Download Lesson 23, Section 4

Lesson 23, Section 4.1 (part 1)
Inequalities
Ex 1: Which of the following numbers would be solutions of the inequality shown?
4 x  5  2( x  1)
0, -5, -3, 12
4(0)  5  2(0  1)
0  5  2(1)
true
5  2
4(5)  5  2(5  1)
 20  5  2(6)
false
 15  12
4(3)  5  2(3  1)
 12  5  2(4)
true
 7  8
4(12)  5  2(12  1)
48  5  2(11)
true
53  22
Solutions include 0, -3, and 12.
There are three ways to represent an inequality; using the inequality symbol (set-builder
notation), a number-line graph, and using interval notation. The examples below
represent equivalent forms of all three ways.
inequality symbol
number-line graph
interval notation
{x | x  5}
(,5)
5
{x | x  2}
[2, )
2
{x | 0  x  3}
(0, 3)
0
3
{x | 4  x  10}
[4, 10]
4
10
{x | 3  x  1}
(3,1]
-3
1
There is a summary of these ways on the course webpage (other information,
inequalities).
*Do not confuse interval notation with an ordered pair (point). The context in which
each is used will make the meanings clear.
{ y | y  2}
a)
Write in interval notation and graph on the number line.
b)
Write using set-builder notation and using interval notation.
-5
c)
Write using set-builder notation and graph on the number line.
[0,5)
Begin with the following inequality: 20 > 12
Do the following operations to both
sides of the inequality and determine if the result is true or false.
add 4
subtract 3
multiply by 4
divide by 2
multiply by -2
divide by -4
20 + 4 > 12 + 4
20 - 3 > 12 - 3
4(20) > 4(12)
20 12

2
2
-2(20) > -2(12)
20 12

4 4
true
true
true
true
?
?
Solving Inequalities: When solving an inequality you may add, subtract, multiply
by a positive number, or divide by a positive number on both sides and the result is
true. However, if you multiply or divide by a negative number, the inequality sign
must be reversed!
Solve these inequalities. Write the answer using both set-builder notation and interval
notation, then graph the solution.
Ex 2:
x  12  5
4
10
x
5
11
Ex 3:

Ex 4:
3a  1
 7
2
Ex 5: 6(2 y  8)  3( y  10)
Ex 6: 7(b  2)  6b  3(3  6b)
Ex 7: 13  (2c  2)  2(c  2)  3c
Ex 8:
1
1
 (6 x  24)  20  (12 x  72)
3
4
Ex 9:
2
3
(5 x  1)  (4 x  2)  2
3
4