Problem Solving

duahsolo wrote:If x^2 = y^2, is it true that x > 0?

(1) x = 2y + 1

(2) y <= -1

Statement 1: x = 2y+1
Substituting x = 2y+1 into x²=y², we get:
(2y+1)² = y²
4y² + 4y + 1 = y²
3y² + 4y + 1 = 0
(3y+1)(y+1) = 0.

Case 1: 3y+1 = 0, implying that y=(-1/3) and that x = 2(-1/3) + 1 = 1/3.
Case 2: y+1 = 0, implying that y=-1 and that x = 2(-1) + 1 = -1.

Since x>0 in Case 1 but x<0 in Case 2, INSUFFICIENT.

Statement 2: y≤-1
Substituting y=-1 into x²=y², we get:
x² = (-1)²
x² = 1
x = ±1.
Since it's possible that x>0 or that x<0, INSUFFICIENT.

Statements combined:
Of the two cases yielded in Statement 1, only Case 2 satisfies Statement 2.
In Case 2, x=-1.
Since x<0, the answer to the question stem is NO.
SUFFICIENT.

The correct answer is C.