# SIMPLE EQUATIONS MATHEMATICS JSS1 THIRD TERM WEEK TWO

Reference Materials

• Scheme of Work
• Online Information
• Textbooks
• Workbooks
• 9 Year Basic Education Curriculum

Previous Knowledge :

The pupils have previous knowledge of

Revision

Behavioural Objectives :  At the end of the lesson, the pupils should be able to

• define simple equation
• solve questions on simple equation
• enumerate or simplify some mental sums on algebra

Content :

WEEK TWO

SIMPLE EQUATIONS:

An algebraic equation is the combination of letters and figures in an equation. Algebraic expressions separated by an equal sign. The left hand side is equal to the right hand side (LHS = RHS)

e.g     7 + 3 = 10,

20 – 16 = 4,

4 x B = 20,

35/7 = 5

Translation of algebraic equations into words: Any letter of the alphabet can be used to represent the unknown number.

Translate the following equations into words:

1. X + 9 = 12; means ‘a certain number plus nine is equal to twelve’
2. 15= 7 – 2x; means ‘fifteen is equal to seven minus twice a certain number’
3. 4x5 = 6; means ‘four-fifth of a number equal to six’
4. 3k+8=20; ‘three times a certain number plus eight is equal to twenty’

Evaluation

Translate the following equations into words:

1. 16 = 9 – 2x   2. 9 + 5x = 23    3.  X + 5 = seventy      4. 3x4=9

Translation of algebraic sentences into equations:

Example: Translate the following into equations:

1. Three times a certain number plus 20 is equal to the number plus 12.
2. A woman is p years old. In seven years’time, she will be 45 years old.
3. The result of taking 10 from the product of a certain number and 7 is the same as taking 4 from twice the number.

Solution:

1. Let the number be m

3 x m + 20 = m + 12

i.e 3m + 20 = m + 12

1. Woman is p years old;

7 years’ time, she will be (p + 7) years

i.e p + 7 = 45

1. Let the number be a,

Product of a and 7 = 7a

Taking 10 from 7a = 7a – 10

Taking 4 from twice the number = 2a – 4

Then, 7a – 10 = 2a – 4

Evaluation:Translate to algebraic equations

1. A certain number is added to 15, the result is six minus the same number.
2. Ayo is y years old, 7 years ago, she was 15 years old.

Second Lesson

Use of Balancing or See saw Method

This is very easy and convenient way of solving linear equations. An equation can be compared to a balance. To maintain balance, whatever is done to the LHS of the scale must be done to the RHS every time.

Examples:

Solve the following equations using the balancing method.

1. X + 4 = 9    (b) x – 9 = 15  (c) 5x = 35   (d) x3=7

Solution

1. X + 4 = 9

To eliminate 4 from the LHS and RHS of the equation, subtract 4 from both sides

X + 4 -4 = 9 – 4

X = 5

1. X – 9 = 15

Add 9 to both sides of the equation to eliminate -9

X – 9 + 9 = 15 + 9

X = 24

1. 5x = 35

Divide both sides by 5 to balance the equation

5x5 =355

X = 7

1. x3=7

Multiply both sides by 3 to eliminate 3 from the LHS

x3 ×3=7×3

X = 21

Evaluation: Solve the following equations using the balancing equation method

1. 4x = 25    (2) x + 16 = -19   (3) –x -3 = -9   (4)  x2=1.4

Presentation

The topic is presented step by step

Step 1:

The class teacher revises the previous topics

Step 2.

He introduces the new topic

Solving Linear Simple Equations Involving Collection of Like Terms

Simple equations can be solved by collecting like terms. That is taking the unknown like terms to one side and the known to the other side.

Example:

Solve the following equations:

1. 2y + 3 = y + 1           (b) 4c – 8 = 10 – 5c

Solution

1. 2y + 3 = y + 1

Subtract y from both sides to eliminate y from RHS

2y – y + 3 = y – y + 1

y + 3 = 1

Subtract 3 from both sides to eliminate 3 from LHS

y + 3 – 3 = 1 – 3

y = -2

1. 4c – 8 = 10 – 5c

Collect like terms by adding 5c to both sides to eliminate 5c from the RHS

4c + 5c – 8 = 10 – 5c + 5c

9c – 8 = 10

Add 8 to both sides to eliminate 8 from LHS

9c – 8 + 8 = 10 + 8

9c = 18

Divide both sides by 9

9c9= 189

C = 2

Evaluation: Solve the following equations by using the balancing method:

1. 17a – 11 = 10a + 3   (2) 7d – 6 = 30 – 2d   (3) -6 – 2x = 5 – 7x

Solving Linear Simple Equations Involving Fractions

To solve equations involving fractions, the first thing is to clear the fractions and then collect

like terms.

Example: Solve the following equations;

1. x+43=3         (b) x225= 45

Solution:

1. x+45=3

Multiply both sides by the LCM 5

(x+45)×5=3 ×5

X + 4 = 15

Subtract 4 from both sides

X + 4 – 4 = 15 -4

X = 9

1. x225=45

Multiply both sides by 10, the LCM

x2 ×10-25×10= 45 ×10

5x – 4 = 8

5x – 4 + 4 = 8 + 4

5x = 12

Divide both sides by 5

5x5=125

X = 2.4

Step 3:

The class teacher allows the pupils to give their own examples and he corrects them when the needs arise

Evaluation

Solve the following equations using the balancing method:

1. x5+14=1720(2) x+72=1

General Evaluation:

1. Solve using the balancing method: (a) 14 – x -5 = -5x + 3 (b) 12y – 4 = 2   (c) y3-4=1
2. Twice a certain numberis added to 10. If the result is minus fourteen, find the number.
3. Two thirds of a certain number plus five equals ten less than the same number. What is the number?

Essential Mathematics for Junior Secondary Schools 1. Page 144- 154

Weekend Assignment:

1. If 8 is added to a number, the result is 27, What is the number? (a) 25 (b) 35 (c) 19 (d) -27
2. Solve 4x6=5    (a) 30   (b) 7.5     (c)   15    (d) 26
3. Solve 3y + 4 = 22   (a) 6     (b) 263      (c) 18    (d) 54
4. Solve x + 0.4  = 0.6    (a) 0.10    (b)  0.2    (c)  – 0.2   (d)  -1.0
5. Solve -3x + 5 –x = 14 – 6x    (a) 4.5    (b) -4.5   (c) 4.75   (d)   9

Conclusion :

The class teacher wraps up or conclude the lesson by giving out short note to summarize the topic that he or she has just taught.

The class teacher also goes round to make sure that the notes are well copied or well written by the pupils.

He or she does the necessary corrections when and where  the needs arise.

: