Step 1: Convert all decimals to fractions

$$\begin{array}{ cccc }4~ x&+~~3~ y&+~~2~ z&~=~95\\5~ x&+~~2~ y&+~~3~ z&~=~98\\12~ x&+~~9~ y&+~~6~ z&~=~285\end{array}$$

Step 2: Divide the third equation by $ 3 $. The result is:

$$\begin{array}{ cccc }4~ x&+~~3~ y&+~~2~ z&~=~95\\5~ x&+~~2~ y&+~~3~ z&~=~98\\4~ x&+~~3~ y&+~~2~ z&~=~95\end{array}$$

Step 3: Multiply the first equation by $ -\frac{ 5 }{ 4 } $ and add the result to the second equation. The result is:

$$\begin{array}{ cccc }4~ x&+~~3~ y&+~~2~ z&~=~95\\&-~~~\dfrac{ 7 }{ 4 }~ y&+~~\dfrac{ 1 }{ 2 }~ z&~=~-\dfrac{ 83 }{ 4 }\\4~ x&+~~3~ y&+~~2~ z&~=~95\end{array}$$

Step 4: Multiply the first equation by $ -1 $ and add the result to the third equation. The result is:

$$\begin{array}{ cccc }4~ x&+~~3~ y&+~~2~ z&~=~95\\&-~~~\dfrac{ 7 }{ 4 }~ y&+~~\dfrac{ 1 }{ 2 }~ z&~=~-\dfrac{ 83 }{ 4 }\\&&0&~=~0\end{array}$$

Step 5: Get rid of the fractions by multiplying the second equation by 4. The result is:

$$\begin{array}{ cccc }4~ x&+~~3~ y&+~~2~ z&~=~95\\&-~~~7~ y&+~~2~ z&~=~-83\\&&0&~=~0\end{array}$$

Step 6: All the elements in the last row are zero, so the variable $ z $ can take any value, let

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

Step 7: Supstitute $ \color{blue}{ z = t } $ into second equation and solve for $ y $:

$$ \color{red}{y} = \frac{ - 2 z - 83 } { -7 } = \color{red}{ \frac{ 2 }{ 7 } t + \frac{ 83 }{ 7 } } $$

Step 8: Supstitute $ y = \frac{ 2 }{ 7 } t + \frac{ 83 }{ 7 } $ and $ z = t $ into first equation to find $ x $:

$$ \color{green}{x} = \frac{ 95 - 3 \cdot y - 2 \cdot z } { 4 } = \color{green}{ - \frac{ 5 }{ 7 } t + \frac{ 104 }{ 7 } }$$

System 3x3 calculator