\(\sqrt{\text{(Negative number)}}\text{ is not possible. Think about what makes }6x-3\text{ a negative number.}\)
Nope, it is already solved.
\((x,y,z) = \left(\dfrac{8}{7},\dfrac{19}{7}, \dfrac{-11}{7}\right)\).
See the parts I have encircled with a box.
It's because you posted the same problem twice.
More accurately, it's every real number except -8.
\(\begin{pmatrix} 4&2&|&23\\ -1&3&|&4\\ \end{pmatrix}\\\sim\begin{pmatrix} 0&14&|&39\\ -1&3&|&4\\ \end{pmatrix}\\ y = \dfrac{39}{14}\\ -x + 3\left(\dfrac{39}{14}\right) = 4\\ x = \dfrac{61}{14}\)
Yes. Then the domain is every real number except -8.
\(\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ -2&4&2&|1 \end{pmatrix}\\ \sim\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ 0&6&6&|3\\ \end{pmatrix}\\ \sim\begin{pmatrix} 1&1&2&|1\\ 0&1&1&|1\\ 0&1&1&|\dfrac{1}2\\ \end{pmatrix}\\\)
No solution.
\(\begin{pmatrix} 5&0&3|1\\ -1&1&1|0\\ 0&3&2|5\\ \end{pmatrix}\\ \sim \begin{pmatrix} 5&0&3&|1\\ 0&1&\dfrac{8}5&|\dfrac{1}{5}\\ 0&3&2&|5\\ \end{pmatrix}\\ \sim \begin{pmatrix} 5&0&3&|1\\ 0&1&\dfrac{8}5&|\dfrac{1}{5}\\ 0&0&-\dfrac{14}{5}&|\dfrac{22}{5}\\ \end{pmatrix}\\ -\dfrac{14}{5}z = \dfrac{22}5\\ \boxed{z = \dfrac{-11}{7}}\\ y + \dfrac{8}{5}z = \dfrac{1}{5}\\ \boxed{y = \dfrac{1}{5} + \dfrac{8}{5}\cdot \dfrac{11}{7} = \dfrac{19}{7}}\\ 5x + 3z = 1\\ \boxed{x = \dfrac{1-3(\dfrac{-11}{7})}{5} = \dfrac{8}{7}}\)
Let's call the first equation (1) and the second equation (2).
Now, consider (1) - 2*(2) and you are done.
You are spamming.
