The document discusses inequalities and comparisons of numbers on the real number line. It explains that numbers to the right are greater than numbers to the left. Various inequality symbols like <, >, ≤ are introduced to represent relationships between numbers. Intervals on the real number line, such as -3 ≤ x < 2, are defined. Basic algebra rules for retaining inequality signs under addition, subtraction, multiplication, and division are also covered.

1. Inequalities

2. Inequalities
We recall that comparison of two numbers is based on their
locations on the real line.

3. Inequalities
We recall that comparison of two numbers is based on their
locations on the real line. Specifically, given two numbers
L and R corresponding to two points on the real line as shown,
L R
– +

4. Inequalities
We recall that comparison of two numbers is based on their
locations on the real line. Specifically, given two numbers
L and R corresponding to two points on the real line as shown,
we define the number to the right to be greater than the
number to the left.
L R
– +

5. Inequalities
We recall that comparison of two numbers is based on their
locations on the real line. Specifically, given two numbers
L and R corresponding to two points on the real line as shown,
we define the number to the right to be greater than the
number to the left. Hence R is greater than L or that L is
smaller than R .
L R
– +

6. Inequalities
We recall that comparison of two numbers is based on their
locations on the real line. Specifically, given two numbers
L and R corresponding to two points on the real line as shown,
we define the number to the right to be greater than the
number to the left. Hence R is greater than L or that L is
smaller than R .
–
L < R
+
We write this as L < R or as R > L.

7. Inequalities
We recall that comparison of two numbers is based on their
locations on the real line. Specifically, given two numbers
L and R corresponding to two points on the real line as shown,
we define the number to the right to be greater than the
number to the left. Hence R is greater than L or that L is
smaller than R .
–
L < R
+
We write this as L < R or as R > L.
For example, 2 < 4, –3 < –2, 0 > –1 are true statements

8. Inequalities
We recall that comparison of two numbers is based on their
locations on the real line. Specifically, given two numbers
L and R corresponding to two points on the real line as shown,
we define the number to the right to be greater than the
number to the left. Hence R is greater than L or that L is
smaller than R .
–
L < R
+
We write this as L < R or as R > L.
For example, 2 < 4, –3 < –2, 0 > –1 are true statements and
–2 < –5 , 5 < 3 are false statements.

9. Inequalities
We recall that comparison of two numbers is based on their
locations on the real line. Specifically, given two numbers
L and R corresponding to two points on the real line as shown,
we define the number to the right to be greater than the
number to the left. Hence R is greater than L or that L is
smaller than R .
–
L < R
+
We write this as L < R or as R > L.
For example, 2 < 4, –3 < –2, 0 > –1 are true statements and
–2 < –5 , 5 < 3 are false statements.
We use inequalities to represent segments of the real line, e.g.
“all the numbers x that are greater than 5" as “5 < x".

10. Inequalities
We recall that comparison of two numbers is based on their
locations on the real line. Specifically, given two numbers
L and R corresponding to two points on the real line as shown,
we define the number to the right to be greater than the
number to the left. Hence R is greater than L or that L is
smaller than R .
–
L < R
+
We write this as L < R or as R > L.
For example, 2 < 4, –3 < –2, 0 > –1 are true statements and
–2 < –5 , 5 < 3 are false statements.
We use inequalities to represent segments of the real line, e.g.
“all the numbers x that are greater than 5" as “5 < x".
5 x
– +

11. Inequalities
In general, we write "a < x" for all In picture,

12. Inequalities
In general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,
a a<x +
–
open dot

13. Inequalities
In general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,
a a<x +
–
open dot
If we want all the numbers x greater than or equal to a which
includes a, we write it as a < x.

