Inequalities Involving Absolute Values

To solve solve inequalities that involve absolute values, we use following definition of absolute values:

`|f(x)|={(f(x) if f(x)>=0),(-f(x) if f(x)<0):}`.

Besides, we can use method of squaring of both sides of inequality, that is based on the following fact:

Fact. If expressions f(x) and g(x) for all x take only non-negative values, then inequalities `f(x)>g(x)` and `(f(x))^2>(g(x))^2` are equivalent.

Example 1. Solve inequality `|x-1|<2`.

Solution 1. Quantity |x-1| can be considered as distance on coordinate line between points x and 1. Therefore, we need to find all points x, such that distance between x and 1 is less than 2. With the help of coordinate line it can be found that set of solutions of inequality is interval (-1,3).

Solution 2. Since |x-1| and 2 are always non-negative (in fact 2 is always positive) we can use above fact: `|x-1|<2` is equivalent to `(|x-1|)^2>2^2` or `x^2-2x+1>4`. From this we have that `x^2-2x-3<0`, whose set of solutions is (-1,3). Thus, set of solutions of initial inequality is also (-1,3).

Solution 3. By definition of absolute value: `|x-1|={(x-1 if x-1>=0),(-(x-1) if x-1<0):}`.

Therefore, we have following set of systems:

`{(x-1<2),(x-1>=0):}` or `{(-(x-1)<2),(x-1<0):}`.

From first system it follows that `1<=x<3`, from second system it follows that `-1<x<1`. Taking union of these intervals, we obtain final answer: `-1<x<3` or in interval notation (-1,3).

Example 2. Solve inequaltiy `|2x+4|<=3x+2`.

If `|2x+4|>=0` then `|2x+4|=2x+4`, and inequality can be rewritten as `2x+4<=3x+2`.

If `|2x+4|<0` then `|2x+4|=-(2x+4)`, and inequality can be rewritten as `-(2x+4)<=3x+2`.

Therefore, we obtain set of systems:

`{(2x+4<=3x+2),(2x+4>=0):}` or `{(-(2x+4)<=3x+2),(2x+4<0):}`.

This set can be rewritten as

`{(x>=2),(x>=-2):}` or `{(5x>=-6),(x<-2):}`.

From first system we obtain that `x>=2`, second system doesn't have solutions.

Therefore, solution of the initial inequality is `[2,+oo)`.

Example 3. Solve inequaltiy `|1-2x|>8-x`.

If `1-2x>=0` then `1-2x=1-2x`, and inequality can be rewritten as `1-2x>8-x`.

If `1-2x<0` then `|1-2x|=-(1-2x)`, and inequality can be rewritten as `-(1-2x)>8-x`.

Therefore, we obtain set of systems:

`{(1-2x>8-x),(1-2x>=0):}` or `{(-(1-2x)>8-x),(1-2x<0):}`.

This set can be rewritten as

`{(x<-7),(x<=1/2):}` or `{(x>3),(x>1/2):}`.

From first system we obtain that `x<-7`, from second we have that `x>3`.

Therefore, solution of the initial inequality is `(-oo,-7)uu(3,+oo)`.