Answer
Desired system is $y=-2+\frac{x}{3}$, $y=2+\frac{x}{3}$.
Work Step by Step
The provided set is not the solution of the equation if it does not satisfy the equation.
So, put the values $x=6$ and $y=2$ in the equation:
$\begin{align}
& x-3y=6 \\
& 6-3\cdot 2=6 \\
& 6-6=6 \\
& 0\ne 6
\end{align}$
So, $\left( 6,2 \right)$ is not the solution set of the equation $x-3y=6$
And the solution is $\varnothing $,
And for the slope–intercept form write the equation in the form of $y=mx+c$
Add -x to both sides in the equation $x-3y=6$:.
$\begin{align}
& x-3y-x=6-x \\
& -3y=6-x
\end{align}$
And multiply $-\frac{1}{3}$ on both sides of the equation $x-3y=6$:
$\begin{align}
& -3\cdot \frac{-1}{3}y=-\frac{1}{3}\cdot 6+\left( \frac{1}{3} \right)\cdot x \\
& y=-2+\frac{x}{3}
\end{align}$
Again, substitute the values of $x=6$ and $y=2$ in the equation
$\begin{align}
& x-3y=-6 \\
& 6-3\cdot \left( 2 \right)=-6 \\
& 6-6=-6 \\
& 0\ne -6
\end{align}$
So, $\left( 6,2 \right)$ is not the solution set of the equation $x-3y=-6$.
And for the slope–intercept form, write the equation in the form of $y=mx+c$
Add $-x$ to both sides in the equation $x-3y=-6$
$\begin{align}
& x-3y-x=-6-x \\
& -3y=-6-x
\end{align}$
Multiply $-\frac{1}{3}$ on both sides of the equation $x-3y=-6$:
$\begin{align}
& -3\cdot -\frac{1}{3}y=-6\cdot \left( -\frac{1}{3} \right)-x\cdot \left( -\frac{1}{3} \right) \\
& y=2+\frac{x}{3}
\end{align}$
But $x-y=4$ and $x+3y=12$ are satisfied by the solution set $\left( 6,2 \right)$.
Hence, the equations in slope–intercept form that do not satisfy the solution set $\left( 6,2 \right)$ are:
$y=-2+\frac{x}{3}$, $y=2+\frac{x}{3}$