14. Inequalities
In general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,
a a<x +
–
open dot
If we want all the numbers x greater than or equal to a which
includes a, we write it as a < x. In picture
a a≤x
– solid dot
+

15. Inequalities
In general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,
a a<x +
–
open dot
If we want all the numbers x greater than or equal to a which
includes a, we write it as a < x. In picture
a a≤x
– solid dot
+
The numbers x such that a < x < b where a < b are all the
numbers x between a and b.
a a<x≤b b
– +

16. Inequalities
In general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,
a a<x +
–
open dot
If we want all the numbers x greater than or equal to a which
includes a, we write it as a < x. In picture
a a≤x
– solid dot
+
The numbers x such that a < x < b where a < b are all the
numbers x between a and b. A line segment as such is called
an interval.
a a<x≤b b
– +

17. Inequalities
In general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,
a a<x +
–
open dot
If we want all the numbers x greater than or equal to a which
includes a, we write it as a < x. In picture
a a≤x
– solid dot
+
The numbers x such that a < x < b where a < b are all the
numbers x between a and b. A line segment as such is called
an interval.
a a<x≤b b
– +
–3 2
Hence –3 ≤ x < 2 is

18. Inequalities
In general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,
a a<x +
–
open dot
If we want all the numbers x greater than or equal to a which
includes a, we write it as a < x. In picture
a a≤x
– solid dot
+
The numbers x such that a < x < b where a < b are all the
numbers x between a and b. A line segment as such is called
an interval.
a a<x≤b b
– +
–3 2
Hence –3 ≤ x < 2 is
Expressions such as 2 < x > 3 or 2 < x < –3 do not have any
solution.

19. Inequalities
We note the following algebra about inequalities.

20. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.

21. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign,

22. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.

23. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.

24. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
b. Multiplying or dividing a positive number c to both sides
retains the inequality sign,

25. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
b. Multiplying or dividing a positive number c to both sides
retains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.

26. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
b. Multiplying or dividing a positive number c to both sides
retains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.
Given that 6 < 12, if we multiply –1 to both sides

27. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
b. Multiplying or dividing a positive number c to both sides
retains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.
Given that 6 < 12, if we multiply –1 to both sides then
(–1)6 > (–1)12
– 6 > –12 is true.

28. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
b. Multiplying or dividing a positive number c to both sides
retains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.
Given that 6 < 12, if we multiply –1 to both sides then
(–1)6 > (–1)12
– 6 > –12 is true.
Multiplying by –1 switches the relation of two numbers.

29. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
b. Multiplying or dividing a positive number c to both sides
retains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.
Given that 6 < 12, if we multiply –1 to both sides then
(–1)6 > (–1)12
– 6 > –12 is true.
Multiplying by –1 switches the relation of two numbers.
– 0 +
6 < 12

30. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
b. Multiplying or dividing a positive number c to both sides
retains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.
Given that 6 < 12, if we multiply –1 to both sides then
(–1)6 > (–1)12
– 6 > –12 is true.
Multiplying by –1 switches the relation of two numbers.
– 0 +
–6 6 < 12

31. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
b. Multiplying or dividing a positive number c to both sides
retains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.
Given that 6 < 12, if we multiply –1 to both sides then
(–1)6 > (–1)12
– 6 > –12 is true.
Multiplying by –1 switches the relation of two numbers.
– 0 +
–12 < –6 6 < 12

32. Inequalities
We note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
a. Adding or subtracting the same quantity to both sides
retains the inequality sign, i.e. if a < b, then a ± c < b ± c.
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
b. Multiplying or dividing a positive number c to both sides
retains the inequality sign, i.e. if 0 < c and a < b, then ac < bc.
Given that 6 < 12, if we multiply –1 to both sides then
(–1)6 > (–1)12
– 6 > –12 which is true.
Multiplying by –1 switches the relation of two numbers.
– 0 +
–12 < –6 6 < 12
c. Multiplying or dividing by an negative number c reverses
the inequality sign, i.e. if c < 0 and a < b then ca > cb

33. Inequalities
Following are the steps for solving linear inequalities.

34. Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.

35. Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.
2. Gather the x-terms to one side and the number-terms to the
other sides (use the “change side-change sign” rule).

36. Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.
2. Gather the x-terms to one side and the number-terms to the
other sides (use the “change side-change sign” rule).
3. Multiply or divide to get x. If we multiply or divide by
negative numbers to both sides, the inequality sign must be
turned around.

37. Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.
2. Gather the x-terms to one side and the number-terms to the
other sides (use the “change side-change sign” rule).
3. Multiply or divide to get x. If we multiply or divide by
negative numbers to both sides, the inequality sign must be
turned around. (This rule can be avoided by keeping the
x-term positive.)

38. Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.
2. Gather the x-terms to one side and the number-terms to the
other sides (use the “change side-change sign” rule).
3. Multiply or divide to get x. If we multiply or divide by
negative numbers to both sides, the inequality sign must be
turned around. (This rule can be avoided by keeping the
x-term positive.)
Example A. a. Solve –3x + 2 < 11. Draw the solution.

39. Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.
2. Gather the x-terms to one side and the number-terms to the
other sides (use the “change side-change sign” rule).
3. Multiply or divide to get x. If we multiply or divide by
negative numbers to both sides, the inequality sign must be
turned around. (This rule can be avoided by keeping the
x-term positive.)
Example A. a. Solve –3x + 2 < 11. Draw the solution.
–3x + 2 < 11 move the x to the other side to make it positive

40. Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.
2. Gather the x-terms to one side and the number-terms to the
other sides (use the “change side-change sign” rule).
3. Multiply or divide to get x. If we multiply or divide by
negative numbers to both sides, the inequality sign must be
turned around. (This rule can be avoided by keeping the
x-term positive.)
Example A. a. Solve –3x + 2 < 11. Draw the solution.
–3x + 2 < 11 move the x to the other side to make it positive
2 – 11 < 3x

41. Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.
2. Gather the x-terms to one side and the number-terms to the
other sides (use the “change side-change sign” rule).
3. Multiply or divide to get x. If we multiply or divide by
negative numbers to both sides, the inequality sign must be
turned around. (This rule can be avoided by keeping the
x-term positive.)
Example A. a. Solve –3x + 2 < 11. Draw the solution.
–3x + 2 < 11 move the x to the other side to make it positive
2 – 11 < 3x
–9 < 3x

42. Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.
2. Gather the x-terms to one side and the number-terms to the
other sides (use the “change side-change sign” rule).
3. Multiply or divide to get x. If we multiply or divide by
negative numbers to both sides, the inequality sign must be
turned around. (This rule can be avoided by keeping the
x-term positive.)
Example A. a. Solve –3x + 2 < 11. Draw the solution.
–3x + 2 < 11 move the x to the other side to make it positive
2 – 11 < 3x
–9 < 3x divide by 3, no change with the inequality
–9 3x
3 < 3
–3 < x

43. Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.
2. Gather the x-terms to one side and the number-terms to the
other sides (use the “change side-change sign” rule).
3. Multiply or divide to get x. If we multiply or divide by
negative numbers to both sides, the inequality sign must be
turned around. (This rule can be avoided by keeping the
x-term positive.)
Example A. a. Solve –3x + 2 < 11. Draw the solution.
–3x + 2 < 11 move the x to the other side to make it positive
2 – 11 < 3x
–9 < 3x divide by 3, no change with the inequality
–9 3x
3 < 3
–3 < x
-3
In picture, 0

44. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.

45. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,

46. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9

47. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,

48. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
–6 –3x 9
–3 > –3 ≥ –3

49. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
–6 –3x 9
–3 > –3 ≥ –3
–3 –2
–2 > x ≥ –3 or

50. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
–6 –3x 9
–3 > –3 ≥ –3
–3 –2
–2 > x ≥ –3 or
Signs and Inequalities

51. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
–6 –3x 9
–3 > –3 ≥ –3
–3 –2
–2 > x ≥ –3 or
Signs and Inequalities
One of the most important association to inequalities is
the sign of a given expression.

52. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
–6 –3x 9
–3 > –3 ≥ –3
–3 –2
–2 > x ≥ –3 or
Signs and Inequalities
One of the most important association to inequalities is
the sign of a given expression. Let f be an expression,
following adjectives translate into the following inequalities.

53. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
–6 –3x 9
–3 > –3 ≥ –3
–3 –2
–2 > x ≥ –3 or
Signs and Inequalities
One of the most important association to inequalities is
the sign of a given expression. Let f be an expression,
following adjectives translate into the following inequalities.
i. “f is positive” ↔ 0 < f or f > 0

54. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
–6 –3x 9
–3 > –3 ≥ –3
–3 –2
–2 > x ≥ –3 or
Signs and Inequalities
One of the most important association to inequalities is
the sign of a given expression. Let f be an expression,
following adjectives translate into the following inequalities.
i. “f is positive” ↔ 0 < f or f > 0
“f is non–negative” ↔ 0 ≤ f or f ≥ 0

55. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
–6 –3x 9
–3 > –3 ≥ –3
–3 –2
–2 > x ≥ –3 or
Signs and Inequalities
One of the most important association to inequalities is
the sign of a given expression. Let f be an expression,
following adjectives translate into the following inequalities.
i. “f is positive” ↔ 0 < f or f > 0
“f is non–negative” ↔ 0 ≤ f or f ≥ 0
ii. “f is negative” ↔ f < 0 or 0 > f

56. Inequalities
b. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
–6 –3x 9
–3 > –3 ≥ –3
–3 –2
–2 > x ≥ –3 or
Signs and Inequalities
One of the most important association to inequalities is
the sign of a given expression. Let f be an expression,
following adjectives translate into the following inequalities.
i. “f is positive” ↔ 0 < f or f > 0
“f is non–negative” ↔ 0 ≤ f or f ≥ 0
ii. “f is negative” ↔ f < 0 or 0 > f
“f is non–positive” ↔ f ≤ 0 or 0 ≥ f

57. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.

58. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
Being positive translates into the inequality
3x + 12 > 0

59. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
Being positive translates into the inequality
3x + 12 > 0
3x > –12
x > –4

60. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
Being positive translates into the inequality
3x + 12 > 0
3x > –12
–4 x
x > –4 or 0

61. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
Being positive translates into the inequality
3x + 12 > 0
3x > –12
–4 x
x > –4 or 0
b. Let x be the price of a movie ticket. Translate the sentence
“$20 minus the cost of two movie tickets is non–positive”
into an inequality. Solve it and draw the solution.

62. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
Being positive translates into the inequality
3x + 12 > 0
3x > –12
–4 x
x > –4 or 0
b. Let x be the price of a movie ticket. Translate the sentence
“$20 minus the cost of two movie tickets is non–positive”
into an inequality. Solve it and draw the solution.
Two tickets cost $2x,

63. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
Being positive translates into the inequality
3x + 12 > 0
3x > –12
–4 x
x > –4 or 0
b. Let x be the price of a movie ticket. Translate the sentence
“$20 minus the cost of two movie tickets is non–positive”
into an inequality. Solve it and draw the solution.
Two tickets cost $2x, so “$20 minus the cost of two tickets”
is (20 – 2x).

64. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
Being positive translates into the inequality
3x + 12 > 0
3x > –12
–4 x
x > –4 or 0
b. Let x be the price of a movie ticket. Translate the sentence
“$20 minus the cost of two movie tickets is non–positive”
into an inequality. Solve it and draw the solution.
Two tickets cost $2x, so “$20 minus the cost of two tickets”
is (20 – 2x). The inequality for “(20 – 2x) is non–positive” is
20 – 2x ≤ 0

65. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
Being positive translates into the inequality
3x + 12 > 0
3x > –12
–4 x
x > –4 or 0
b. Let x be the price of a movie ticket. Translate the sentence
“$20 minus the cost of two movie tickets is non–positive”
into an inequality. Solve it and draw the solution.
Two tickets cost $2x, so “$20 minus the cost of two tickets”
is (20 – 2x). The inequality for “(20 – 2x) is non–positive” is
20 – 2x ≤ 0
20 ≤ 2x
10 ≤ x

66. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
Being positive translates into the inequality
3x + 12 > 0
3x > –12
–4 x
x > –4 or 0
b. Let x be the price of a movie ticket. Translate the sentence
“$20 minus the cost of two movie tickets is non–positive”
into an inequality. Solve it and draw the solution.
Two tickets cost $2x, so “$20 minus the cost of two tickets”
is (20 – 2x). The inequality for “(20 – 2x) is non–positive” is
20 – 2x ≤ 0
20 ≤ 2x
10 x
10 ≤ x or 0

67. Inequalities
Example B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
Being positive translates into the inequality
3x + 12 > 0
3x > –12
–4 x
x > –4 or 0
b. Let x be the price of a movie ticket. Translate the sentence
“$20 minus the cost of two movie tickets is non–positive”
into an inequality. Solve it and draw the solution.
Two tickets cost $2x, so “$20 minus the cost of two tickets”
is (20 – 2x). The inequality for “(20 – 2x) is non–positive” is
20 – 2x ≤ 0
20 ≤ 2x
10 x
10 ≤ x or 0
So the price of a ticket has to be $10 or more.

68. Inequalities
Intersection and Union (∩ & U)

69. Inequalities
Intersection and Union (∩ & U)
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
1 3
I:

70. Inequalities
Intersection and Union (∩ & U)
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
1 3
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.

71. Inequalities
Intersection and Union (∩ & U)
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
1 3
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Let J = {2 < x < 4 } be another interval as shown.
2 4
J:

72. Inequalities
Intersection and Union (∩ & U)
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
1 3
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Let J = {2 < x < 4 } be another interval as shown.
2 4
J:
The common portion of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:

73. Inequalities
Intersection and Union (∩ & U)
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
1 3
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Let J = {2 < x < 4 } be another interval as shown.
2 4
J:
The common portion of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
2 3

74. Inequalities
Intersection and Union (∩ & U)
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
1 3
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Let J = {2 < x < 4 } be another interval as shown.
2 4
J:
The common portion of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
2 3
I ∩ J:
is called the intersection of I and J and it’s denoted as I ∩ J.

75. Inequalities
Intersection and Union (∩ & U)
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
1 3
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Let J = {2 < x < 4 } be another interval as shown.
2 4
J:
The common portion of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
2 3
I ∩ J:
is called the intersection of I and J and it’s denoted as I ∩ J.
In this case I ∩ J = {2 < x ≤ 3 }.

76. Inequalities
The merge of the two intervals I and J shown here

77. Inequalities
The merge of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
1 2 3 4

78. Inequalities
The merge of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
1 2 3 4

79. Inequalities
The merge of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
1 2 3 4
I U J:
is called the union of I and J and it’s denoted as I U J.

80. Inequalities
The merge of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
1 2 3 4
I U J:
is called the union of I and J and it’s denoted as I U J.
In this case I U J: = {1 ≤ x < 4 }.

81. Inequalities
The merge of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
1 2 3 4
I U J:
is called the union of I and J and it’s denoted as I U J.
In this case I U J: = {1 ≤ x < 4 }.
The Interval Notation

82. Inequalities
The merge of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
1 2 3 4
I U J:
is called the union of I and J and it’s denoted as I U J.
In this case I U J: = {1 ≤ x < 4 }.
The Interval Notation
We use the infinity symbol “∞” to mean “continues onward and
surpasses all numbers”.

83. Inequalities
The merge of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
1 2 3 4
I U J:
is called the union of I and J and it’s denoted as I U J.
In this case I U J: = {1 ≤ x < 4 }.
The Interval Notation
We use the infinity symbol “∞” to mean “continues onward and
surpasses all numbers”. We label the left of the real line with –
∞ and to the right with ∞ as shown below when needed.
–∞ 0 ∞

84. Inequalities
The merge of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
1 2 3 4
I U J:
is called the union of I and J and it’s denoted as I U J.
In this case I U J: = {1 ≤ x < 4 }.
The Interval Notation
We use the infinity symbol “∞” to mean “continues onward and
surpasses all numbers”. We label the left of the real line with –
∞ and to the right with ∞ as shown below when needed.
–∞ 0 ∞
Infinity “∞” is not a number but we say that –∞ < x < ∞
for all real number x.

85. Inequalities
The merge of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
1 2 3 4
I U J:
is called the union of I and J and it’s denoted as I U J.
In this case I U J: = {1 ≤ x < 4 }.
The Interval Notation
We use the infinity symbol “∞” to mean “continues onward and
surpasses all numbers”. We label the left of the real line with –
∞ and to the right with ∞ as shown below when needed.
–∞ 0 ∞
Infinity “∞” is not a number but we say that –∞ < x < ∞
for all real number x. In fact, the set of all real numbers R is
R = {–∞ < x < ∞ }.

86. Inequalities
The merge of the two intervals I and J shown here
1 2 3
I:
2 3 4
J:
1 2 3 4
I U J:
is called the union of I and J and it’s denoted as I U J.
In this case I U J: = {1 ≤ x < 4 }.
The Interval Notation
We use the infinity symbol “∞” to mean “continues onward and
surpasses all numbers”. We label the left of the real line with –
∞ and to the right with ∞ as shown below when needed.
–∞ 0 ∞
Infinity “∞” is not a number but we say that –∞ < x < ∞
for all real number x. In fact, the set of all real numbers R is
R = {–∞ < x < ∞ }. However, we do not write x ≤ ∞ because ∞ is
not a number in particular x can’t be ∞.

87. Inequalities
Using the “∞” symbols, we write the line segment
a
∞ or a ≤ x

88. Inequalities
Using the “∞” symbols, we write the line segment
a
∞ or a ≤ x as [a, ∞),

89. Inequalities
Using the “∞” symbols, we write the line segment
a
∞ or a ≤ x as [a, ∞),
We use ] or [ to include the end points,

90. Inequalities
Using the “∞” symbols, we write the line segment
a
∞ or a ≤ x as [a, ∞),
a
∞ or a < x as (a, ∞),
We use ] or [ to include the end points, ) and ( to exclude the
end points.

91. Inequalities
Using the “∞” symbols, we write the line segment
a
∞ or a ≤ x as [a, ∞),
a
∞ or a < x as (a, ∞),
–∞ a
or x ≤ a as (–∞, a],
–∞ a
or x < a as (–∞, a),
We use ] or [ to include the end points, ) and ( to exclude the
end points.

92. Inequalities
Using the “∞” symbols, we write the line segment
a
∞ or a ≤ x as [a, ∞),
a
∞ or a < x as (a, ∞),
–∞ a
or x ≤ a as (–∞, a],
–∞ a
or x < a as (–∞, a),
Let a, b be two numbers such that a < b, we write the intervals
a b
or a ≤ x ≤ b as [a, b],
a b
or a < x < b as (a, b),
a b
or a ≤ x < b as [a, b),
a b
or a < x ≤ b as (a, b],
We use ] or [ to include the end points, ) and ( to exclude the
end points.

93. Inequalities
Using the “∞” symbols, we write the line segment
a
∞ or a ≤ x as [a, ∞),
a
∞ or a < x as (a, ∞),
–∞ a
or x ≤ a as (–∞, a],
–∞ a
or x < a as (–∞, a),
Let a, b be two numbers such that a < b, we write the intervals
a b
or a ≤ x ≤ b as [a, b],
a b
or a < x < b as (a, b),
a b
or a ≤ x < b as [a, b),
a b
or a < x ≤ b as (a, b],
We use ] or [ to include the end points, ) and ( to exclude the
end points. Note the interval notation (a, b) is the same as the
coordinate of a point so its interpretation depends on the context.

94. Inequalities
The interval [a, a] consists of exactly one point {x = a}.

95. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.

96. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.

97. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.

98. Inequalities
The interval [a, a] consists of exactly one point {x = a}.

99. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.

100. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.

101. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.
Example C. Given intervals I, J , and K, do the set operation.
Draw. Put the answer in the interval and the inequality notation.
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1

102. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.
Example C. Given intervals I, J , and K, do the set operation.
Draw. Put the answer in the interval and the inequality notation.
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
a. K U J
b. K ∩ I

103. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.
Example C. Given intervals I, J , and K, do the set operation.
Draw. Put the answer in the interval and the inequality notation.
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
a. K U J
–3 1
0
K
b. K ∩ I

104. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.
Example C. Given intervals I, J , and K, do the set operation.
Draw. Put the answer in the interval and the inequality notation.
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
a. K U J
–3 –2 1
0
K
b. K ∩ I

105. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.
Example C. Given intervals I, J , and K, do the set operation.
Draw. Put the answer in the interval and the inequality notation.
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
a. K U J
–3 –2 1 –3
0 so K U J is 0
K
b. K ∩ I

106. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.
Example C. Given intervals I, J , and K, do the set operation.
Draw. Put the answer in the interval and the inequality notation.
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
a. K U J
–3 –2 1 –3
0 so K U J is 0
K
Hence K U J is (–3, ∞) or {–3 < x}.
b. K ∩ I

107. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.
Example C. Given intervals I, J , and K, do the set operation.
Draw. Put the answer in the interval and the inequality notation.
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
a. K U J
–3 –2 1 –3
0 so K U J is 0
Hence K U J is (–3, ∞) or {–3 < x}.
b. K ∩ I
–3 K
1
0

108. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.
Example C. Given intervals I, J , and K, do the set operation.
Draw. Put the answer in the interval and the inequality notation.
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
a. K U J
–3 –2 1 –3
0 so K U J is 0
Hence K U J is (–3, ∞) or {–3 < x}.
b. K ∩ I
–3 1
–4 –1 0
K
I

109. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.
Example C. Given intervals I, J , and K, do the set operation.
Draw. Put the answer in the interval and the inequality notation.
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
a. K U J
–3 –2 1 –3
0 so K U J is 0
Hence K U J is (–3, ∞) or {–3 < x}.
b. K ∩ I
–3 1
–4 –1 0
K
I
The common or overlapping portion is shown.

110. Inequalities
The interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
The interval (a, a) or (a, a] or [a, a) = Φ.
Example C. Given intervals I, J , and K, do the set operation.
Draw. Put the answer in the interval and the inequality notation.
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
a. K U J
–3 –2 1 –3
0 so K U J is 0
Hence K U J is (–3, ∞) or {–3 < x}.
b. K ∩ I
–3 1
–4 –1 0
K
I
The common or overlapping portion is shown.
Hence K ∩ I is (–3, –1) or {–3 < x < –1}

111. Inequalities
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
c. (K ∩ J) U I

112. Inequalities
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
c. (K ∩ J) U I
We do the parenthesis first.

113. Inequalities
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
c. (K ∩ J) U I
We do the parenthesis first.
–3 –2 1
K∩J: 0

114. Inequalities
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
c. (K ∩ J) U I
We do the parenthesis first.
–3 –2 1
K∩J: 0
–2 1
so K ∩ J is 0

115. Inequalities
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
c. (K ∩ J) U I
We do the parenthesis first.
–3 –2 1
K∩J: 0
–2 1
so K ∩ J is 0
Therefore (K ∩ J) U I is
–4 –2
–1 1
0

116. Inequalities
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
c. (K ∩ J) U I
We do the parenthesis first.
–3 –2 1
K∩J: 0
–2 1
so K ∩ J is 0
Therefore (K ∩ J) U I is
–4 –2
–1 1
0
so (K ∩ J) U I = {–4 < x ≤ 1} or (–4, 1]

117. Inequalities
–4 –1
I: 0 J: x > –2 K: –3 < x ≤ 1
c. (K ∩ J) U I
We do the parenthesis first.
–3 –2 1
K∩J: 0
–2 1
so K ∩ J is 0
Therefore (K ∩ J) U I is
–4 –2
–1 1
0
so (K ∩ J) U I = {–4 < x ≤ 1} or (–4, 1]
Your turn. Find K ∩ (J U I ). Is this the same as (K ∩ J) U I?