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PHILIPPINE NORMAL UNIVERSITY VISAYAS
The National Center for Teacher Education
The Environment and Green Technology Education Hub
Cadiz City, Negros Occidental
PHILIPPINE NORMAL UNIVERSITY VISAYAS
The National Center for Teacher Education
The Environment and Green Technology Education Hub
Cadiz City, Negros Occidental
After this lesson, one is expected to solve
equations and inequalities.
An equation is a statement that two expressions are equal.
Example 1.4 The following are examples of equation.
2π‘₯ βˆ’ 6 = 0 π‘₯2
βˆ’ 5π‘₯ + 6 = 0
To solve an equation means to find all the values of the variables
that will make the statement true. The values of the variables that
make the statement true are called solutions or roots of the
equation.
Example 1.5 The equation 2π‘₯ βˆ’ 6 = 0 is a true statement for
π‘₯ = 3 but it is false for any other number. The root or solution
of this equation is 3.
Properties of Equations
Let π‘Ž, 𝑏, 𝑐 ∈ ℝ.
1. Reflexive Property
π‘Ž = π‘Ž
Example 2 = 2
2. Symmetric Property (or Replacement Property)
If π‘Ž = 𝑏, then 𝑏 = π‘Ž.
If π‘Ž = 𝑏, then π‘Ž can replace 𝑏 in any instance.
Example If 5 = π‘₯, then π‘₯ = 5.
3. Transitive Property
If π‘Ž = 𝑏, and 𝑏 = 𝑐, then π‘Ž = 𝑐.
Example If π‘₯ = 𝑦 and 𝑦 = 5, then π‘₯ = 5.
4. Addition Property
If π‘Ž = 𝑏, then π‘Ž + 𝑐 = 𝑏 + 𝑐.
Example If π‘₯ = 𝑦 then π‘₯ + 3 = 𝑦 + 3.
5. Multiplication Property
If π‘Ž = 𝑏, then π‘Žπ‘ = 𝑏𝑐.
Example If π‘₯ = 𝑦, then 5π‘₯ = 5𝑦.
Example 1.6 Use the properties of equations and properties of real numbers
to solve the equation
2
3
π‘₯ βˆ’ 1 = π‘₯ βˆ’ 3.
Solution.
3
2
3
π‘₯ βˆ’ 1 = π‘₯ βˆ’ 3 3 Multiplication Property
2 π‘₯ βˆ’ 1 = π‘₯ βˆ’ 3 3
2π‘₯ βˆ’ 2 = 3π‘₯ βˆ’ 9 Distributive Property
2π‘₯ βˆ’ 2 + 2 βˆ’ 3π‘₯ = 3π‘₯ βˆ’ 9 + (2 βˆ’ 3π‘₯) Addition Property
(2π‘₯ βˆ’ 3π‘₯) + βˆ’2 + 2 = (3π‘₯ βˆ’ 3π‘₯) + (βˆ’9 + 2) Associative Property
βˆ’π‘₯ + 0 = 0 βˆ’ 7 Additive Inverse
βˆ’π‘₯ = βˆ’7 Additive Identity
Thus, π‘₯ = 7 Multiplication Property
Solve the equation
2
3
π‘₯ βˆ’ 1 = π‘₯ βˆ’ 3.
Solution: π‘₯ = 7
To check, using replacement property,
2
3
π‘₯ βˆ’ 1 = π‘₯ βˆ’ 3
2
3
7 βˆ’ 1 = 7 βˆ’ 3;
4 = 4 which is true by reflexive property of equality.
An equation that is true for all values of the variable is called an identity.
An equation that is true for some values of the variables but not true for other
values is called a conditional equation. An equation that has no solutions is
called a contradiction.
Example 1.7
Classify each equation as an identity, a conditional equation, or a
contradiction.
a) π‘₯2
+ 2π‘₯ = π‘₯(π‘₯ + 2)
b) 5π‘₯ βˆ’ 2 = π‘₯ βˆ’ 10
c) 3π‘₯ + 2 = 1 + 3π‘₯
Classify each equation as an identity, a conditional equation, or a
contradiction.
a) π‘₯2
+ 2π‘₯ = π‘₯(π‘₯ + 2)
Solution
Getting the product of the right side of π‘₯2
+ 2π‘₯ = π‘₯(π‘₯ + 2),
we have
π‘₯2
+ 2π‘₯ = π‘₯2
+ 2π‘₯
Since the left and right expressions are the same, the equation will
hold true for whatever value of π‘₯, thus we have an identity.
