**SUBJECT: MATHEMATICS**

**CLASS: JSS 1**

**DATE:**

**TERM: 3rd TERM**

**REFERENCE TEXTBOOKS **

- New General Mathematics, Junior Secondary School Book 1
- Essential Mathematics for Junior Secondary School Book 1

WEEK TWO

**TOPIC: SIMPLE EQUATIONS:**

An algebraic equation is two 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 -6 = 14, 4 x 5 = 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:

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

**Evaluation**

Translate the following equations into words:

- 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:

- Three times a certain number plus 20 is equal to the number plus 12.
- A woman is p years old. In seven years’time, she will be 45 years old.
- 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:

- Let the number be m

3 x m + 20 = m + 12

i.e 3m + 20 = m + 12

- Woman is p years old;

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

i.e p + 7 = 45

- 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**:

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

**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.

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

Solution

- 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

- X - 9 = 15

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

X – 9 + 9 = 15 + 9

X = 24

- 5x = 35

Divide both sides by 5 to balance the equation

5x5 =355

X = 7

- 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

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

**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:

- 2y + 3 = y + 1 (b) 4c – 8 = 10 – 5c

Solution

- 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

- 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:

- 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;

- x+43=3 (b) x2- 25= 45

**Solution:**

- 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

- x2-25=45

Multiply both sides by 10, the LCM

x2 ×10-25×10= 45 ×10

5x – 4 = 8

Add 4 to both sides

5x – 4 + 4 = 8 + 4

5x = 12

Divide both sides by 5

5x5=125

X = 2.4

**Evaluation**

Solve the following equations using the balancing method:

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

**General Evaluation:**

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

**Reading Assignment**

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

**Weekend Assignment:**

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

**Theory**

- Solve the linear equations (a) x – 2 = 2x + 1 (b) 19x – 12 = 11x + 4
- Subtracting nine from a certain number gives thirteen.

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