Linear Inequalities and Interval Notation Section 1.5 Equations and Inequalities.
41 interval notation and review on inequalities
Transcript of 41 interval notation and review on inequalities
Inequalities
We recall that comparison of two numbers is based on their
locations on the real line.
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,
Inequalities
+–RL
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.
Inequalities
+–RL
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 .
Inequalities
+–RL
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 .
Inequalities
+–R
We write this as L < R or as R > L.
L <
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 .
Inequalities
+–R
We write this as L < R or as R > L.
For example, 2 < 4, –3 < –2, 0 > –1 are true statements
L <
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 .
Inequalities
+–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.
L <
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 .
Inequalities
+–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.
L <
We use inequalities to represent segments of the real line, e.g.
“all the numbers x that are greater than 5" as “5 < x".
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 .
Inequalities
+–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.
L <
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
InequalitiesIn general, we write "a < x" for all In picture,
InequalitiesIn general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,+
–a
open dot
a < x
InequalitiesIn general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,+
–a
open dot
If we want all the numbers x greater than or equal to a which
includes a, we write it as a < x.
a < x
InequalitiesIn general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,+
–a
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
solid dot
a < x
a ≤ x
InequalitiesIn general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,+
–a
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
solid dot
a < x
a ≤ x
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
InequalitiesIn general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,+
–a
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
solid dot
a < x
a ≤ x
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
InequalitiesIn general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,+
–a
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
solid dot
a < x
a ≤ x
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
Hence –3 ≤ x < 2 is2–3
InequalitiesIn general, we write "a < x" for all the numbers x greater than a
but excluding a. In picture,+
–a
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
solid dot
a < x
a ≤ x
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
Expressions such as 2 < x > 3 or 2 < x < –3 do not have any
solution.
Hence –3 ≤ x < 2 is2–3
InequalitiesWe note the following algebra about inequalities.
Given that 6 < 12, then 6 + 3 < 12 + 3 is still true.
InequalitiesWe 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,
InequalitiesWe 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.
InequalitiesWe 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.
InequalitiesWe 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.
b. Multiplying or dividing a positive number c to both sides
retains the inequality sign,
Given that 6 < 12, then multiplying by 3*6 < 3*12 is still true.
InequalitiesWe 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.
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, then multiplying by 3*6 < 3*12 is still true.
InequalitiesWe 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.
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, then multiplying by 3*6 < 3*12 is still true.
InequalitiesWe note the following algebra about inequalities.
Given that 6 < 12, if we multiply –1 to both sides
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.
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, then multiplying by 3*6 < 3*12 is still true.
InequalitiesWe note the following algebra about inequalities.
Given that 6 < 12, if we multiply –1 to both sides then
(–1)6 > (–1)12
– 6 > –12 is true.
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.
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, then multiplying by 3*6 < 3*12 is still true.
InequalitiesWe note the following algebra about inequalities.
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.
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.
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, then multiplying by 3*6 < 3*12 is still true.
InequalitiesWe note the following algebra about inequalities.
60 +–
12<
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.
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.
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, then multiplying by 3*6 < 3*12 is still true.
InequalitiesWe note the following algebra about inequalities.
60 +–
12–6 <
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.
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.
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, then multiplying by 3*6 < 3*12 is still true.
InequalitiesWe note the following algebra about inequalities.
60 +–
12–6–12 <<
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.
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.
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, then multiplying by 3*6 < 3*12 is still true.
InequalitiesWe note the following algebra about inequalities.
c. Multiplying or dividing by an negative number c reverses
the inequality sign, i.e. if c < 0 and a < b then ca > cb
60 +–
12–6–12 <<
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.
Following are the steps for solving linear inequalities.
Inequalities
Following are the steps for solving linear inequalities.
1. Simplify both sides of the inequalities.
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).
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.
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.)
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.)
Inequalities
Example A. a. Solve –3x + 2 < 11. Draw the solution.
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.)
Inequalities
Example A. a. Solve –3x + 2 < 11. Draw the solution.
–3x + 2 < 11 move the x to the other side to make it positive
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.)
Inequalities
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
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.)
Inequalities
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
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.)
Inequalities
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
<3x–9
3 3 –3 < x
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.)
