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## Comment on

Which Quadrant?## What about x= -2 and y=-3?

## Are you trying to show that

Are you trying to show that the combined statements are not sufficient? If so, I should point out that x= -2 and y= -3 does not meet the condition in statement 1 (xy < 0), because (-2)(-3) = 6, and 6 is not less than zero.

## I think he's suggesting

## Ahh, good point

Ahh, good point/interpretation!

So, statement 2 is not sufficient because the point could lie in THREE possible quadrants.

## Hi Brent,

In statement 2, X-Y>0, if we picked up a negative number for Y, would not the inequality X-(-Y), which would result in a positive number to rely on quadrant 1? Thank you!

## Not quite.

Not quite.

If y is negative, then there are many possible cases that satisfy the inequality x - y > 0:

CASE A) x = 3 and y = -1. In this case, (x,y) is in quadrant IV.

CASE B) x = -1 and y = -3. In this case, (x,y) is in quadrant III.

Also note that there's nothing in statement 1 to suggest that y is negative. So, we must also consider positive values for y such as:

CASE C) x = 3 and y = 1. In this case, (x,y) is in quadrant I.

Does that help?

Cheers,

Brent

## It does, thank you Brent for

## Hi Brent. Would statement 2

## You're correct to say that,

You're correct to say that, in Quadrant IV, x-y is always positive.

So, the point COULD be in Quadrant IV.

However, there are points in other quadrants where x-y is also positive.

For example, the point (3,2) lies in Quadrant I, and here x - y = 3 - 2 = 1, which is positive.

So, the point COULD be in Quadrant I.

Likewise, the point (-1,-5) lies in Quadrant III, and here x - y = (-1) - (-5) = 4, which is positive.

So, the point COULD be in Quadrant III.

Since we can't be certain which quadrant the point lies, statement 2 is not sufficient.

Does that help?

Cheers,

Brent