CAMANISHPARMAR wrote:
If xyz ≠ 0, xy > 0, and x^2 + xy + xz < 0, then which of the following must be true?
I. x(y + z) < 0
II. x + y + z < 0
III. If x < 0, then z > 0.
A) I only
B) I and II only
C) I and III only
D) II and III only
E) I, II, and III
Fun problem! The key with these is to do as much work as you can using logic and simplification, before you start testing cases. That's especially true on this problem, because there are three variables. If there are more than two variables in a problem, I usually try to avoid testing cases as much as possible - it just takes too long to consider all of the possibilities.
The first thing I noticed is that
x and y have the same sign.
I also noticed that the inequality, \(x^2 + xy + xz < 0\), can be simplified like this: \(x(x+y+z) < 0\). This means that
either x is negative, or x+y+z is negative, but not both.Now, go look at options I, II, and III.
Option II stood out to me the most, because I already know something about x+y+z from my reasoning above. It doesn't HAVE to be negative, since the alternative is that x could just be negative. Then, if y and z were big enough, x+y+z would be positive.
Eliminate B, D, and E, because they include option II.
This would be a good point to guess! Or, we could keep going if we have enough time.
Notice that the difference between (A) and (C) is that (A) doesn't include III, while (C) does. They both include I! So, at this point,
we know that I has to be true without testing it. We only have to worry about III. That's a good reason to use the process of elimination!
III says that if x < 0, z > 0. Well, if x < 0, from the above reasoning, we know that x + y + z > 0. We also know that x and y have the same sign. So, here's what we have:
If x is negative, y is also negative, but x + y + z is positive.
This should only happen if z is positive! You can't add three negative numbers together and get a positive result. So, z definitely has to be positive. III is true.
Eliminate (A) and choose (C).
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