Classify each equation as an identity, a conditional equation, or a
contradiction.
𝑏. 5π‘₯ βˆ’ 2 = π‘₯ βˆ’ 10
Solution
Adding both sides of 5π‘₯ βˆ’ 2 = π‘₯ βˆ’ 10 by (2 βˆ’ π‘₯),
we have
5π‘₯ βˆ’ 2 + 2 βˆ’ π‘₯ = π‘₯ βˆ’ 10 + (2 βˆ’ π‘₯)
4π‘₯ = βˆ’8.
Multiplying both sides of 4π‘₯ = βˆ’8 by
1
4
,
we get π‘₯ = βˆ’2
The original equation is satisfied by π‘₯ = βˆ’2 but not by other values.
Thus, the equation is a conditional equation.
Classify each equation as an identity, a conditional equation, or a
contradiction.
𝑐. 3π‘₯ + 2 = 1 + 3π‘₯
Solution
Subtracting 3π‘₯ from both sides of the equation
3π‘₯ + 2 βˆ’ 3π‘₯ = 1 + 3π‘₯ βˆ’ 3π‘₯
we get 2 = 1
which is a false statement.
The equation is a false statement no matter what the value of π‘₯ is
and thus, it has no solution. Therefore, it is a contradiction.
Real numbers are shown on the number line with larger
numbers written to the right. For any two real numbers, the
one to the left is less than the one to the right. The symbol <
means β€œis less than” and the symbol > means β€œis greater
than”.
A statement that one quantity is greater than or less than
another quantity is called an inequality.
β–ͺ If π‘Ž and 𝑏 are real numbers,
π‘Ž is greater than 𝑏
or π‘Ž > 𝑏
means π‘Ž is to the right of 𝑏 on the number line,
and π‘Ž βˆ’ 𝑏 is positive.
β–ͺ If π‘Ž is less than 𝑏,
or π‘Ž < 𝑏,
this means π‘Ž is to the left of 𝑏 on the number line,
and π‘Ž βˆ’ 𝑏 is negative
Note the following:
π‘Ž > 𝑏 and 𝑏 < π‘Ž have the same meaning.
π‘Ž > 0 means that π‘Ž is positive,
π‘Ž < 0 means that π‘Ž is negative.
Lesson 1.2 NT (Equation and Inequalities).pdf
The following statements are also inequalities:
π‘Ž ≀ 𝑏 means β€œπ‘Ž is less than or equal to 𝑏”
π‘Ž β‰₯ 𝑏 means β€œπ‘Ž is greater than or equal to 𝑏
If π‘Ž ≀ 𝑏 and π‘Ž β‰₯ 𝑏 hold simultaneously, then π‘Ž = 𝑏.
For any real number π‘Ž,
π‘Ž ≀ 0 means π‘Ž is not positive, and
π‘Ž β‰₯ 0 means π‘Ž is not negative.
Example 1.9 The following are examples of inequalities:
a) 2π‘₯ + 3 < 12
b) π‘₯2
βˆ’ 2π‘₯ βˆ’ 8 ≀ 0
c) π‘₯ + 2𝑦 > 1 + 𝑦
Properties of Inequalities
Let π‘Ž, 𝑏, 𝑐 ∈ ℝ.
1. Trichotomy Property. If π‘Ž, 𝑏 ∈ ℝ, then one and only one of the following
relations holds: π‘Ž < 𝑏, π‘Ž = 𝑏, π‘œπ‘Ÿ π‘Ž > 𝑏.
2. Transitive Property. If π‘Ž, 𝑏, 𝑐 ∈ ℝ, such that π‘Ž > 𝑏 and 𝑏 > 𝑐, then π‘Ž > 𝑐.
3. Addition Property. If π‘Ž, 𝑏, 𝑐 ∈ ℝ, such that π‘Ž > 𝑏 , then π‘Ž + 𝑐 > 𝑏 + 𝑐.
4. Multiplication Property. Let π‘Ž, 𝑏, 𝑐 ∈ ℝ.
𝑖. If π‘Ž > 𝑏 and 𝑐 > 0, then π‘Žπ‘ > 𝑏𝑐.
