Alright, let's solve the given equations for the respective variables and classify them into the correct categories under [tex]\( x=3 \)[/tex] and [tex]\( x \neq 3 \)[/tex].
1. Solving [tex]\(-6 + x = -9\)[/tex]:
[tex]\[
-6 + x = -9 \implies x = -9 + 6 = -3
\][/tex]
[tex]\(\( x = -3\)[/tex])
2. Solving [tex]\(x - 5 = -2\)[/tex]:
[tex]\[
x - 5 = -2 \implies x = -2 + 5 = 3
\][/tex]
[tex]\(\( x = 3\)[/tex])
3. Solving [tex]\(\frac{z}{3} = 9\)[/tex]:
[tex]\[
\frac{z}{3} = 9 \implies z = 9 \times 3 = 27
\][/tex]
[tex]\(\( z = 27 \)[/tex])
4. Solving [tex]\(-\frac{3}{5} + x = \frac{12}{5}\)[/tex]:
[tex]\[
-\frac{3}{5} + x = \frac{12}{5} \implies x = \frac{12}{5} + \frac{3}{5} = \frac{15}{5} = 3
\][/tex]
[tex]\(\( x = 3\)[/tex])
5. Solving [tex]\(-14x = 0\)[/tex]:
[tex]\[
-14x = 0 \implies x = \frac{0}{-14} = 0
\][/tex]
[tex]\(\( x = 0\)[/tex])
6. Solving [tex]\(\frac{z}{4} = \frac{6}{8}\)[/tex]:
[tex]\[
\frac{z}{4} = \frac{6}{8} \implies z = (\frac{6}{8}) \times 4 = 3
\][/tex]
[tex]\(\( z = 3 \)[/tex])
Now, organizing the equations into the table:
[tex]\[
\begin{tabular}{|l|l|}
\hline
\textcolor{blue}{$x = 3$} & \textcolor{blue}{$x \neq 3$} \\
\hline
$x - 5 = -2$ & \(-6 + x = -9\) \\
$-\frac{3}{5} + x = \frac{12}{5}$ & \(\frac{z}{3} = 9\) \\
& $-14x = 0$ \\
& $\frac{z}{4} = \frac{6}{8}$ \\
\hline
\end{tabular}
\][/tex]
Here is the correctly sorted table of equations.
