To determine which point is a solution to the system of inequalities:

$y < 3x + 8 \quad \text{and} \quad y > -2x + 3$

we will test each of the given points by substituting the coordinates $(x, y)$ into both inequalities and check if both inequalities are satisfied.

1. Point $(-6, 0)$

• For $y < 3x + 8$: $0 < 3(-6) + 8 = -18 + 8 = -10$ This is false because $0 \not< -10$.

2. Point $(-12, -10)$

• For $y < 3x + 8$: $-10 < 3(-12) + 8 = -36 + 8 = -28$ This is false because $-10 \not< -28$.

3. Point $(-12, 4)$

• For $y < 3x + 8$: $4 < 3(-12) + 8 = -36 + 8 = -28$ This is false because $4 \not< -28$.

4. Point $(4, 10)$

• For $y < 3x + 8$: $10 < 3(4) + 8 = 12 + 8 = 20$ This is true because $10 < 20$.

• For $y > -2x + 3$: $10 > -2(4) + 3 = -8 + 3 = -5$ This is true because $10 > -5$.

Conclusion:

The point $(4, 10)$ satisfies both inequalities, so it is the solution.

Answer: $(4, 10)$

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Do you want further details on any part of this solution, or have any questions?

Here are 5 related questions:

1. How do you graph the system of inequalities $y < 3x + 8$ and $y > -2x + 3$?
2. Can a system of inequalities have more than one solution?
3. How would the solution region of this system change if the inequalities were non-strict (i.e., $\leq$ and $\geq$)?
4. How do you solve a system of inequalities with more than two constraints?
5. Can the method used here apply to nonlinear inequalities?

Tip: Always graph the system of inequalities to visualize the solution set clearly.