The distance interpretation for absolute value provides a good basis for solving some inequalities that involve absolute value. Consider the following examples.
Solve |x|<2, and graph the solution set.
The number, x, must be less than 2 units away from zero. Thus |x|<2 is equivalent to
x>-2 and x<2
The solution set is (-2,2), and its graph:
Solve |x+3|<1, and graph the solutions.
x+3>-1 and x+3<1
Solving each equation of the disjunction yields
The solution set is (-4,-2) and its graph:
The following general property should seem reasonable.
|ax+b|<k is equivalent to ax+b>-k and ax+b<k, where k is a positive number.