# 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).