DrDelMath

Solving equations or inequalities involving absolute values is based on the definition of absolute value.

Notice the precise definition of absolute value has two cases:
    Case 1: The expression inside the absolute value symbol is positive or zero.
    Case 2: The expression inside the absolute value symbol is negative.

Procedure: To solve an equation or an inequality involving absolute values of variable expressions, it is necessary to consider the two equations which naturally result from the definition of absolute value.

Every equation or inequality involving absolute value is solved by considering the two cases as in this general example:
Generic Example: To solve an equation or inequality involving |something|, two cases must be considered. The two cases arise from the definition of absolute value. Therefore to solve any equation or inequality involving |something|, we consider:
    Case 1: The equation that results from relacing |something| with (something)
    Case 2: The equation that results from relacing |something| with (the opposite of something).

The following two properties of equations and three properties of inequalities are also important tools when solving equations involving absolute value.

These five properties might be used to:
    Simplify the equation or inequality before considering the two cases.
    Solve the equations or inequalities in Case 1 and Case 2.

In this section of the textbook the expression inside the absolute value symbol is always a linear expresion in one variable. Each of these equations or inequalities can be solved by directly using the two cases that arise from the definition of absolute value together with the two properties for generating equivalent equations and the three properties for generating equivalent inequalities. HOWEVER, there is an easier method which will be proven in the next chapter. For the time being we will simply explain and demonstrate the easier method.

Solving Equations and Inequalities of the form |ax + b| = k, |ax + b| < k, and |ax + b| > k

The Law of Trichotomy informs us that the constant k on the right side is negative, zero, or positive.
If k is negative, we easily observe:

• |ax + b| = k has no solutions because the absolute value is never negative.
• |ax + b| < k has no solutions because the absolute value is never negative.
• The solution set for |ax + b| > k is R because the absolute value is always positive.

The only remaining case is when k is positive. Therefore in the remaining discussion we will assume that the constant k on the right is positive.

Fact:When k is positive:

• the solution set for |ax + b | < k is the set of numbers in an interval on the number line
  
• the solution set for |ax + b | = k is the set containing only the endpoints of that interval
  
• the solution set for |ax + b | > k is the set of numbers on the rest of the number line
  

The word equivalent as used in the above fact means | X | < k has the same solution set as -k < X < k.

Fact: Neither the corresponding "greater than" inequality nor the corresponding equality can be written in such a compact form.

The equation is called the boundary equation because its graph forms a boundary between the graphs of the two inequalities.

The fact that the "less than" inequality is equivalent to a compact compound inequality makes it the easiest to solve because the compact form incorporates both cases into one computational structure.

The Law of Trichotomy informs us that when we consider any one of |ax + b| = k, |ax + b| < k, or |ax + b| > k we should, in fact, consider all three of them.

The Law of Trichotomy informs us that each real number is a solution to one of |ax + b| = k, |ax + b| < k, or |ax + b| > k.
Therefore the union of the three solution sets is R.
Therefore when all three are graphed on the same number line, the entire number line is used.

The Law of Trichotomy informs us that each real number is a solution to only one of the three.
Therefore the intersection of any two of the solution sets is the empty set.
Therefore when all three are graphed on the same number line, none of the graphs overlap.

The Process

1. Whether solving |ax + b| = k, |ax + b| < k, or |ax + b| > k always solve |ax + b| < k computationally and then solve the other two by using deductive reasoning without any additional computation.
   
   
2. Convert |ax + b| < k to its compact compound inequality form - k < ax + b < k.
   
   
3. Solve the compact compound inequality by using the three methods to generate equivalent inequalities until simplest inequalities are obtained. The solution set will be an interval, call it S.
4. Use Deductive Reasoning to conclude the solution set for the equality is the set containing the endpoints of S.
   
   
5. Use Deductive Reasoning to conclude the solution set for the "greater than" inequality is everything else. So the solution set for the "greater than" inequality will be the two rays to the left of and to the right of S.