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: 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

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: 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

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] 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

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: 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: 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 ( )

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:

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

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