Question Bank - Equations, Inequalities, & VIC's #6

[Guest]

Is statement 2 correct? What if x = 3 and y= -2. The statement is still true.


Is x + y > 0?

(1) x - y > 0

(2) x2 - y2 > 0

We can rephrase the question by subtracting y from both sides of the inequality: Is x > -y?

(1) INSUFFICIENT: If we add y to both sides, we see that x is greater than y. We can use numbers here to show that this does not necessarily mean that x > -y. If x = 4 and y = 3, then it is true that x is also greater than -y. However if x = 4 and y = -5, x is greater than y but it is NOT greater than -y.

(2) INSUFFICIENT: If we factor this inequality, we come up (x + y)(x - y) > 0. For the product of (x + y) and (x - y) to be greater than zero, the must have the same sign, i.e. both negative or both positive. This does not help settle the issue of the sign of x + y.

(1) AND (2) SUFFICIENT: From statement 2 we know that (x + y) and (x - y) must have the same sign, and from statement 1 we know that (x - y) is positive, so it follows that (x + y) must be positive as well.

The correct answer is C.

[brian]

Guest,

Can you be more specific about your question? I'm trying to understand what you are looking for.

Thanks!

-Brian Lange
Instructor & Forum Moderator

[RonPurewal]

Is statement 2 correct? What if x = 3 and y= -2. The statement is still true.

(2) INSUFFICIENT: If we factor this inequality, we come up (x + y)(x - y) > 0. For the product of (x + y) and (x - y) to be greater than zero, the must have the same sign, i.e. both negative or both positive. This does not help settle the issue of the sign of x + y.

i think the poster has misunderstood the meaning of 'they'.

when we say 'THEY must have the same sign', we mean that the things actually being multiplied - i.e., x+y and x-y - must have the same sign. i think the poster misinterpreted the statement as referring to x and y themselves.

in the poster's example, x and y have opposite signs, but x + y (= 1) and x - y (= 5) have the same sign.

hth

[Guest]

Great. Thanks

[rfernandez]

You're welcome.