𝑖𝑖. If π‘Ž > 𝑏 and 𝑐 < 0, then π‘Žπ‘ < 𝑏𝑐.
Example 1.10 Solve the inequality
2
3
π‘₯ βˆ’ 1 β‰₯ π‘₯ βˆ’ 3.
Solution.
3
2
3
π‘₯ βˆ’ 1 β‰₯ π‘₯ βˆ’ 3 3 Multiplication Property (𝑖)
2 π‘₯ βˆ’ 1 β‰₯ π‘₯ βˆ’ 3 3
2π‘₯ βˆ’ 2 β‰₯ 3π‘₯ βˆ’ 9 Distributive Property
2π‘₯ βˆ’ 2 + 2 βˆ’ 3π‘₯ β‰₯ 3π‘₯ βˆ’ 9 + (2 βˆ’ 3π‘₯) Addition Property
(2π‘₯ βˆ’ 3π‘₯) + βˆ’2 + 2 β‰₯ (3π‘₯ βˆ’ 3π‘₯) + (βˆ’9 + 2) Associative Property
βˆ’π‘₯ + 0 β‰₯ 0 βˆ’ 7 Additive Inverse
βˆ’π‘₯ β‰₯ βˆ’7 Additive Identity
Thus, π‘₯ ≀ 7 Multiplication Property (𝑖𝑖)
Writing the solution in a set,
π‘₯ȁπ‘₯ ≀ 7
Read: The set of all elements π‘₯ such that π‘₯ is less than or equal to 7.
To illustrate the solution in a number line,
To check, choose one value in the solution, say π‘₯ = βˆ’2 .
Substituting this value for π‘₯,
2
3
βˆ’2 βˆ’ 1 β‰₯ βˆ’2 βˆ’ 3
2
3
βˆ’3 β‰₯ βˆ’5 and
βˆ’2 β‰₯ βˆ’5 which is true.
Compound Inequalities
A compound inequality is formed by joining two
inequalities with the connective word and or or.
The solution set of a compound inequality with the
connective word π‘œπ‘Ÿ is the union of the solution sets of the two
inequalities.
The solution set of a compound inequality with the
connective word π‘Žπ‘›π‘‘ is the intersection of the solution sets of the
two inequalities.
Example 1.11 Solve the following of compound inequalities:
a) 2π‘₯ < 8 π‘œπ‘Ÿ π‘₯ βˆ’ 3 > 4
b) π‘₯ + 3 > 5 π‘Žπ‘›π‘‘ 3π‘₯ βˆ’ 7 < 14
Solution
a) 2π‘₯ < 8 π‘œπ‘Ÿ π‘₯ βˆ’ 3 < 4
Solving each inequality, we have
2π‘₯ < 8 π‘₯ βˆ’ 3 > 4
π‘₯ < 4 π‘₯ > 7
π‘₯ȁπ‘₯ < 4 π‘₯ȁπ‘₯ > 7
Write the union of the solution sets.
π‘₯ȁπ‘₯ < 4 βˆͺ π‘₯ȁπ‘₯ > 7 = π‘₯ȁπ‘₯ < 4 π‘œπ‘Ÿ π‘₯ > 7
Illustrating the solution in a number line,
π‘₯ȁπ‘₯ < 4 βˆͺ π‘₯ȁπ‘₯ > 7 = π‘₯ȁπ‘₯ < 4 π‘œπ‘Ÿ π‘₯ > 7
b) π‘₯ + 3 > 5 π‘Žπ‘›π‘‘ 3π‘₯ βˆ’ 10 > 14
Solving each inequality, we have
π‘₯ + 3 > 5
π‘₯ > 2
π‘₯ȁπ‘₯ > 2
3π‘₯ βˆ’ 10 > 14
3π‘₯ > 24
π‘₯ > 8
π‘₯ȁπ‘₯ > 8
Write the intersection of the solution sets.
π‘₯ȁπ‘₯ > 2 ∩ π‘₯ȁπ‘₯ > 8 = π‘₯ȁ π‘₯ > 8 .
The inequality given by π‘Ž < 𝑏 < 𝑐, is equivalent to the compound
inequality π‘Ž < 𝑏 π‘Žπ‘›π‘‘ 𝑏 < 𝑐.
