Inequalities and multiplying/dividing by a variable

Question

The book says that when you multiply/divide both sides of an inequality, the relation switches if what you're multiplying/dividing by is negative. That's fine if I'm multiplying/dividing by a constant, but what happens when I'm multiplying/dividing by a variable or an expression containing variables? Does the inequality then branch into two forms?

For example, let's say I have the inequality 5 < 10 and I want to multiply both sides by x-1. Does the inequality then branch out to multiple inequalities?

* 5\*(x-1) < 10\*(x-1), if x-1 > 0
* 5\*(x-1) > 10\*(x-1), if x-1 < 0
* ????, if x-1 = 0

Then what happens when you keep taking on stuff to the sides? Now lets say I want to divide both sides by y. Do the branches keep exploding out?

* (5\*(x-1))/y < (10\*(x-1))/y, if x-1 > 0 and y > 0
* (5\*(x-1))/y > (10\*(x-1))/y, if x-1 > 0 and y < 0
* (5\*(x-1))/y > (10\*(x-1))/y, if x-1 < 0 and y > 0
* (5\*(x-1))/y < (10\*(x-1))/y, if x-1 < 0 and y < 0
* ... cases where x-1 and y are 0 ...

Answer 1

Yes, though the case where the multiplicand is 0 is trivial: you just get 0 on both sides and `0 = 0`. And you can't divide by 0 so that case doesn't apply.

Answer 2

Yes. Inequalities compare real numbers (for most purposes anyway) instead of general expressions/functions. So if you see an inequality in the real variable x,

f(x) < g(x)

we are actually evaluating the expressions one value of x at a time. It's easy to come up with an example wherein both

f(x_0) < g(x_0)

and

f(x_1) ≥ g(x_1)

with x_0 distinct from x_1, are true.

 So an inequality involving a real variable x has an accompanying range of values for x for the inequality to be true.