To find a point that would be a solution to the system of linear inequalities, we will first work with the two inequalities:

1. $y \geq \frac{3}{2}x - 8$
2. $y < \frac{1}{2}x + 3$

These inequalities represent two lines, and the solution to the system is the region where the shaded areas of both inequalities overlap.

Step 1: Graph the inequalities

To graph both inequalities, we first express them as equations (turning the inequalities into equalities), and then graph each one.

1. The first inequality is $y \geq \frac{3}{2}x - 8$. This is a line with a slope of $\frac{3}{2}$ and a y-intercept of $-8$.
2. The second inequality is $y < \frac{1}{2}x + 3$. This is a line with a slope of $\frac{1}{2}$ and a y-intercept of $3$.

Step 2: Find the intersection point

The point where the two lines intersect is the solution to the system of equations when the inequalities are treated as equalities. We can solve the system of equations:

$y = \frac{3}{2}x - 8$ $y = \frac{1}{2}x + 3$

Setting these two equations equal to each other:

$\frac{3}{2}x - 8 = \frac{1}{2}x + 3$

Now, solve for $x$:

$\frac{3}{2}x - \frac{1}{2}x = 3 + 8$ $\frac{2}{2}x = 11$ $x = 11$

Now, substitute $x = 11$ into either equation to find $y$. Let's use $y = \frac{1}{2}x + 3$:

$y = \frac{1}{2}(11) + 3 = \frac{11}{2} + 3 = \frac{11}{2} + \frac{6}{2} = \frac{17}{2}$

Thus, the intersection point is $(11, \frac{17}{2})$ or $(11, 8.5)$.

Step 3: Test a point to determine if it satisfies both inequalities

To find a point that satisfies both inequalities, we need a point that is below the line $y = \frac{3}{2}x - 8$ and above the line $y = \frac{1}{2}x + 3$, because of the $\geq$ and $<$ symbols.

A simple approach is to test the point $(0, 0)$, which is often easy to check:

1. For $y \geq \frac{3}{2}x - 8$, substitute $x = 0$ and $y = 0$: $0 \geq \frac{3}{2}(0) - 8 \quad \Rightarrow \quad 0 \geq -8 \quad \text{(True)}$
2. For $y < \frac{1}{2}x + 3$, substitute $x = 0$ and $y = 0$: $0 < \frac{1}{2}(0) + 3 \quad \Rightarrow \quad 0 < 3 \quad \text{(True)}$

Thus, the point $(0, 0)$ satisfies both inequalities.

Conclusion:

The point $(0, 0)$ is a solution to the system of inequalities.

Would you like further details or have any questions about this process? Here's some related exploration for you:

1. How would the solution change if the inequalities were reversed (e.g., $y < \frac{3}{2}x - 8$ and $y \geq \frac{1}{2}x + 3$)?
2. How do you graph inequalities that involve slopes and intercepts?
3. What happens if you test a point that lies on one of the boundary lines?
4. How can you verify the solution using substitution into the original inequalities?
5. How can you use systems of linear inequalities in real-life scenarios?

Tip: Always test a point in the region where you think the inequalities intersect to confirm if it lies within the feasible solution region!