Can I have help with this? Please I would really appreciate it!

5x-y+4z=14

-5x-y-2z=-10

-5x+3y-6z=-18

Solving this by linear combination?

$\begin{array}{|lrcll|} \hline (1): & 5x - y + 4z & = & 14 \\ (2): & -5x - y - 2z & = & -10 \\ (3): & -5x +3y - 6z & = & -18 \\\\ \hline \\ (1) + (2): & 5x - y + 4z -5x - y - 2z &=& 14 - 10 \\ & - 2y + 2z &=& 4 \quad & | \quad :2 \\ & - y + z &=& 2 \\ (I): & z &=& 2 + y \\\\ (1) + (3): & 5x - y + 4z -5x +3y - 6z &=& 14 - 18 \\ & 2y - 2z &=& -4 \quad & | \quad : (-2) \\ & -y + z &=& 2 \\ (II): & z &=& 2 + y \quad & | \quad \mathbf{(II)= (I)\ !!!} \\\\ \hline \\ (1): & 5x - y + 4z & = & 14 \quad & | \quad z=2+y \\ & 5x - y + 4\cdot(2+y) & = & 14 \\ (III): & 5x +3y & = & 6 \\\\ (2): & -5x - y - 2z & = & -10 \quad & | \quad z=2+y \\ & -5x - y - 2\cdot(2+y) & = & -10 \\ & -5x -3y & = & -6 \quad & | \quad \cdot(-1) \\ (IV): & 5x +3y & = & 6 \quad & | \quad \mathbf{(IV)= (III)\ !!!} \\\\ \hline \end{array}$

Solution:

$\text{Let } t \in R \text{ and we set } z = t$

$\begin{array}{|lrcll|} \hline (I): & z &=& 2 + y \quad & | \quad z=t \\ & t &=& 2 + y \\ & y &=& t-2 \\\\ (III): & 5x +3y & = & 6 \quad & | \quad y = t-2 \\ & 5x +3\cdot (t-2) & = & 6 \\ & 5x +3t - 6 & = & 6 \\ & 5x +3t & = & 12 \\ & 5x & = & 12 - 3t \\ & 5x & = & 12 - 3t \quad & | \quad : 5 \\ & x & = & \frac{12}{5} - \frac{3}{5}t \\ \hline \end{array}$

summary:
$\begin{array}{|rcll|} \hline t \in R \\ x &=& \frac{12}{5} - \frac{3}{5}t \\ y &=& t-2\\ z &=& t \\ \hline \end{array}$

are you sure thats right