Is 2x - 3y ? (2x - 3y = -2 x > 2 and y > 0...

Understanding the Question

We need to determine: Is \(2\mathrm{x} - 3\mathrm{y} < \mathrm{x}^2\)?

This is a yes/no question. To have sufficiency, we need to be able to answer definitively YES (the inequality is always true) or NO (the inequality is always false).

What We Need to Determine

We're comparing a linear expression (\(2\mathrm{x} - 3\mathrm{y}\)) with a quadratic expression (\(\mathrm{x}^2\)). The key insight: quadratic terms grow much faster than linear terms, especially as x increases.

Key Insights

Remember that \(\mathrm{x}^2\) is always non-negative (\(\geq 0\)) for any real value of x. This simple fact alone might be enough to answer our question in certain scenarios. Additionally, when x takes on larger values, \(\mathrm{x}^2\) will increasingly dominate the linear term \(2\mathrm{x}\).

Analyzing Statement 1

Statement 1: \(2\mathrm{x} - 3\mathrm{y} = -2\)

What Statement 1 Tells Us

This directly gives us the value of the left side of our inequality. So our question becomes:

Is \(-2 < \mathrm{x}^2\)?

Logical Analysis

Since \(\mathrm{x}^2\) is always non-negative for any real value of x:

- If \(\mathrm{x} \neq 0\), then \(\mathrm{x}^2 > 0\), which means \(-2 < \mathrm{x}^2\) ✓

- Even if \(\mathrm{x} = 0\), then \(\mathrm{x}^2 = 0\), and we still have \(-2 < 0\) ✓

The answer is always YES because a negative number (-2) is always less than a non-negative number (\(\mathrm{x}^2\)).

Conclusion

Statement 1 is sufficient to answer YES.

[STOP - Sufficient!]

This eliminates choices B, C, and E.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2: \(\mathrm{x} > 2\) and \(\mathrm{y} > 0\)

What Statement 2 Provides

We have constraints on both variables:

- x must be greater than 2

- y must be positive

Logical Analysis

When \(\mathrm{x} > 2\), we enter the region where quadratic dominance becomes crystal clear:

At the boundary (\(\mathrm{x} = 2\)):

- \(\mathrm{x}^2 = 4\)

- \(2\mathrm{x} = 4\)

- They're exactly equal

For any \(\mathrm{x} > 2\):

- \(\mathrm{x}^2\) grows faster than \(2\mathrm{x}\)

- The gap widens as x increases

Now consider our expression \(2\mathrm{x} - 3\mathrm{y}\):

- We start with \(2\mathrm{x}\) (which already loses to \(\mathrm{x}^2\) when \(\mathrm{x} > 2\))

- We subtract a positive quantity \(3\mathrm{y}\) (since \(\mathrm{y} > 0\))

- This makes the left side even smaller

Example to visualize: If \(\mathrm{x} = 3\), then:

- \(\mathrm{x}^2 = 9\)

- \(2\mathrm{x} = 6\)

- Even before subtracting \(3\mathrm{y}\), we have \(6 < 9\)

- After subtracting any positive \(3\mathrm{y}\), the gap only increases

The critical question: Could \(2\mathrm{x} - 3\mathrm{y}\) ever catch up to \(\mathrm{x}^2\)?

- This would require \(2\mathrm{x} - 3\mathrm{y} \geq \mathrm{x}^2\)

- In the best case for the left side (\(\mathrm{y} \to 0\)), we'd need \(2\mathrm{x} \geq \mathrm{x}^2\)

- This means \(\mathrm{x}^2 - 2\mathrm{x} \leq 0\), or \(\mathrm{x}(\mathrm{x} - 2) \leq 0\)

- This only happens when \(0 \leq \mathrm{x} \leq 2\)

- But we're told \(\mathrm{x} > 2\)!

Conclusion

Statement 2 is sufficient to answer YES - the quadratic dominance guarantees the inequality.

[STOP - Sufficient!]

The Answer: D

Both statements independently guarantee that \(2\mathrm{x} - 3\mathrm{y} < \mathrm{x}^2\), making each statement alone sufficient.

Answer Choice D: "Each statement alone is sufficient."