Example 1.12 Solve the inequality 5 < 2π‘₯ βˆ’ 3 < 13.
Solution. This can be solved by either of the following methods.
Method 1 5 < 2π‘₯ βˆ’ 3 < 13 is equivalent to 5 < 2π‘₯ βˆ’ 3 π‘Žπ‘›π‘‘ 2π‘₯ βˆ’ 3 < 13.
5 < 2π‘₯ βˆ’ 3 π‘Žπ‘›π‘‘ 2π‘₯ βˆ’ 3 < 13
8 < 2π‘₯ 2π‘₯ < 16
4 < π‘₯ π‘œπ‘Ÿ π‘₯ > 4 π‘₯ < 8
π‘₯ȁπ‘₯ > 4 ∩ ȁ
π‘₯ π‘₯ < 8 = { ȁ
π‘₯ 4 < π‘₯ < 8}
Method 2.
Solve 5 < 2π‘₯ βˆ’ 3 < 13
5 < 2π‘₯ βˆ’ 3 < 13 Add 3 to all parts of the inequality.
8 < 2π‘₯ < 16 Multiply the parts by Β½.
4 < π‘₯ < 8
Thus, the solution set is
{ ȁ
π‘₯ 4 < π‘₯ < 8}.
Example 1.13 Solve the inequality 2π‘₯ βˆ’ 5 < 9.
Solution
The definition of the absolute value of a number π‘Ž, denoted by
π‘Ž , is the real number such that
π‘Ž = π‘Ž when π‘Ž β‰₯ 0.
π‘Ž = βˆ’π‘Ž when π‘Ž < 0.
From the definition, we have
2π‘₯ βˆ’ 5 < 9 and βˆ’ 2π‘₯ βˆ’ 5 < 9
Note that βˆ’ 2π‘₯ βˆ’ 5 < 9 when multiplied by βˆ’1, becomes
2π‘₯ βˆ’ 5 > βˆ’9
Thus, we can write the inequalities as
2π‘₯ βˆ’ 5 < 9 and 2π‘₯ βˆ’ 5 > βˆ’9
Thus, we can write the inequalities as
2π‘₯ βˆ’ 5 < 9 and 2π‘₯ βˆ’ 5 > βˆ’9
Or
βˆ’9 < 2π‘₯ βˆ’ 5 < 9
Add 5,
βˆ’4 < 2π‘₯ < 14
Multiply by
1
2
(or divide by 2),
βˆ’2 < π‘₯ < 7
Thus, the solution set is { ȁ
π‘₯ βˆ’ 2 < π‘₯ < 7}.
Solving the two inequalities,
2π‘₯ βˆ’ 5 < 9 and βˆ’ 2π‘₯ βˆ’ 5 < 9 π‘œπ‘Ÿ 2π‘₯ βˆ’ 5 > βˆ’9. Or
βˆ’9 < 2π‘₯ βˆ’ 5 < 9 Add 5.
βˆ’4 < 2π‘₯ < 14 Multiply the parts by Β½.
βˆ’2 < π‘₯ < 7. Thus, the solution set is { ȁ
π‘₯ βˆ’
2 < π‘₯ < 7}.
Solving the two inequalities,
2π‘₯ βˆ’ 5 < 9 and βˆ’ 2π‘₯ βˆ’ 5 < 9 π‘œπ‘Ÿ 2π‘₯ βˆ’ 5 > βˆ’9. Or
βˆ’9 < 2π‘₯ βˆ’ 5 < 9 Add 5.
βˆ’4 < 2π‘₯ < 14 Multiply the parts by Β½.
βˆ’2 < π‘₯ < 7. Thus, the solution set is { ȁ
π‘₯ βˆ’
2 < π‘₯ < 7}.
An inequality that is true for all values of the variable is
called an absolute inequality.
An inequality that is true for some values of the variables
but not true for other values is called a conditional inequality.
Example 1.14
The inequality π‘₯ + 5 < 8 is true only if π‘₯ < 3 and thus, is
a conditional inequality.
In the inequality π‘₯2
+ 1 > 0, the lowest possible value of
π‘₯2
is 0, that is, when π‘₯ = 0. For all other real values of π‘₯, π‘₯2
will always be positive. Thus, π‘₯2
+ 1 > 0 is true for all real
values of π‘₯ and is an absolute inequality.