Inequalities
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
0-3
<3x–9
3 3 –3 < x
In picture,
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9
Inequalitiesb. 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,
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
–6 –3
–3x –3>
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
≥9 –3
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
–6 –3
–3x –3>
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
≥9 –3
–2 > x ≥ –3 or–2–3
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
–6 –3
–3x –3>
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
≥9 –3
–2 > x ≥ –3 or–2–3
Signs and Inequalities
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
–6 –3
–3x –3>
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
≥9 –3
–2 > x ≥ –3 or–2–3
Signs and Inequalities
One of the most important association to inequalities is
the sign of a given expression.
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
–6 –3
–3x –3>
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
≥9 –3
–2 > x ≥ –3 or–2–3
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.
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
–6 –3
–3x –3>
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
≥9 –3
–2 > x ≥ –3 or–2–3
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
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
–6 –3
–3x –3>
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
≥9 –3
–2 > x ≥ –3 or–2–3
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
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
–6 –3
–3x –3>
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
≥9 –3
–2 > x ≥ –3 or–2–3
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
ii. “f is negative” ↔ f < 0 or 0 > f
“f is non–negative” ↔ 0 ≤ f or f ≥ 0
Inequalitiesb. Solve the interval inequality 5 < –3x + 1 ≤ 10.
Draw the solution.
–6 –3
–3x –3>
5 < –3x + 1 ≤ 10 subtract 1 from all entries,
–6 < –3x ≤ 9 divide by –3, reverse the inequality signs,
≥9 –3
–2 > x ≥ –3 or–2–3
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
ii. “f is negative” ↔ f < 0 or 0 > f
“f is non–negative” ↔ 0 ≤ f or f ≥ 0
“f is non–positive” ↔ f ≤ 0 or 0 ≥ f
InequalitiesExample B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
InequalitiesExample 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
InequalitiesExample B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
3x > –12
x > –4
Being positive translates into the inequality
3x + 12 > 0
InequalitiesExample B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
3x > –12
x > –4
Being positive translates into the inequality
3x + 12 > 0
or0
x–4
InequalitiesExample B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
3x > –12
x > –4
Being positive translates into the inequality
3x + 12 > 0
or0
x–4
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.
InequalitiesExample B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
3x > –12
x > –4
Being positive translates into the inequality
3x + 12 > 0
or0
x–4
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,
InequalitiesExample B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
3x > –12
x > –4
Being positive translates into the inequality
3x + 12 > 0
or0
x–4
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).
InequalitiesExample B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
3x > –12
x > –4
Being positive translates into the inequality
3x + 12 > 0
or0
x–4
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
InequalitiesExample B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
3x > –12
x > –4
Being positive translates into the inequality
3x + 12 > 0
or0
x–4
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.
20 ≤ 2x
10 ≤ x
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
InequalitiesExample B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
3x > –12
x > –4
Being positive translates into the inequality
3x + 12 > 0
or0
x–4
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.
20 ≤ 2x
10 ≤ x
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
or 0x10
InequalitiesExample B. Translate the sentence “3x + 12 is positive” into
inequalities and solve for x. Draw the solution.
3x > –12
x > –4
Being positive translates into the inequality
3x + 12 > 0
or0
x–4
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.
20 ≤ 2x
10 ≤ x
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
or 0x10
So the price of a ticket has to be $10 or more.
InequalitiesIntersection and Union (∩ & U)
Inequalities
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
31
Intersection and Union (∩ & U)
I:
Inequalities
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
31
Intersection and Union (∩ & U)
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Inequalities
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
31
Intersection and Union (∩ & U)
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Let J = {2 < x < 4 } be another interval as shown.
42J:
Inequalities
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
31
Intersection and Union (∩ & U)
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Let J = {2 < x < 4 } be another interval as shown.
42J:
The common portion of the two intervals I and J shown here
31I:
42J:
2
3
Inequalities
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
31
Intersection and Union (∩ & U)
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Let J = {2 < x < 4 } be another interval as shown.
42J:
The common portion of the two intervals I and J shown here
31I:
42J:
2
3
32
Inequalities
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
31
Intersection and Union (∩ & U)
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Let J = {2 < x < 4 } be another interval as shown.
42J:
The common portion of the two intervals I and J shown here
31I:
42J:
2
3
3I ∩ J:
is called the intersection of I and J and it’s denoted as I ∩ J.
2
Inequalities
Let I be the interval consists of 1 ≤ x ≤ 3 as shown,
31
Intersection and Union (∩ & U)
I:
and in the set notation we write I = {1 ≤ x ≤ 3 }.
Let J = {2 < x < 4 } be another interval as shown.
42J:
The common portion of the two intervals I and J shown here
In this case I ∩ J = {2 < x ≤ 3 }.
