Download PPT for Section 2.8, 2.9

WEEK 2
SECTION 2.8 AND SECTION
2.9
SRABASTI DUTTA
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.
IMPORTANT CONCEPT 1
 Number Line:
<---------------Decreasing
-1




Increasing --------------->
O
1
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.
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 2 + 1 < 3
+ 1 is true.
•Similarly, a – c < b – c is true because 2 – 1 < 3 – 1 is true.
Example: 5 < 10. Let c = 90  5 + 90 < 10 + 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).
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.
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.
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 – 5 + 5 < 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.
WRITING THE SOLUTION USING DIFFERENT NOTATIONS.
 x < 3 is the solution.
 Set-Theory Notation: {x: x < 3} or {x| x < 3}
 Graphical:
The red-line is the solution
0
)
3
 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)
ANOTHER EXAMPLE:
 3x – 5 ≤ 6 – 2x
 Add 5 to both sides: 3x – 5 + 5 ≤ 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]
The red-line is the solution
0
]
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.
ANOTHER EXAMPLE
Do this problem yourself; then, look at the solution in the next
slide.
13 – 7x ≥ 10x – 4
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 + -13 ≥ 10x – 4 + -13
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 – 17+ -10x
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:
17 x 17

 x 1
17 17
NOTE: how we have reversed the inequality sign since we are dividing
both sides by a negative number (-17 in this case).
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
PREVIOUS EXAMPLE CONTINUES
 x lies between -4 and 1. Graphically:
(
-4
O
)
1
 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).
ANOTHER EXAMPLE (LINEAR INEQUALITY)
4x  3
2x 1
Look at the example with fractions:
2
6
12
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:
 4x  3

 2x 1 
12
 2   12

 6

 12 
 4x  3 
 12
  12(2)  2 x  1
 6 
 24 x  3  24  2 x  1
PREVIOUS EXAMPLE CONTINUES
Step 3: Continue with the algebraic simplifications:
24 x  3  24  2 x  1
 8 x  6  24  2 x  1
 8 x  18  2 x  1
 8 x  18  18  2 x  1  18
 8 x  2 x  19
 8 x  2 x  2 x  19  2 x
 6 x  19
 19
x
6
PREVIOUS EXAMPLE CONTINUES
Interval-Notation: [-19/6, ∞)
Set-Theory Notation: {x|x ≥ -19/6}
Graphically:
[
-19/6
O