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Lesson 1.2 NT (Equation and Inequalities).pdf

  • 1. PHILIPPINE NORMAL UNIVERSITY VISAYAS The National Center for Teacher Education The Environment and Green Technology Education Hub Cadiz City, Negros Occidental
  • 2. PHILIPPINE NORMAL UNIVERSITY VISAYAS The National Center for Teacher Education The Environment and Green Technology Education Hub Cadiz City, Negros Occidental
  • 3. After this lesson, one is expected to solve equations and inequalities.
  • 4. An equation is a statement that two expressions are equal. Example 1.4 The following are examples of equation. 2π‘₯ βˆ’ 6 = 0 π‘₯2 βˆ’ 5π‘₯ + 6 = 0 To solve an equation means to find all the values of the variables that will make the statement true. The values of the variables that make the statement true are called solutions or roots of the equation. Example 1.5 The equation 2π‘₯ βˆ’ 6 = 0 is a true statement for π‘₯ = 3 but it is false for any other number. The root or solution of this equation is 3.
  • 5. Properties of Equations Let π‘Ž, 𝑏, 𝑐 ∈ ℝ. 1. Reflexive Property π‘Ž = π‘Ž Example 2 = 2 2. Symmetric Property (or Replacement Property) If π‘Ž = 𝑏, then 𝑏 = π‘Ž. If π‘Ž = 𝑏, then π‘Ž can replace 𝑏 in any instance. Example If 5 = π‘₯, then π‘₯ = 5.
  • 6. 3. Transitive Property If π‘Ž = 𝑏, and 𝑏 = 𝑐, then π‘Ž = 𝑐. Example If π‘₯ = 𝑦 and 𝑦 = 5, then π‘₯ = 5. 4. Addition Property If π‘Ž = 𝑏, then π‘Ž + 𝑐 = 𝑏 + 𝑐. Example If π‘₯ = 𝑦 then π‘₯ + 3 = 𝑦 + 3. 5. Multiplication Property If π‘Ž = 𝑏, then π‘Žπ‘ = 𝑏𝑐. Example If π‘₯ = 𝑦, then 5π‘₯ = 5𝑦.
  • 7. Example 1.6 Use the properties of equations and properties of real numbers to solve the equation 2 3 π‘₯ βˆ’ 1 = π‘₯ βˆ’ 3. Solution. 3 2 3 π‘₯ βˆ’ 1 = π‘₯ βˆ’ 3 3 Multiplication Property 2 π‘₯ βˆ’ 1 = π‘₯ βˆ’ 3 3 2π‘₯ βˆ’ 2 = 3π‘₯ βˆ’ 9 Distributive Property 2π‘₯ βˆ’ 2 + 2 βˆ’ 3π‘₯ = 3π‘₯ βˆ’ 9 + (2 βˆ’ 3π‘₯) Addition Property (2π‘₯ βˆ’ 3π‘₯) + βˆ’2 + 2 = (3π‘₯ βˆ’ 3π‘₯) + (βˆ’9 + 2) Associative Property βˆ’π‘₯ + 0 = 0 βˆ’ 7 Additive Inverse βˆ’π‘₯ = βˆ’7 Additive Identity Thus, π‘₯ = 7 Multiplication Property
  • 8. Solve the equation 2 3 π‘₯ βˆ’ 1 = π‘₯ βˆ’ 3. Solution: π‘₯ = 7 To check, using replacement property, 2 3 π‘₯ βˆ’ 1 = π‘₯ βˆ’ 3 2 3 7 βˆ’ 1 = 7 βˆ’ 3; 4 = 4 which is true by reflexive property of equality.
  • 9. An equation that is true for all values of the variable is called an identity. An equation that is true for some values of the variables but not true for other values is called a conditional equation. An equation that has no solutions is called a contradiction. Example 1.7 Classify each equation as an identity, a conditional equation, or a contradiction. a) π‘₯2 + 2π‘₯ = π‘₯(π‘₯ + 2) b) 5π‘₯ βˆ’ 2 = π‘₯ βˆ’ 10 c) 3π‘₯ + 2 = 1 + 3π‘₯
  • 10. Classify each equation as an identity, a conditional equation, or a contradiction. a) π‘₯2 + 2π‘₯ = π‘₯(π‘₯ + 2) Solution Getting the product of the right side of π‘₯2 + 2π‘₯ = π‘₯(π‘₯ + 2), we have π‘₯2 + 2π‘₯ = π‘₯2 + 2π‘₯ Since the left and right expressions are the same, the equation will hold true for whatever value of π‘₯, thus we have an identity.
