ASSESSMENT 3

RPH

Class   : 3A1 (32 students)

Date    : 10^th August 2011

Day     : Wednesday

Week  : 12^th Week

Time   : 80 minutes (7.35-8.55 a.m.)

Topic   : Linear Inequalities

Learning Area : Linear Inequalities

Learning Outcome      : 1. Identify relationship

a.       greater than

b.      less than

c.       using a symbol / sign “> “or “ <” and “≥ “or” ≤

2. Identify the relationship between 2 given number

a. greater than or equal to

b. less than or equal to base on given situation

Learning Objectives    : 1. Students be able to understand and use the concept of inequalities

                                     2. Students be able to understand and use the concept of linear 

      Inequalities in one unknown

 3. Understand the sign “> “or “ <” and “≥ “or” ≤

Learning Activities     :

Let a, b and c be real numbers.

1. Transitive Property
   If a < b and b < c then a < c 
2. Addition Property
   If a < b then a + c < b + c 
3. Subtraction Property
   If a < b then a - c < b - c 
4. Multiplication Property
1. If a < b and c is positive then c*a < c*b 
2. If a < b and c is negative c*a > c*b

Note:

3. If each inequality sign is reversed in the above properties, we obtain similar properties.
4. If the inequality sign < is replaced by <= ( less than or equal) or the sign > is replaced by >= ( greater than or equal ), we also obtain similar properties.

• Solve 2x < 9.

If  given "2x = 9", should have divided the 2 from each side. can do the same thing here:

                      

Then the solution is: x < ⁹/₂

...or, if you prefer decimals (and if your instructor will accept decimal equivalents instead of fractions):

x < 4.5

• Solve ^x/₄ > ¹/₂.

If they had given " ^x/₄ = ¹/₂ ",  have multiplied both sides by 4. I can do the same thing here:

Then the solution is: x > 2

• Solve –2x < 5.

Remember how I said that solving linear inequalities is "almost" exactly like solving linear equations? Well, this is the one place where it's different. To explain what I'm about to do, consider the following:   

3 > 2

What happens to the above inequality when I multiply through by –1? The temptation is to say that the answer will be "–3 > –2". But –3 is not greater than –2; it is in actuality smaller. That is, the correct inequality is actually the following:

–3 < –2

As you can see, multiplying by a negative ("–1", in this case) flipped the inequality sign from "greater than" to "less than". This is the new wrinkle for solving inequalities:

When solving inequalities, if multiply or divide through by a negative, must also flip the inequality sign.

To solve "–2x < 5", I need to divide through by a negative ("–2"), so I will need to flip the inequality:

Then the solution is: x > ^–5/₂

• Solve ^{(2x – 3)}/₄  < 2.

First, I'll multiply through by 4. Since the "4" is positive, I don't have to flip the inequality sign:

^{(2x – 3)}/₄   < 2
(4) × ^{(2x – 3)}/₄  < (4)(2)
2x – 3 < 8
2x < 11
x < ¹¹/₂  = 5.5

Achievement   : 1. Student must know the difference of the sign

                          2. Ask the selected student to answer the question one by one

                          3. Student is able to answer the question

Reflection       : Student knows how to state the numbers by one unknown in linear inequalities.