If x > 3 and y < 2, then x² − 2y > 5

Suppose x > 3 and y < 2. Then x² − 2y > 5.

Let’s break this down step by step.

We’re told:

• $x > 3$
• $y < 2$

We want to see if these imply:

$$ x^2 - 2y > 5 $$

---

Step 1: Lower bound for $x^2$

If $x > 3$, then:

$$ x^2 > 9 $$

---

Step 2: Lower bound for $-2y$

If $y < 2$, then multiplying both sides by $-2$ (which reverses the inequality):

$$ -2y > -4 $$

So the smallest possible value for $-2y$ is just greater than $-4$. It can be much larger if $y$ is negative (or even very large if $y$ is very negative).

---

Step 3: Combine

We have:

$$ x^2 - 2y > 9 - 4 = 5 $$

That’s already enough to conclude $x^2 - 2y > 5$, because:

• $x^2$ is strictly greater than 9,
• $-2y$ is strictly greater than $-4$, so their sum is strictly greater than $9 - 4 = 5$.

---

✅ Conclusion: Yes — from $x > 3$ and $y < 2$, it follows that:

$$ x^2 - 2y > 5 $$

---

If you’d like, I can also show you the geometric interpretation of this inequality in the $(x, y)$-plane, so you can see exactly which region these conditions describe.