  • 11. Classify each equation as an identity, a conditional equation, or a contradiction. 𝑏. 5π‘₯ βˆ’ 2 = π‘₯ βˆ’ 10 Solution Adding both sides of 5π‘₯ βˆ’ 2 = π‘₯ βˆ’ 10 by (2 βˆ’ π‘₯), we have 5π‘₯ βˆ’ 2 + 2 βˆ’ π‘₯ = π‘₯ βˆ’ 10 + (2 βˆ’ π‘₯) 4π‘₯ = βˆ’8. Multiplying both sides of 4π‘₯ = βˆ’8 by 1 4 , we get π‘₯ = βˆ’2 The original equation is satisfied by π‘₯ = βˆ’2 but not by other values. Thus, the equation is a conditional equation.
  • 12. Classify each equation as an identity, a conditional equation, or a contradiction. 𝑐. 3π‘₯ + 2 = 1 + 3π‘₯ Solution Subtracting 3π‘₯ from both sides of the equation 3π‘₯ + 2 βˆ’ 3π‘₯ = 1 + 3π‘₯ βˆ’ 3π‘₯ we get 2 = 1 which is a false statement. The equation is a false statement no matter what the value of π‘₯ is and thus, it has no solution. Therefore, it is a contradiction.
  • 13. Real numbers are shown on the number line with larger numbers written to the right. For any two real numbers, the one to the left is less than the one to the right. The symbol < means β€œis less than” and the symbol > means β€œis greater than”. A statement that one quantity is greater than or less than another quantity is called an inequality.
  • 14. β–ͺ If π‘Ž and 𝑏 are real numbers, π‘Ž is greater than 𝑏 or π‘Ž > 𝑏 means π‘Ž is to the right of 𝑏 on the number line, and π‘Ž βˆ’ 𝑏 is positive. β–ͺ If π‘Ž is less than 𝑏, or π‘Ž < 𝑏, this means π‘Ž is to the left of 𝑏 on the number line, and π‘Ž βˆ’ 𝑏 is negative
  • 15. Note the following: π‘Ž > 𝑏 and 𝑏 < π‘Ž have the same meaning. π‘Ž > 0 means that π‘Ž is positive, π‘Ž < 0 means that π‘Ž is negative.
  • 17. The following statements are also inequalities: π‘Ž ≀ 𝑏 means β€œπ‘Ž is less than or equal to 𝑏” π‘Ž β‰₯ 𝑏 means β€œπ‘Ž is greater than or equal to 𝑏 If π‘Ž ≀ 𝑏 and π‘Ž β‰₯ 𝑏 hold simultaneously, then π‘Ž = 𝑏. For any real number π‘Ž, π‘Ž ≀ 0 means π‘Ž is not positive, and π‘Ž β‰₯ 0 means π‘Ž is not negative.
  • 18. Example 1.9 The following are examples of inequalities: a) 2π‘₯ + 3 < 12 b) π‘₯2 βˆ’ 2π‘₯ βˆ’ 8 ≀ 0 c) π‘₯ + 2𝑦 > 1 + 𝑦
  • 19. Properties of Inequalities Let π‘Ž, 𝑏, 𝑐 ∈ ℝ. 1. Trichotomy Property. If π‘Ž, 𝑏 ∈ ℝ, then one and only one of the following relations holds: π‘Ž < 𝑏, π‘Ž = 𝑏, π‘œπ‘Ÿ π‘Ž > 𝑏. 2. Transitive Property. If π‘Ž, 𝑏, 𝑐 ∈ ℝ, such that π‘Ž > 𝑏 and 𝑏 > 𝑐, then π‘Ž > 𝑐. 3. Addition Property. If π‘Ž, 𝑏, 𝑐 ∈ ℝ, such that π‘Ž > 𝑏 , then π‘Ž + 𝑐 > 𝑏 + 𝑐. 4. Multiplication Property. Let π‘Ž, 𝑏, 𝑐 ∈ ℝ. 𝑖. If π‘Ž > 𝑏 and 𝑐 > 0, then π‘Žπ‘ > 𝑏𝑐. 𝑖𝑖. If π‘Ž > 𝑏 and 𝑐 < 0, then π‘Žπ‘ < 𝑏𝑐.
