2. Week 2
Section 2.8 and section 2.9
Srabasti dutta

3. Definition
An inequality is a mathematical expression involving the symbols <, ≤, > or ≥.
< is read as “less than.”
≤ is read a “less than or equal to.”
> is read as “greater than.”
≥ is read as “greater than or equal to.”
Examples: 3 is less than 5 can be written as 3 < 5
Example: 3 is greater than 2 can be written as 3 > 2.
Example: x ≤ y means x can be either less than y or equal to y. We don’t know since both x and y are variables.
Example: Similarly, x ≥ y means x can be either greater than y or equal to y.

4. Important Concept 1 Number Line:
Example: 2 < 3 (which can also be equivalently written as 3 > 2)
Example: -2 > -3 (which can also be written as -3 < -2)
Thus, - 1 < 0 and 0 < 1
So, keep in mind: numbers are increasing to the right of zero and numbers are decreasing to the left of zero. O
Increasing >
< Decreasing -1 1

5. Important Concept 2
Addition Principle of Inequalities: Let a, b and c be some numbers. If a < b is true, then a + c < b + c and a – c < b – c are also true. That is, adding or subtracting a number does not affect the inequality sign.
Example: Let a = 2, b = 3, c = 1. Then a < b is true because 2 < 3. Then, adding c to both sides will still keep the inequality true, a + c < b + c because < is true.
Similarly, a – c < b – c is true because 2 – 1 < 3 – 1 is true.
Example: 5 < 10. Let c = 90  < since left side value is 95 and the right side value is 100.
5 – 90 < 10 – 90 because we have -85 < -80 (this is true because negative numbers decreases).

6. Important Concept 3
Multiplication Principle for Inequality: a, b, c are some numbers. Let c > 0. If a < b is true, then ac < bc is also true. However, if c < 0, then a <b true implies that ac > bc
The above principle can be remembered as “multiplying an inequality by a negative number results in the inequality sign being reversed.”
Example: a = 2, b = 3, c = 4. Then, a < b is true because 2 < 3 is true. Thus, ac < bc is also true because 2*4 < 3*4 is true.
Let c = -4. Then, note 2*-4 = -8 and 3*-4 = -12 and thus it is no longer ac < bc. Rather, it is -8 > -12 or ac > bc.

7. Definition
Solution of an inequality are those numbers that satisfy the inequality.
All those numbers form the solution set.
Solution set are denoted using either graphical, set-theory or interval notations.
Linear Inequality are the expressions involving only inequality sign. Example: 2x – 3 < 5
Compound Inequality are those expressions that involve more than one inequality sign. Example: -3 < 2x + 5 < 7. Example: 3x + 2 < 8 < x – 1.

8. Solving Linear Inequality
Solve 3x – 5 < 4 and write down the solution using all the notations.
Solution:
3x – 5 < 4.
Add 5 to both sides: 3x – < 4 + 5
3x < 9
Divide both sides by 3 to get x < 3. Thus, any number
less than 3 will satisfy the equation.
Check: Let x = 2. Then, 3(2) – 5 = 6- 5 = 1 and thus 1 < 4 is
true.

9. Writing the Solution Using Different Notations.
x < 3 is the solution.
Set-Theory Notation: {x: x < 3} or {x| x < 3}
Graphical:
Note the parenthesis on 3. It is known as an “open-parenthesis” and is used only when we have either strictly less-than or strictly greater-than inequality sign.
Interval-Notation: (-∞, 3) 3 )
The red-line is the solution

10. Another Example: 3x – 5 ≤ 6 – 2x
Add 5 to both sides: 3x – ≤ 6 – 2x + 5
3x ≤ 11 – 2x
Add 2x to both sides: 3x + 2x ≤ 11 – 2x + 2x
5x ≤ 11
Divide both sides by 5 to get the solution: x ≤ 11/5
Set-Theory: {x| x ≤ 11/5}
Interval: (-∞, 11/5]
Note the parenthesis. The square parenthesis/bracket, [], is used whenever we are dealing with “less than equal to” or “greater than equal to.” Mathematically, square bracket signifies that 11/5 is also included in the solution set.
11/5 ]
The red-line is the solution

11. Another Example
Do this problem yourself; then, look at the solution in the next slide.
13 – 7x ≥ 10x – 4

12. Solution
13 – 7x ≥ 10x – 4
Step 1: to have all the variables on the left side, we need to get rid of that 13. Since it is positive 13, we add the opposite: 13 – 7x ≥ 10x –
Step 2: Cancelling the 13’s gives us -7x ≥ 10x – 17
Step 3: Let’s get rid off 10x from the left side by adding opposite: -7x + -10x ≥ 10x – x
Step 4: We get -17x ≥ -17 after cancelling out the 10x’s on the right side
Step 5: Our goal is to make x stand alone. Since x has -17 multiplied to it, we divide both sides by -17 to get:
NOTE: how we have reversed the inequality sign since we are dividing both sides by a negative number (-17 in this case).

13. Solving Compound Inequality
Compound inequalities form when two inequalities are joined by words and/or.
Example – 3 < 2x + 5 and 2x + 5 <7 can be written as
-3 < 2x + 5 < 7
We solve it using the same procedure:
Step 1: Subtract 5 from all the sides:
-3 – 5 < 2x + 5 – 5 < 7 – 5
-8 < 2x < 2
Step 2: Divide everything by 2:
-8/2 < 2x/2 < 2/2
-4 < x < 1
Step3: Thus, x lies between -4 and 1 and that is our solution

14. Previous Example Continues
x lies between -4 and 1. Graphically:
Set-notation: {x|-4<x<1}
Interval-notation: (-4,1)
NOTE: 1. Interval notations are always written using numbers only. It is always a pair of numbers. We do not include the variable x in it.
NOTE: 2. The smallest number is written first and then the larger number. Thus, we write (-4,1) but not (1,-4). -4 O 1 ( )

15. Another Example (Linear Inequality)
Look at the example with fractions:
Step 1: The denominators are 6 and 12. The smallest number divisible by both 6 and 12 is 12.
Step 2: Multiply the whole inequality by 12:

16. Previous Example Continues
Step 3: Continue with the algebraic simplifications:

17. Previous Example Continues
Interval-Notation: [-19/6, ∞)
Set-Theory Notation: {x|x ≥ -19/6}
Graphically:
-19/6 O [