Answer
Desired linear system is $y=x-4,y=-\frac{1}{3}x+4$.
Work Step by Step
We know that if the equation satisfies the value of $x$ and $y$ then the set is the solution of the equation.
Therefore, put the value of $x=6$ and $y=2$ in the equation
$\begin{align}
& x-y=4 \\
& 6-2=4 \\
& 4=4
\end{align}$
So, $\left( 6,2 \right)$ is the solution set of the equation $x-y=4$.
Then, for the slope–intercept form write the equation in the form of $y=mx+c$:
And add –x both sides in the equation $x-y=4$:
$\begin{align}
& x-y-x=4-x \\
& -y=4-x
\end{align}$
And multiply -1 both sides in the equation $x-y=4$:
$\begin{align}
& \left( -1\cdot -y \right)=-1\cdot 4-\left( -1\cdot x \right) \\
& y=x-4
\end{align}$
And put the value of $x=6$ and $y=2$ in the equation:
$\begin{align}
& x+3y=12 \\
& 6+3\cdot \left( 2 \right)=12 \\
& 12=12
\end{align}$
So, $\left( 6,2 \right)$ is the solution set of the equation $x+3y=12$.
And for the slope–intercept form write the equation in the form of $y=mx+c$:
And multiply $\frac{1}{3}$ both sides in the equation $x+3y=12$
$\begin{align}
& \frac{1}{3}\cdot x+\frac{1}{3}\cdot 3y=\frac{1}{3}\cdot 12 \\
& \frac{x}{3}+y=4
\end{align}$
And add $-\frac{x}{3}$ both sides in the equation $x+3y=12$:
$\begin{align}
& \frac{x}{3}+y-\frac{x}{3}=4-\frac{x}{3} \\
& y=-\frac{x}{3}+4
\end{align}$
But $x-3y=-6$ and $x-3y=6$ does not satisfy the solution set $\left( 6,2 \right)$.
Hence, the equation in slope–intercept form that satisfies the solution set $\left( 6,2 \right)$
$y=x-4,\ y=-\frac{1}{3}x+4$.
