How to solve absolute value inequality?

Question

$\frac{|x-B|}{|x|}<C$

I've tried many ways without getting far so I decided not to post my steps.

Can you please explain me how it's done?

Answer 1

We have $$\left|\frac{x-B}{x}\right|<C$$ By the definition of absolute value, this means $$-C< \frac{x-B}{x}<C$$ Now we just have to clear the denominator by multiplying through by $x$. There are two cases. When $x>0$ the inequality signs don't change. When $x<0$ the signs are reversed.

Can you take it from here?

Answer 2

There is no solution if $C \leq 0.$ (Think about it.)

We must have $x \neq 0.$ With that in mind you get $\lvert x-B\rvert < C\lvert x\rvert.$

There are then various approaches. One is to split into cases as much as you need in order to proceed, then put it back together.

To find the cases for an inequality with absolute values, one way is to go systematically through the inequality, creating two cases for the first absolute value (the absolute value is either just the expression inside, or the negative of that expression), and within each of those two cases two subcases for the second absolute value (so now you have four cases), and within each of those cases two subcases for the third absolute value, and so forth. In this problem you start with two absolute values (or just one if you recall that $\lvert p\rvert/\lvert q\rvert = \lvert p/q\rvert$), so at worst you have four cases to work out.

Another approach is, since both sides must be positive, you don't lose solutions or introduce any spurious solutions if you square both sides. This gets rid of the absolute values and now you just need to solve a quadratic inequality.

Answer 3

I mostly agree with J.W. Tanner's idea, except that it is unknown whether B > 0. I recommend a more formal approach as follows:

First of all, it is easiest to interpret |an expression| as depending on whether the expression is >=0 or < 0 (two cases) rather than making the expression = 0 a third case to consider.

Thus, since your equation has two occurrences of |an expression|, you have to deal with 2x2 = 4 cases. There are three qualifications to this approach.

1. Sometimes some of the cases may be impossible, or may dovetail into each other. For instance the case of x < -5 would dovetail into the case of x < -3, if those two cases showed up (which they don't in your problem).

2. The solution may depend on whether B >= 0. A formal approach would be for you to give one solution based on the assumption that B >=0, and another solution based on the assumption that B < 0.

3. In your specific problem, you are obviously going to have to look at the special case of x=0 separately, because of the denominator.