  • 20. Example 1.10 Solve the inequality 2 3 π‘₯ βˆ’ 1 β‰₯ π‘₯ βˆ’ 3. Solution. 3 2 3 π‘₯ βˆ’ 1 β‰₯ π‘₯ βˆ’ 3 3 Multiplication Property (𝑖) 2 π‘₯ βˆ’ 1 β‰₯ π‘₯ βˆ’ 3 3 2π‘₯ βˆ’ 2 β‰₯ 3π‘₯ βˆ’ 9 Distributive Property 2π‘₯ βˆ’ 2 + 2 βˆ’ 3π‘₯ β‰₯ 3π‘₯ βˆ’ 9 + (2 βˆ’ 3π‘₯) Addition Property (2π‘₯ βˆ’ 3π‘₯) + βˆ’2 + 2 β‰₯ (3π‘₯ βˆ’ 3π‘₯) + (βˆ’9 + 2) Associative Property βˆ’π‘₯ + 0 β‰₯ 0 βˆ’ 7 Additive Inverse βˆ’π‘₯ β‰₯ βˆ’7 Additive Identity Thus, π‘₯ ≀ 7 Multiplication Property (𝑖𝑖)
  • 21. Writing the solution in a set, π‘₯ȁπ‘₯ ≀ 7 Read: The set of all elements π‘₯ such that π‘₯ is less than or equal to 7. To illustrate the solution in a number line, To check, choose one value in the solution, say π‘₯ = βˆ’2 . Substituting this value for π‘₯, 2 3 βˆ’2 βˆ’ 1 β‰₯ βˆ’2 βˆ’ 3 2 3 βˆ’3 β‰₯ βˆ’5 and βˆ’2 β‰₯ βˆ’5 which is true.
  • 22. Compound Inequalities A compound inequality is formed by joining two inequalities with the connective word and or or. The solution set of a compound inequality with the connective word π‘œπ‘Ÿ is the union of the solution sets of the two inequalities. The solution set of a compound inequality with the connective word π‘Žπ‘›π‘‘ is the intersection of the solution sets of the two inequalities.
  • 23. Example 1.11 Solve the following of compound inequalities: a) 2π‘₯ < 8 π‘œπ‘Ÿ π‘₯ βˆ’ 3 > 4 b) π‘₯ + 3 > 5 π‘Žπ‘›π‘‘ 3π‘₯ βˆ’ 7 < 14 Solution a) 2π‘₯ < 8 π‘œπ‘Ÿ π‘₯ βˆ’ 3 < 4 Solving each inequality, we have 2π‘₯ < 8 π‘₯ βˆ’ 3 > 4 π‘₯ < 4 π‘₯ > 7 π‘₯ȁπ‘₯ < 4 π‘₯ȁπ‘₯ > 7 Write the union of the solution sets. π‘₯ȁπ‘₯ < 4 βˆͺ π‘₯ȁπ‘₯ > 7 = π‘₯ȁπ‘₯ < 4 π‘œπ‘Ÿ π‘₯ > 7
  • 24. Illustrating the solution in a number line, π‘₯ȁπ‘₯ < 4 βˆͺ π‘₯ȁπ‘₯ > 7 = π‘₯ȁπ‘₯ < 4 π‘œπ‘Ÿ π‘₯ > 7
  • 25. b) π‘₯ + 3 > 5 π‘Žπ‘›π‘‘ 3π‘₯ βˆ’ 10 > 14 Solving each inequality, we have π‘₯ + 3 > 5 π‘₯ > 2 π‘₯ȁπ‘₯ > 2 3π‘₯ βˆ’ 10 > 14 3π‘₯ > 24 π‘₯ > 8 π‘₯ȁπ‘₯ > 8 Write the intersection of the solution sets. π‘₯ȁπ‘₯ > 2 ∩ π‘₯ȁπ‘₯ > 8 = π‘₯ȁ π‘₯ > 8 .
