Step 1: Convert all decimals to fractions

$$\begin{array}{ cccc }300~ x&+~~200~ y&+~~200~ z&~=~3500\\40~ x&+~~60~ y&+~~40~ z&~=~660\\120~ x&+~~150~ y&+~~108~ z&~=~1806\end{array}$$

Step 2: Divide the first equation by $ 100 $. The result is:

$$\begin{array}{ cccc }3~ x&+~~2~ y&+~~2~ z&~=~35\\40~ x&+~~60~ y&+~~40~ z&~=~660\\120~ x&+~~150~ y&+~~108~ z&~=~1806\end{array}$$

Step 3: Divide the second equation by $ 20 $. The result is:

$$\begin{array}{ cccc }3~ x&+~~2~ y&+~~2~ z&~=~35\\2~ x&+~~3~ y&+~~2~ z&~=~33\\120~ x&+~~150~ y&+~~108~ z&~=~1806\end{array}$$

Step 4: Divide the third equation by $ 6 $. The result is:

$$\begin{array}{ cccc }3~ x&+~~2~ y&+~~2~ z&~=~35\\2~ x&+~~3~ y&+~~2~ z&~=~33\\20~ x&+~~25~ y&+~~18~ z&~=~301\end{array}$$

Step 5: Multiply the first equation by $ -\frac{ 2 }{ 3 } $ and add the result to the second equation. The result is:

$$\begin{array}{ cccc }3~ x&+~~2~ y&+~~2~ z&~=~35\\&\dfrac{ 5 }{ 3 }~ y&+~~\dfrac{ 2 }{ 3 }~ z&~=~\dfrac{ 29 }{ 3 }\\20~ x&+~~25~ y&+~~18~ z&~=~301\end{array}$$

Step 6: Multiply the first equation by $ -\frac{ 20 }{ 3 } $ and add the result to the third equation. The result is:

$$\begin{array}{ cccc }3~ x&+~~2~ y&+~~2~ z&~=~35\\&\dfrac{ 5 }{ 3 }~ y&+~~\dfrac{ 2 }{ 3 }~ z&~=~\dfrac{ 29 }{ 3 }\\&\dfrac{ 35 }{ 3 }~ y&+~~\dfrac{ 14 }{ 3 }~ z&~=~\dfrac{ 203 }{ 3 }\end{array}$$

Step 7: Get rid of the fractions by multiplying the second equation by 3 and the third equation by 3. The result is:

$$\begin{array}{ cccc }3~ x&+~~2~ y&+~~2~ z&~=~35\\&5~ y&+~~2~ z&~=~29\\&35~ y&+~~14~ z&~=~203\end{array}$$

Step 8: Divide the third equation by $ 7 $. The result is:

$$\begin{array}{ cccc }3~ x&+~~2~ y&+~~2~ z&~=~35\\&5~ y&+~~2~ z&~=~29\\&5~ y&+~~2~ z&~=~29\end{array}$$

Step 9: Multiply the second equation by $ -1 $ and add the result to the third equation. The result is:

$$\begin{array}{ cccc }3~ x&+~~2~ y&+~~2~ z&~=~35\\&5~ y&+~~2~ z&~=~29\\&&0&~=~0\end{array}$$

Step 10: All the elements in the last row are zero, so the variable $ z $ can take any value, let

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

Step 11: Supstitute $ \color{blue}{ z = t } $ into second equation and solve for $ y $:

$$ \color{red}{y} = \frac{ - 2 z + 29 } { 5 } = \color{red}{ - \frac{ 2 }{ 5 } t + \frac{ 29 }{ 5 } } $$

Step 12: Supstitute $ y = - \frac{ 2 }{ 5 } t + \frac{ 29 }{ 5 } $ and $ z = t $ into first equation to find $ x $:

$$ \color{green}{x} = \frac{ 35 - 2 \cdot y - 2 \cdot z } { 3 } = \color{green}{ - \frac{ 2 }{ 5 } t + \frac{ 39 }{ 5 } }$$

System 3x3 calculator