We are tasked with finding the solution to the following system of inequalities:

1. $2y - x < -2$
2. $y + 3x < 2x^2 - 5$

To determine which points lie in the solution region of this system, we will evaluate each inequality at the given points and check if both inequalities are satisfied. The points to check are:

• $(7, -6)$
• $(-3, 4)$
• $(-6, 8)$
• $(0, 5)$

Let's evaluate both inequalities for each point.

Step 1: Evaluate at $(7, -6)$

Inequality 1:

$2(-6) - 7 < -2$ $-12 - 7 < -2$ $-19 < -2$ (True)

Inequality 2:

$-6 + 3(7) < 2(7)^2 - 5$ $-6 + 21 < 2(49) - 5$ $15 < 98 - 5$ $15 < 93$ (True)

Since both inequalities hold true, $(7, -6)$ is a solution.

Step 2: Evaluate at $(-3, 4)$

Inequality 1:

$2(4) - (-3) < -2$ $8 + 3 < -2$ $11 < -2$ (False)

Since the first inequality is false, $(-3, 4)$ is not a solution.

Step 3: Evaluate at $(-6, 8)$

Inequality 1:

$2(8) - (-6) < -2$ $16 + 6 < -2$ $22 < -2$ (False)

Since the first inequality is false, $(-6, 8)$ is not a solution.

Step 4: Evaluate at $(0, 5)$

Inequality 1:

$2(5) - 0 < -2$ $10 < -2$ (False)

Since the first inequality is false, $(0, 5)$ is not a solution.

Conclusion:

The only point that satisfies both inequalities is $(7, -6)$.

Would you like a graph for this system of inequalities?

---

5 Related Questions:

1. How do you graph a system of linear inequalities?
2. Can systems of inequalities have multiple solutions?
3. What are boundary lines in inequality graphs?
4. How do quadratic terms like $2x^2$ affect inequality regions?
5. How can inequalities be solved algebraically?

Tip: When graphing inequalities, always use a dashed line for strict inequalities ("<" or ">") and check which side of the boundary contains solutions.