Precalculus (6th Edition) Blitzer

Desired system is $y=-2+\frac{x}{3}$, $y=2+\frac{x}{3}$.
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: $y=-2+\frac{x}{3}$, $y=2+\frac{x}{3}$