To determine which point would not be a solution to the given system of inequalities, let’s examine each inequality with the provided points:

Inequalities:

1. $y \geq \frac{3}{2}x + 4$
2. $y < \frac{1}{2}x + 5$

For each point, we will substitute the $x$ and $y$ values and check if they satisfy both inequalities.

Testing the Points:

1. Point $(-8, -4)$:

   • First inequality: $-4 \geq \frac{3}{2}(-8) + 4 \Rightarrow -4 \geq -12 + 4 \Rightarrow -4 \geq -8$ (True)
   • Second inequality: $-4 < \frac{1}{2}(-8) + 5 \Rightarrow -4 < -4 + 5 \Rightarrow -4 < 1$ (True)

   This point satisfies both inequalities.

2. Point $(-8, -2)$:

   • First inequality: $-2 \geq \frac{3}{2}(-8) + 4 \Rightarrow -2 \geq -12 + 4 \Rightarrow -2 \geq -8$ (True)
   • Second inequality: $-2 < \frac{1}{2}(-8) + 5 \Rightarrow -2 < -4 + 5 \Rightarrow -2 < 1$ (True)

   This point satisfies both inequalities.

3. Point $(-12, -6)$:

   • First inequality: $-6 \geq \frac{3}{2}(-12) + 4 \Rightarrow -6 \geq -18 + 4 \Rightarrow -6 \geq -14$ (True)
   • Second inequality: $-6 < \frac{1}{2}(-12) + 5 \Rightarrow -6 < -6 + 5 \Rightarrow -6 < -1$ (True)

   This point satisfies both inequalities.

4. Point $(8, 0)$:

   • First inequality: $0 \geq \frac{3}{2}(8) + 4 \Rightarrow 0 \geq 12 + 4 \Rightarrow 0 \geq 16$ (False)

   Since this point does not satisfy the first inequality, it does not satisfy the system.

Conclusion:

The point $(8, 0)$ would not be a solution to the system of inequalities.

Would you like further details on this solution, or do you have any additional questions?

Related Questions:

1. How do you determine if a point is a solution to a system of inequalities?
2. What are common strategies for solving systems of linear inequalities graphically?
3. How do you graph inequalities with different inequality symbols (e.g., $\geq$ vs. $>$)?
4. What applications do systems of inequalities have in real-life situations?
5. How can we interpret the intersection region of inequalities?

Tip:

When testing points in a system of inequalities, always substitute each point into each inequality and verify if the point satisfies all inequalities in the system.