Originally Posted by

**Soroban** Hello, Morphayne!

o_O has the best method.

I'll show you another . . .

We have: .$\displaystyle \begin{array}{cccc}3x - y + 4z &=& 2 & {\color{blue}[1]} \\ x + 6y + 10z &=& \text{-}8 & {\color{blue}[2]} \end{array}$

$\displaystyle \begin{array}{cccc}\text{Multiply {\color{blue}[1]} by 6:} & 18x - 6y + 24z &=& 12 \\ \text{Add {\color{blue}[2]}:} & x + 6y + 10z &=& -8 \\

\text{and we have:} & 19x + 34z &=& 4\end{array}$

. . Hence: .$\displaystyle x \:=\:\frac{4}{19} - \frac{34}{19}z$

Substitute into [2]: .$\displaystyle \left(\frac{4}{19} - \frac{34}{19}z\right) + 6y + 10z \:=\:\text{-}8 \quad\Rightarrow\quad y \:=\:\text{-}\frac{2}{19} - \frac{26}{19}z$

We have: .$\displaystyle \begin{Bmatrix}x &=& \frac{4}{19} - \frac{34}{19}z \\ \\[-3mm] y &=& \text{-}\frac{2}{19} - \frac{26}{19}z \\ \\[-3mm] z &=& z \end{Bmatrix} $

On the right side, replace $\displaystyle z$ with the parameter $\displaystyle t.$

And we have: . $\displaystyle \begin{Bmatrix}x &=& \frac{4}{19} - \frac{34}{19}t \\ \\[-3mm] y &=& \text{-}\frac{2}{19} - \frac{26}{19}t \\ \\[-3mm] z &=& 0 + t \end{Bmatrix} $

Therefore: .$\displaystyle [x,y,z] \;=\;\left[\frac{4}{19},\:\text{-}\frac{2}{19},\:0\right] + t\left[\text{-}\frac{34}{19},\:\text{-}\frac{26}{19},\:1\right] $