Which point would be a solution to the system of linear inequalities shown below? \[ y \leq-2 x-6 \quad y \geq-\frac{5}{2} x-8 \] Answer \( \begin{array}{ll}(12,-4) & (-2,-3) \\ (12,-6) & (2,6)\end{array} \)

To find the solutions to the system of inequalities, we can substitute each given point into both inequalities. For example, the point (-2, -3) can be checked against both inequalities. Substituting in the first inequality \(y \leq -2x - 6\) gives \(-3 \leq -2(-2) - 6\), which simplifies to \(-3 \leq -2\), not true. Moving on to the second inequality \(y \geq -\frac{5}{2}x - 8\) gives \(-3 \geq -\frac{5}{2}(-2) - 8\), simplifying to \(-3 \geq -2.5 - 8\), which is true. Therefore, -2, -3 is not a solution to the system. Next, let’s try the point (12, -4). For \(y \leq -2x - 6\): \(-4 \leq -2(12) - 6\) simplifies to \(-4 \leq -30\), which is not true, so this point does not satisfy the first inequality. After testing all the options, we find that (-2, -3) and (12, -6) fail one or both inequalities. The key is finding a point that lies within the overlapping region defined by both inequalities. By graphing or testing points systematically, we can identify points that work. For example, if we find a point like (0, -5) or (0, -7), they might lead us to identify a correct solution that satisfies both inequalities effectively. A common mistake in these types of problems is to only evaluate the inequalities without considering their overlapping regions. Remember that the inequalities need to be true simultaneously for a point to be a solution!