Answer
The linear system of equations is:
\[\begin{align}
& y=x-4 \\
& y=-\frac{1}{3}x+4
\end{align}\]
Work Step by Step
Since the solution set is\[\left\{ \left( 6,2 \right) \right\}\], it implies the two equations intersect at the point\[\left( 6,2 \right)\].
According to the given graph the equations that intersect at\[\left( 6,2 \right)\]are,
\[\begin{align}
& x-y=4 \\
& x+3y=12
\end{align}\]
Solve the above equations to verify.
Subtract the two equations as follows:
\[\underline{\begin{align}
& x-y=4 \\
& -x-3y=-12
\end{align}}\]
\[\begin{align}
& \ \ \ \ \ \ -4y=-8 \\
& y=2
\end{align}\]
Put \[y=2\]in any equation to get,
\[\begin{align}
& x-2=4 \\
& x=6
\end{align}\]
To verify put\[y=2\]and\[x=6\]in\[x+3y=12\]as follows:
\[\begin{align}
& 6+3\left( 2 \right)=12 \\
& 6+6=12 \\
& 12=12
\end{align}\]
RHS\[=\]LHS
Hence verified.
Hence, the linear system of equations is:
\[\begin{align}
& y=x-4 \\
& y=-\frac{1}{3}x+4
\end{align}\]
With the solution set as\[\left\{ \left( 6,2 \right) \right\}\].