31I:
42J:
2
3
3I ∩ J:
is called the intersection of I and J and it’s denoted as I ∩ J.
2
InequalitiesThe merge of the two intervals I and J shown here
InequalitiesThe merge of the two intervals I and J shown here
31I:
42J:
2
3
2 41 3
InequalitiesThe merge of the two intervals I and J shown here
31I:
42J:
2
3
2 41 3
InequalitiesThe merge of the two intervals I and J shown here
31I:
42J:
2
3
I U J:
is called the union of I and J and it’s denoted as I U J.
2 41 3
InequalitiesThe merge of the two intervals I and J shown here
In this case I U J: = {1 ≤ x < 4 }.
31I:
42J:
2
3
I U J:
is called the union of I and J and it’s denoted as I U J.
2 41 3
InequalitiesThe merge of the two intervals I and J shown here
31I:
42J:
2
3
I U J:
is called the union of I and J and it’s denoted as I U J.
2 41
The Interval Notation
3
In this case I U J: = {1 ≤ x < 4 }.
InequalitiesThe merge of the two intervals I and J shown here
31I:
42J:
2
3
I U J:
is called the union of I and J and it’s denoted as I U J.
2 41
The Interval Notation
We use the infinity symbol “∞” to mean “continues onward and
surpasses all numbers”.
3
In this case I U J: = {1 ≤ x < 4 }.
InequalitiesThe merge of the two intervals I and J shown here
31I:
42J:
2
3
I U J:
is called the union of I and J and it’s denoted as I U J.
2 41
The Interval Notation
–∞ ∞0
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.
3
In this case I U J: = {1 ≤ x < 4 }.
InequalitiesThe merge of the two intervals I and J shown here
31I:
42J:
2
3
I U J:
is called the union of I and J and it’s denoted as I U J.
2 41
The Interval Notation
–∞ ∞0
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.
Infinity “∞” is not a number but we say that –∞ < x < ∞
for all real number x.
3
In this case I U J: = {1 ≤ x < 4 }.
InequalitiesThe merge of the two intervals I and J shown here
31I:
42J:
2
3
I U J:
is called the union of I and J and it’s denoted as I U J.
2 41
The Interval Notation
–∞ ∞0
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.
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 < ∞ }.
3
In this case I U J: = {1 ≤ x < 4 }.
InequalitiesThe merge of the two intervals I and J shown here
31I:
42J:
2
3
I U J:
is called the union of I and J and it’s denoted as I U J.
2 41
The Interval Notation
–∞ ∞0
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.
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 ∞.
3
In this case I U J: = {1 ≤ x < 4 }.
InequalitiesUsing the “∞” symbols, we write the line segment
∞aor a ≤ x
InequalitiesUsing the “∞” symbols, we write the line segment
∞aor a ≤ x as [a, ∞),
InequalitiesUsing the “∞” symbols, we write the line segment
∞aor a ≤ x as [a, ∞),
We use ] or [ to include the end points,
InequalitiesUsing the “∞” symbols, we write the line segment
∞aor a ≤ x as [a, ∞),
∞aor a < x as (a, ∞),
We use ] or [ to include the end points, ) and ( to exclude the
end points.
InequalitiesUsing the “∞” symbols, we write the line segment
∞aor a ≤ x as [a, ∞),
–∞ a
or x ≤ a as (–∞, a],
∞aor a < x as (a, ∞),
–∞ aor x < a as (–∞, a),
We use ] or [ to include the end points, ) and ( to exclude the
end points.
InequalitiesUsing the “∞” symbols, we write the line segment
∞aor a ≤ x as [a, ∞),
–∞ a
or x ≤ a as (–∞, a],
∞aor a < x as (a, ∞),
–∞ aor x < a as (–∞, a),
We use ] or [ to include the end points, ) and ( to exclude the
end points.
Let a, b be two numbers such that a < b, we write the intervals
or a ≤ x ≤ b as [a, b],ba
or a < x < b as (a, b),ba
or a ≤ x < b as [a, b),ba
or a < x ≤ b as (a, b],ba
InequalitiesUsing the “∞” symbols, we write the line segment
∞aor a ≤ x as [a, ∞),
–∞ a
or x ≤ a as (–∞, a],
∞aor a < x as (a, ∞),
–∞ aor x < a as (–∞, a),
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.
