Algebra Honors ( Period 4 & 7)

Inequalities Involving Absolute Value 5-5

The inequality │x│< 3 means that the distance between x and 0 is less than 3

Graph:

so x > -3 and x <  3

The set builder notation or solution set is {x │ -3 < x < 3}

When solving absolute value inequalities there are two cases to consider:

Case 1 The expression inside the absolute value symbols is nonnegative

Case 2  The expression inside the absolute value symbol is negative

The solution is the intersection of these two cases.

When the absolute value inequality is less than… I think of “less thAND” … it is the intersection of two parts. It is the “YO”  we talked about in class

│m + 2 │ < 11

Rewrite  │m + 2 │ < 11       for both the above cases

m + 2 < 11                   and            -(m+2)  < 11

m  < ; 9                   this one  becomes m + 2 >; -11

m < ; 9     and     m >  -13

or think m + 2 < 11   AND  -11 < m + 2

The solution set is { m │ -13 < m <; 9}

│y -1│ <  -2    WAIT… this can NEVER be TRUE. There is no solution The solution set is the empty set.  {  }  or

When the absolute value inequality is a greater than… I think of  “greatOR”  it is the union
of two parts..  It is the “DORKY DANCER we talked about in class… going one way and then the other.
│x│ > 3 means that the distance between x and 0 is greater than 3…

Graph:

so x <  -3 OR  x >; 3 The solution set is { x │ x < -3 or x >  3}

We must consider two cases always
Case 1 The expression inside the absolute value symbols is nonnegative
Case 2 The expression inside the absolute value symbols is negative

Solve      │3n + 6│≥ 12

Case 1
3n + 6 is non negative

3n + 6 ≥ 12              n ≥ 2        That was easy

Case 2
3n + 6 is  negative

The book sets it up  -(3n + 6) ≥ 12

That would mean    3n + 6 ≤ -12

n  ≤ -6     so

 n  ≥2   OR   n  ≤ -6

The solutions set is { n │ n  ≥2   OR   n  ≤ -6}

Graph