Hi all! I have some questions on inequality i need help with. Thanks in advance

Graph the regions:

1. |xy|Greater than or equal to 1 (is x=0 included or excluded in this case?)

2.1/x>1/y

1) Yes, exclude x = 0, because if x is 0, |xy| = 0, which isn't greater than or equal to 1.

The region in question must be symmetric about both coordinate axes. This is because if a point (x,y) satisfies the inequality, so does (+/- x, +/- y), for any independent choices of the +/-. So the region to shade can be found by finding the appropriate region in the first quadrant, and then just flipping about the axes and origin (so get the corresponding pictures in the other quadrants).

So when x, y > 0 (first quadrant), the inequation becomes xy >= 1, so this is the region on and above the hyperbola xy = 1 (which is y = 1/x) in the first quadrant.

So to get the full region, sketch in these hyperbolic branches in each quadrant and shade in the places on the sides of the branches that are "away" from the origin (just like in the first quadrant). The branches themselves are part of the region, since the inequation was ">=", rather than a strict ">".

2) Note that x and y are both non-zero, since otherwise there'll be a zero denominator.

__This means the coordinate axes are excluded from the region to shade.__ Note that any point (x,y) with x > 0 and y < 0 satisfies the inequation, so shade in the whole fourth quadrant.

Clearly no point in the second quadrant satisfies the inequation, since then the LHS (1/x) is negative whilst the RHS (1/y) is positive.

Now, when x and y have the same sign (first or third quadrant), we have 1/x > 1/y if and only if x < y. So in quadrants 1 and 3, shade in the places where y > x (i.e. above the line y = x in these quadrants).