If x^2 = y^2, is true that x > 0 ? x = 2y+1 y leq -1...

Understanding the Question

We need to determine: Is \(\mathrm{x > 0}\)?

Given information: \(\mathrm{x^2 = y^2}\)

This equation tells us something crucial. When two numbers have equal squares, they're either equal or opposites of each other. So \(\mathrm{x^2 = y^2}\) means either \(\mathrm{x = y}\) or \(\mathrm{x = -y}\). There's no other possibility.

To determine whether x is positive, we need to know:

1. Which relationship holds (\(\mathrm{x = y}\) or \(\mathrm{x = -y}\))
2. The sign of the resulting value

For this yes/no question to be sufficient, we need to definitively answer either "yes, \(\mathrm{x > 0}\)" or "no, \(\mathrm{x \leq 0}\)".

Analyzing Statement 1

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

This gives us a specific relationship between x and y. Combined with \(\mathrm{x^2 = y^2}\), let's see what this tells us.

If \(\mathrm{x = 2y + 1}\) and \(\mathrm{x^2 = y^2}\), we can substitute to get:
\(\mathrm{(2y + 1)^2 = y^2}\)

Expanding the left side: \(\mathrm{4y^2 + 4y + 1 = y^2}\)

Simplifying: \(\mathrm{3y^2 + 4y + 1 = 0}\)

Using the quadratic formula: \(\mathrm{y = \frac{-4 \pm \sqrt{16-12}}{6} = \frac{-4 \pm 2}{6}}\)

This gives us \(\mathrm{y = -\frac{1}{3}}\) or \(\mathrm{y = -1}\).

Let's check what happens with x in each case:

• If \(\mathrm{y = -\frac{1}{3}}\), then \(\mathrm{x = 2(-\frac{1}{3}) + 1 = -\frac{2}{3} + 1 = \frac{1}{3} > 0}\) ✓
• If \(\mathrm{y = -1}\), then \(\mathrm{x = 2(-1) + 1 = -2 + 1 = -1 < 0}\) ✗

Since we get different answers about whether \(\mathrm{x > 0}\) (YES when \(\mathrm{y = -\frac{1}{3}}\), NO when \(\mathrm{y = -1}\)), Statement 1 alone is NOT sufficient.

This eliminates choices A and D.

Analyzing Statement 2

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

Statement 2: \(\mathrm{y \leq -1}\)

Remember from the question that \(\mathrm{x^2 = y^2}\), which means \(\mathrm{x = y}\) or \(\mathrm{x = -y}\).

Let's test both cases when \(\mathrm{y \leq -1}\):

• Case 1: If \(\mathrm{x = y}\) and \(\mathrm{y \leq -1}\), then \(\mathrm{x \leq -1}\), so \(\mathrm{x < 0}\) ✗
• Case 2: If \(\mathrm{x = -y}\) and \(\mathrm{y \leq -1}\), then \(\mathrm{x \geq 1}\), so \(\mathrm{x > 0}\) ✓

We get opposite answers depending on which relationship holds. Since we don't know whether \(\mathrm{x = y}\) or \(\mathrm{x = -y}\), Statement 2 alone is NOT sufficient.

This eliminates choice B (and confirms D is already eliminated).

Combining Statements

From Statement 1, we found that y must be either \(\mathrm{-\frac{1}{3}}\) or \(\mathrm{-1}\).
From Statement 2, we know \(\mathrm{y \leq -1}\).

The only value that satisfies both conditions is \(\mathrm{y = -1}\).

When \(\mathrm{y = -1}\), Statement 1 tells us that \(\mathrm{x = 2(-1) + 1 = -1}\).

Therefore, \(\mathrm{x = -1 < 0}\), and we can definitively answer "NO" to whether \(\mathrm{x > 0}\).

[STOP - Sufficient!] Together, the statements are sufficient.

This eliminates choice E.

The Answer: C

Both statements together are sufficient to determine that x is not positive, but neither statement alone is sufficient.

Answer Choice C: "Both statements together are sufficient, but neither statement alone is sufficient."