  • 26. The inequality given by π‘Ž < 𝑏 < 𝑐, is equivalent to the compound inequality π‘Ž < 𝑏 π‘Žπ‘›π‘‘ 𝑏 < 𝑐. Example 1.12 Solve the inequality 5 < 2π‘₯ βˆ’ 3 < 13. Solution. This can be solved by either of the following methods. Method 1 5 < 2π‘₯ βˆ’ 3 < 13 is equivalent to 5 < 2π‘₯ βˆ’ 3 π‘Žπ‘›π‘‘ 2π‘₯ βˆ’ 3 < 13. 5 < 2π‘₯ βˆ’ 3 π‘Žπ‘›π‘‘ 2π‘₯ βˆ’ 3 < 13 8 < 2π‘₯ 2π‘₯ < 16 4 < π‘₯ π‘œπ‘Ÿ π‘₯ > 4 π‘₯ < 8 π‘₯ȁπ‘₯ > 4 ∩ ȁ π‘₯ π‘₯ < 8 = { ȁ π‘₯ 4 < π‘₯ < 8}
  • 27. Method 2. Solve 5 < 2π‘₯ βˆ’ 3 < 13 5 < 2π‘₯ βˆ’ 3 < 13 Add 3 to all parts of the inequality. 8 < 2π‘₯ < 16 Multiply the parts by Β½. 4 < π‘₯ < 8 Thus, the solution set is { ȁ π‘₯ 4 < π‘₯ < 8}.
  • 28. Example 1.13 Solve the inequality 2π‘₯ βˆ’ 5 < 9. Solution The definition of the absolute value of a number π‘Ž, denoted by π‘Ž , is the real number such that π‘Ž = π‘Ž when π‘Ž β‰₯ 0. π‘Ž = βˆ’π‘Ž when π‘Ž < 0. From the definition, we have 2π‘₯ βˆ’ 5 < 9 and βˆ’ 2π‘₯ βˆ’ 5 < 9 Note that βˆ’ 2π‘₯ βˆ’ 5 < 9 when multiplied by βˆ’1, becomes 2π‘₯ βˆ’ 5 > βˆ’9 Thus, we can write the inequalities as 2π‘₯ βˆ’ 5 < 9 and 2π‘₯ βˆ’ 5 > βˆ’9
  • 29. Thus, we can write the inequalities as 2π‘₯ βˆ’ 5 < 9 and 2π‘₯ βˆ’ 5 > βˆ’9 Or βˆ’9 < 2π‘₯ βˆ’ 5 < 9 Add 5, βˆ’4 < 2π‘₯ < 14 Multiply by 1 2 (or divide by 2), βˆ’2 < π‘₯ < 7 Thus, the solution set is { ȁ π‘₯ βˆ’ 2 < π‘₯ < 7}.
  • 30. Solving the two inequalities, 2π‘₯ βˆ’ 5 < 9 and βˆ’ 2π‘₯ βˆ’ 5 < 9 π‘œπ‘Ÿ 2π‘₯ βˆ’ 5 > βˆ’9. Or βˆ’9 < 2π‘₯ βˆ’ 5 < 9 Add 5. βˆ’4 < 2π‘₯ < 14 Multiply the parts by Β½. βˆ’2 < π‘₯ < 7. Thus, the solution set is { ȁ π‘₯ βˆ’ 2 < π‘₯ < 7}. Solving the two inequalities, 2π‘₯ βˆ’ 5 < 9 and βˆ’ 2π‘₯ βˆ’ 5 < 9 π‘œπ‘Ÿ 2π‘₯ βˆ’ 5 > βˆ’9. Or βˆ’9 < 2π‘₯ βˆ’ 5 < 9 Add 5. βˆ’4 < 2π‘₯ < 14 Multiply the parts by Β½. βˆ’2 < π‘₯ < 7. Thus, the solution set is { ȁ π‘₯ βˆ’ 2 < π‘₯ < 7}.
  • 31. An inequality that is true for all values of the variable is called an absolute inequality. An inequality that is true for some values of the variables but not true for other values is called a conditional inequality.
  • 32. Example 1.14 The inequality π‘₯ + 5 < 8 is true only if π‘₯ < 3 and thus, is a conditional inequality. In the inequality π‘₯2 + 1 > 0, the lowest possible value of π‘₯2 is 0, that is, when π‘₯ = 0. For all other real values of π‘₯, π‘₯2 will always be positive. Thus, π‘₯2 + 1 > 0 is true for all real values of π‘₯ and is an absolute inequality.