Answer
see graph
Work Step by Step
Changing the given inequality, $
x-6y\lt12
,$ to equality and then isolating $y$ result to
\begin{array}{l}\require{cancel}
x-6y=12
\\\\
-6y=-x+12
\\\\
y=\dfrac{-1}{-6}x+\dfrac{12}{-6}
\\\\
y=\dfrac{1}{6}x-2
.\end{array}
Use the table of values below to graph this line.
Since the inequality used is "$
\lt
$", use broken lines.
Using the testpoint $(
0,0
)$, then
\begin{array}{l}\require{cancel}
0-6(0)\lt12
\\\\
0-0\lt12
\\\\
0\lt12
\text{ (TRUE)}
.\end{array}
Since the solution above ended with a $\text{
TRUE
}$ statement, then the testpoint is $\text{
part
}$ of the solution set.