Let a, b be two numbers such that a < b, we write the intervals
or a ≤ x ≤ b as [a, b],ba
or a < x < b as (a, b),ba
or a ≤ x < b as [a, b),ba
or a < x ≤ b as (a, b],ba
InequalitiesThe interval [a, a] consists of exactly one point {x = a}.
InequalitiesThe interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
InequalitiesThe 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.
InequalitiesThe 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.
InequalitiesThe interval [a, a] consists of exactly one point {x = a}.
InequalitiesThe interval [a, a] consists of exactly one point {x = a}.
The empty set which contains nothing is denoted as Φ = { }.
InequalitiesThe 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.
InequalitiesThe 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.
I: x > –20–1–4 J: –3 < x ≤ 1K:
InequalitiesThe 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.
I: x > –20–1–4 J: –3 < x ≤ 1K:
a. K U J
b. K ∩ I
InequalitiesThe 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.
I: x > –20–1–4 J: –3 < x ≤ 1K:
a. K U J–3 1
0
b. K ∩ I
K
InequalitiesThe 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.
I: x > –20–1–4 J: –3 < x ≤ 1K:
a. K U J–2–3 1
0
b. K ∩ I
K
InequalitiesThe 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.
I: x > –20–1–4 J: –3 < x ≤ 1K:
a. K U J
so K U J is –2–3 1
0
–30
b. K ∩ I
K
InequalitiesThe 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.
I: x > –20–1–4 J: –3 < x ≤ 1K:
a. K U J
so K U J is –2–3 1
0
–30
Hence K U J is (–3, ∞) or {–3 < x}.
b. K ∩ I
K
InequalitiesThe 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.
I: x > –20–1–4 J: –3 < x ≤ 1K:
a. K U J
so K U J is –2–3 1
0
–30
Hence K U J is (–3, ∞) or {–3 < x}.
10
b. K ∩ I–3 K
InequalitiesThe 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.
I: x > –20–1–4 J: –3 < x ≤ 1K:
a. K U J
so K U J is –2–3 1
0
–30
Hence K U J is (–3, ∞) or {–3 < x}.
10
b. K ∩ I
–1–4 –3 K
I
InequalitiesThe 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.
I: x > –20–1–4 J: –3 < x ≤ 1K:
a. K U J
so K U J is –2–3 1
0
–30
Hence K U J is (–3, ∞) or {–3 < x}.
10
b. K ∩ I
–1–4 –3
The common or overlapping portion is shown.
K
I
InequalitiesThe 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.
I: x > –20–1–4 J: –3 < x ≤ 1K:
a. K U J
so K U J is –2–3 1
0
–30
Hence K U J is (–3, ∞) or {–3 < x}.
10
b. K ∩ I
–1–4 –3
The common or overlapping portion is shown.
Hence K ∩ I is (–3, –1) or {–3 < x < –1}
K
I
Inequalities
c. (K ∩ J) U I
I: x > –20–1–4 J: –3 < x ≤ 1K:
Inequalities
c. (K ∩ J) U I
We do the parenthesis first.
I: x > –20–1–4 J: –3 < x ≤ 1K:
Inequalities
c. (K ∩ J) U I
–2–3 10
We do the parenthesis first.
I: x > –20–1–4 J: –3 < x ≤ 1K:
K ∩ J :
Inequalities
c. (K ∩ J) U I
so K ∩ J is
–2–3 10
–20
1
We do the parenthesis first.
I: x > –20–1–4 J: –3 < x ≤ 1K:
K ∩ J :
Inequalities
c. (K ∩ J) U I
Therefore (K ∩ J) U I is
so K ∩ J is
–2–3 10
–20
1
–20
1–1–4
We do the parenthesis first.
I: x > –20–1–4 J: –3 < x ≤ 1K:
K ∩ J :
Inequalities
c. (K ∩ J) U I
Therefore (K ∩ J) U I is
so K ∩ J is
–2–3 10
–20
1
so (K ∩ J) U I = {–4 < x ≤ 1} or (–4, 1]
–20
1–1–4
We do the parenthesis first.
I: x > –20–1–4 J: –3 < x ≤ 1K:
K ∩ J :
Inequalities
c. (K ∩ J) U I
Therefore (K ∩ J) U I is
so K ∩ J is
–2–3 10
–20
1
so (K ∩ J) U I = {–4 < x ≤ 1} or (–4, 1]
–20
1–1–4
Your turn. Find K ∩ (J U I ). Is this the same as (K ∩ J) U I?
We do the parenthesis first.
I: x > –20–1–4 J: –3 < x ≤ 1K:
K ∩ J :