Determine the solution region for the following system of linear inqeualities by inputting a point in that region. The graph, without shaded solution, is shown on the right.
(Enter an ordered pair $(x,y)$ with integer values, that is, no decimals or fractions.) (Do not enter a point that is actually on one of the lines; keep it in the interior of the solution region.)
$$\begin{array}{r}-3x-3y<24\\ -3x+y<12\end{array}$$
Point in the Solution Region:

Step 1 ：To determine a point in the solution region, we need to check if it satisfies both inequalities.

Step 2 ：First inequality: $-3x-3y<24$ becomes $-3(0)-3(0)<24$, which simplifies to $0<24$. This is true.

Step 3 ：Second inequality: $-3x+y<12$ becomes $-3(0)+0<12$, which simplifies to $0<12$. This is also true.

Step 4 ：Since the origin satisfies both inequalities, it is part of the solution set.

Step 5 ：To graph the solution, we first graph the two inequalities as lines and then shade the appropriate regions.

Step 6 ：First inequality: $y=-x+8$ is a line with slope $-1$ and y-intercept $8$.

Step 7 ：Second inequality: $y=3x+12$ is a line with slope $3$ and y-intercept $12$.

Step 8 ：Since both inequalities are strict inequalities $<$, we use a dashed line for both lines.

Step 9 ：We shade the region below the first line and above the second line.

Step 10 ：One point that lies in the solution set is $(0,0)$, as it is in the shaded region.

Step 11 ：$\boxed{{\displaystyle \text{Solution point: (0,0)}}}$