# Lesson Notes By Weeks and Term - Junior Secondary School 2

SOLUTION OF INEQUALITIES IN ONE VARIABLE

SUBJECT: MATHEMATICS

CLASS:  JSS 2

DATE:

TERM: 2nd TERM

WEEK FOUR

TOPIC: SOLUTION OF INEQUALITIES IN ONE VARIABLE

CONTENT

• Solution of inequalities
• Multiplication and division of negative numbers
• Word problem involving inequalities

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

1. Solve the inequality and show the Solution on a graph.
2. x + 4 < 6
3. <  2x – 1
4. 3x – 3 > 7

Solution

1. x + 4  < 6

x < 6 – 4

x  < 2

x

-2       -1      0      1      2       3

1.   6  2x – 14

7   <   2x

3 ½  <  x

X  >  3 ½

x   >  3 ½

3 ½

1. 3x – 3 > 7

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

1. Solve the following inequalities and show the Solution on a graph:
2. 5x -2   >  8
3. 4x – 2 > 19
4. 3  <  3x   X 5
5. x  +  4   >   10

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

1.   Solve 5 – x > 3
2. Solve 19 >  4 -5x
3. Solve 3 – 2x < 8

Solution

5 – x > 3

-x > 3 – 5

-x > -2

Dividing through by ( -1)

-x   <   - 2

-1        -1

X < 2

1. 19 >  4 – 5x

19 – 4  >  - 5x

15  >  - 5x

Dividing through by -5

15  <  - 5x

-5        – 5

-3  <  -5x

x  >  - 3

1. 3 – 2x < 8

-2x < 8 – 3

-2x  X  5

-2        -2

X >  -5/2

X > -2 ½

Evaluation

Solve the following inequalities

1.   -2x + 5 > 16
2. 10 – 3x <  - 11
3. 2r  6r + 6
4. 9 <  3 – 4t

Show your Solution on a number line

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

1. A triangle has sides xcm, ( x + 4) cm and 11cm, where x is a whole number of cm, if the perimeter of the triangle is less than 32cm. find the possible values of x.

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

1. If 9 is added to a number x, the result is greater than 17. Find the values of x
2. Three times a certain number is not greater than 54. Find the range of values of the number .

REVISION QUESTION

1. If 8 is subtracted from a number, the result is at most 15. Find the range of values of the number
2. If  x  is  subtracted  from  5,  and  the  result  is  greater  than   15.Find  the  range  of  values  of  x.

WEEKEND ASSIGNMENT

1. Find which symbol > or < goes in the box to make the statement  9 + 8           10 true.
2. < B. >     C. none of the above
3. Write the inequality of the statement ‘The cost of meal Nx was over N5” A. x > N5 B. x > N5 C. x .< N 5
4. Solve 5x – 7 > 9 A. x > 32.5 B. x > 3 1/5    C. x , 5 2/3
5. Solve x -2 < 3 A. x < 5     B. x > 5        C. x  > 5
6. If 7.3 is subtracted from y, the result is less than 3.4. find the value of y. A. y < 10.7 B. y > 10.7 C. y <  10.7

THEORY

1. A rectangle is 8cm long and bcm branch find the value of b if the perimeter of the rectangle is not greater than 50cm and not less than 18cm
2. The sides of a rectangle are xcm ( x+ 3cm) and 10cm . if xcm is a whole number , if the perimeter is less than 30cm, find the possible value of x.

Hint: The sum of the length of any two sides of a triangle must be greater than the length of the third side.