Second Term Lesson Note for Week Four.
Class : JSS 2
Subject : Mathematics
Topic : Solving Inequalities and Word Problems on inequalities
Duration : 80 Minutes
Period : Double Period
Reference Book :
- Essential Mathematics for Junior Secondary School, JSS 2 (Basic 8).
- Lagos State Unified schemes of work for Junior Secondary School, JSS 1 – 3.
- Online Resources
Instructional Material : Chart showing the basic rules in solving Inequalities.
Learning Objectives : By the end of the lesson learners will be able to :
i. Explain the term inequalities
ii. Identify the different inequality signs and describes each.
iii. Solve problems involving inequalities and state rules of solving Inequalities.
Content :
SOLVING INEQUALITIES
RULES IN SOLVING INEQUALITIES.
An inequality remains true when the same quantity is added to, or subtracted from both side.
An inequality remains true when both sides are multiplied or divided by the same positive quantity.
An inequality remains true when both sides are multiplied or divided by a negatives quantity provided the inequality sign is reversed.
Example 1 : Find the greatest possible value of a X that satisfies the inequality 8 + 2X > 3 + 5X. if X is an integer.
Solution:
8 + 2X > 3 + 5X
Subtract 8 from both sides
2X > 3 – 8 + 5X
2X > -5 + 5X
Subtract 5X from both sides
2X – 5X > -5 = -3x > -5
Divide both sides by -3 and also reverse the inequality
X < 12/3 [rule three]
The greatest integer value of X is 1.
Example 2 : Find the smallest integer value of X that satisfies the inequality 7X – 2 ≥ 5X – 6
Solution:
7X – 2 ≥ 5X – 6
Add 2 to both sides
7X ≥ 5X – 6 + 2
7X ≥ 5X – 4
Subtract 5X from both sides
7X – 5X ≥ -4 = 2x ≥ -4 , Divide both side by 2
X ≥ -2
WORD PROBLEMS LEADING TO INEQUALITIES
Example 1: David is X years old. In 4 year time his age will still be less than 12 years. (a) Write this information in an inequality in X. (b) Find the maximum age of David to the nearest whole numbers.
Solution:
In 4 year time David age’s will be (X + 4) years.
If at that time his age will be less than 12, then
X + 4 < 12
Subtract 4 from both sides
X < 12 – 4
X < 8.
The maximum age of David is 7 years.
Example 2: A man had #X, out of this, he used #1000 to pay his house rent. The amount he had left is not more than 3500. (a) Write this information in an inequality in X (b) Solve for X.
Solution:
The man used #1000 to pay his house rent out of #X, so the amount left is #(x – 1000).
This amount is not more than 3500
X – 1000 ≤ 500
Add 1000 to both sides
X ≤ 500 + 1000 = X ≤ 1500
Hence the man had less than or equal to #1500
EVALUATION:
PAGE 172 EXERCISE 14.3 No. 3(K,L,M,N and O), PAGE 173 EX 14.4 No. 1, 2 and 3
CONCLUSION : The teacher summarizes the lesson and marks the learners note. He/ she then makes correction.
ASSIGNMENT :
PAGE 172 AND 173 EX 14.3 AND 14.4 No. 5, 6, 7, 8 and 9 IN EX 14.4