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.) \[ \begin{array}{r} -3 x-3 y<24 \\ -3 x+y<12 \end{array} \] 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 ：\(\boxed{\text{Solution point: (0,0)}}\)