Algebra Honors ( Period 6)

Solving Compound  Inequalities 5-4

Solving  Compound  Inequalities 5-4

Inequalities containing AND….  When considered together two inequalities such as

h  ≥  52  AND  h  ≤  72 form a compound inequality.

A compound inequality contains AND is only true if BOTH inequalities are true.

Its graph is where the graphs of the two inequalities overlap. This is called the intersection of the two graphs.

For instance,

x ≥ 3 is graphed

x< 7 is graphed

x ≥ 3  and  x < 7
3 ≤ x <7 o:p="">

The statement   3 ≤ x < 7 can be read as “x is greater than or equal to 3 and less than 7”  or “x is between 3 and 7 including 3” or “3 is less than or equal to x which is less than 7.”

Solve  -2  ≤ x – 3 < 4 Then graph the solution set.

First express   -2  ≤ x – 3 < 4  using AND

-2 ≤ x -3      AND        x -3 < 4   Write the inequalities separately

Solve both of them!

1 ≤ x   and x < 7

so

the solution set is

{ x│1≤ x < 7}

Now graph the solution set

Inequalities containing OR….  A compound inequality containing or is TRUE if AT LEAST one of the inequalities is TRUE. Its graph is the UNION of the graph of the two inequalities 

The human ear can only detect sounds between the frequencies 20 Hertz and 20,000 Hertz. Write and graph a compound inequalities that describes the frequency of sounds humans cannot hear.

Let f = the frequency

f ≤ 20  OR f ≥ 20,000

Graph:

Notice that the graphs do NOT intersect. Humans cannot hear sounds at a frequency less than 20 Hertz or greater than 20,000 Hertz. The compound inequality is  { f│ f ≤ 20 or   f ≥ 20,000}

Intersections and Unions:

The graphs of compound inequalities containing AND will be an intersection. The graphs of compound inequalities containing OR will be a union.

The Triangle Inequality Theorem states that the sum of the measures of any two sides of  a triangle is greater that the measure of the third side!

Write and solve three inequalities to express the relationships among the measure of the sides of the triangle show here

           

x + 9 < 4      x < -5 (wait a minute—it can’t be negative—or even 0)

x + 4 < 9    x < 5

4 + 9 13

Since x > 5 AND x < 13 the solution set includes the whole numbers… 6, 7, 8, 9, 10, 11, 12

The compound inequality is   5< x <13 o:p="">