SUBJECT: MATHEMATICS
CLASS: JSS 2
DATE:
TERM: 2nd TERM
WEEK FOUR
TOPIC: SOLUTION OF INEQUALITIES IN ONE VARIABLE
CONTENT
Solution of inequalities
Inequalities are solved in the same way as simple equation
For Example x = 23 is equation
While x < 3 is an inequalities .
Worked Examples
Solution
x < 6 – 4
x < 2
x
-2 -1 0 1 2 3
7 < 2x
3 ½ < x
X > 3 ½
x > 3 ½
3 ½
3x > 7 + 3
3x > 10
x > 10/3
x > 3
x > 3 ½
-2 -1 0 1 2 3 3 1/3 4 5 6 7
EVALUATION
MULTIPLICATION AND DIVISION BY NEGATIVE NUMBER
When solving an inequality involving negative numbers, the inequality sign must be reversed. For Example if
-2x > 10 is true then on division throughout by -2.
X <- 5 will be true
Worked Examples
Solution
5 – x > 3
-x > 3 – 5
-x > -2
Dividing through by ( -1)
-x < - 2
-1 -1
X < 2
19 – 4 > - 5x
15 > - 5x
Dividing through by -5
15 < - 5x
-5 – 5
-3 < -5x
x > - 3
-2x < 8 – 3
-2x X 5
-2 -2
X > -5/2
X > -2 ½
Evaluation
Solve the following inequalities
Show your Solution on a number line
READING ASSIGNMENT
New General Mathematics UBE Edition, Chapter 22, Nos 1- 2pgs 213-215
Essential Mathematics by A. JS. Oluwasanmi Chapter 23 pgs 40 - 243
WORD PROBLEMS INVOLVING INEQUALITIES
Worked Example
Solution
Perimeter of triangle = x + ( x + 4) + 11
But perimeter < 32cm
x+ 4 +11< 32
2x < 32 – 15
2x < 17
x< 17/2
x < 8 ½
Also in any triangle, the sum of the length of any two sides must be greater than the length of the third side.
Thus,
X + ( x + 4) > 11
2x + 4 > 11
2x > 7
x > 7/2
x > 3 ½
Thus
x< 8 ½ and x > 3 ½ . but x is a whole number of cm therefore, the possible values of x are 4,5,6,7 or 8.
GENERAL EVALUATION
REVISION QUESTION
WEEKEND ASSIGNMENT
THEORY
Hint: The sum of the length of any two sides of a triangle must be greater than the length of the third side.
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