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As we know, \(x\) can be positive or negative, the only required condition to satisfy the inequality is that \(y > 0\).BTGModeratorVI wrote: ↑Fri Apr 03, 2020 9:40 amFor integers x and y, x²y>>0. Which of the following must be true?
I. xy > 0
II. x > 0
III. y > 0
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
Answer: C
Source: Veritas Prep
One approach is to TEST some x and y values that satisfy the given condition that x²y > 0BTGModeratorVI wrote: ↑Fri Apr 03, 2020 9:40 amFor integers x and y, x²y>>0. Which of the following must be true?
I. xy > 0
II. x > 0
III. y > 0
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
Answer: C
Source: Veritas Prep
Since x^2 is always positive, y must be positive as well. So III is true. However, x could be either positive or negative. If x is negative, neither I nor II would be true.BTGModeratorVI wrote: ↑Fri Apr 03, 2020 9:40 amFor integers x and y, x²y>>0. Which of the following must be true?
I. xy > 0
II. x > 0
III. y > 0
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
Answer: C
Source: Veritas Prep
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