~ x&+~~5~ y&+~~~~ z&~=~31\\&-~~~20~ y&&~=~-80\\&&0&~=~0\end{array}$$

Step 6: Divide the second equation by $ 20 $. The result is:

$$\begin{array}{ cccc }2~ x&+~~5~ y&+~~~~ z&~=~31\\&-~~~~~ y&&~=~-4\\&&0&~=~0\end{array}$$

Step 7: All the elements in the last row are zero, so the variable $ z $ can take any value, let

$$ \color{blue}{ z = t } $$

Step 8: Supstitute $ \color{blue}{ z = t } $ into second equation and solve for $ y $:

$$ \color{red}{y} = \frac{ -4 } { -1 } = \color{red}{ 4 } $$

Step 9: Supstitute $ y = 4 $ and $ z = t $ into first equation to find $ x $:

$$ \color{green}{x} = \frac{ 31 - 5 \cdot y - 1 \cdot z } { 2 } = \color{green}{ - \frac{ 1 }{ 2 } t + \frac{ 11 }{ 2 } }$$

System 3x3 calculator