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JSS1: MATHEMATICS - 2ND TERM

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  1. Algebraic Processes | Week 1
    4 Topics
    |
    1 Quiz
  2. Simplification of Algebraic Expressions | Week 2
    4 Topics
    |
    1 Quiz
  3. Simplification of Algebraic Expressions 2 (Use of Brackets) | Week 3
    4 Topics
    |
    1 Quiz
  4. Simple Equations | Week 4
    1 Topic
    |
    1 Quiz
  5. Simple Equations II | Week 5
    3 Topics
    |
    1 Quiz
  6. Plane Shapes I | Week 6
    5 Topics
    |
    2 Quizzes
  7. Plane Shapes II | Week 7
    7 Topics
    |
    1 Quiz
  8. Plane Shapes III | Week 8
    7 Topics
    |
    1 Quiz
  9. Decimal and Percentages I | Week 9
    2 Topics
    |
    1 Quiz
  10. Decimal and Percentages | Week 10
    3 Topics
    |
    1 Quiz



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When solving a problem involving mixed arithmetic operations, we need to deal with the brackets () first, then Division ÷, and Multiplication \( \scriptsize \times \) and finally  Addition +   and Subtraction –  

i.e.  We apply the BODMAS rule,

B – Brackets (   )

O – Of

D – Division

M – Multiplication

A – Addition

S – Subtraction

Example 1

\(\scriptsize 6 \: \times \: \left( 7 \: + \: 4 \right) \)

Using BODMAS rule, we solve the bracket first before multiplying.

\(\scriptsize 6 \: \times \: \left( 7 \: + \: 4 \right) \\ \scriptsize = 6 \: \times \: 11 \\ \scriptsize = 66\)

Similarly in algebra

\(\scriptsize 6 \: \times \: \left( 7x \: + \: 4x \right) \\ \scriptsize = 6 \: \times \: 11x \\ \scriptsize = 66x\)

Example 2

\( \scriptsize 12 \:\: – \: 8 \: \div \: 2 \)

Using BODMAS rule, we solve the division first before subtracting.

\(\scriptsize 12 \:\: – \: 8 \: \div \: 2 \\ \scriptsize 12 \:\: – \: 4 = 8 \)

Similarly in algebra 

\(\scriptsize 12x \; – \: 8x \: \div \: 2 \\ \scriptsize 12x \; – \: 4x = 8x \)

More Examples

Simplify the following expressions using BODMAS rule:

i) \( \scriptsize 3x \: + \: 3 \: \times \: 4x \)

ii) \( \scriptsize 9 \: \times \: 3x \; – \: \left(4x \: + \: 2x \right) \: \times \: 3 \)

iii) \( \scriptsize 8a \: \times \: 12a \: \div \: 2a \; – \: \left(13a \; – \: 3a \right) \)

iv) \(\scriptsize 5 \: \times \: 6y \; – \: 3y \: \times \: 4 \; – \: 4y \: \times \: 2 \)

Solution

i) \( \scriptsize 3x \: + \: 3 \: \times \: 4x \)

Using BODMAS rule, we multiply first, before adding.

1st Step: Multiply, \( \scriptsize 3 \: \times \: 4x = 12x \)

What we now have is

2nd Step: Addition, \( \scriptsize 3x \: + \: 12x \)

Answer = 15x

ii) \( \scriptsize 9 \: \times \: 3x \; – \: \left(4x \: + \: 2x \right) \: \times \: 3 \)

Using BODMAS rule, we solve the bracket first, before multiplying and finally, subtracting.

1st Step: Solve the bracket \( \scriptsize \left(4x \: + \: 2x \right) = 6x \)

\( \scriptsize 9 \: \times \: 3x \; – \: 6x\: \times \: 3 \)

2nd Step: Multiply \( \scriptsize 9\: \times \: 3x \) first, ignore the minus sign for now, then multiply \( \scriptsize 6x \: \times \: 3 \)

\( \scriptsize 9 \: \times \: 3x = 27x \)

\( \scriptsize 6x \: \times \: 3 = 18x \)

= \( \scriptsize 27x \; – \: 18x \)

3rd Step: Finally we subtract \( \scriptsize 27x \; – \: 18x \)

Answer = 9x

iii) \( \scriptsize 8a \: \times \: 12a \: \div \: 2a \; – \: \left(13a \; – \: 3a \right) \)

Using the BODMAS rule, we solve the bracket first, before dividing and then multiplying.

1st Step: Solve the bracket \( \scriptsize \left(13a \; – \: 3a \right) = 10a \)

The equation now becomes;

\( \scriptsize 8a \: \times \: 12a \: \div \: 2a \; – \: 10a \)

2nd Step: Divide, \( \scriptsize 12a \: \div \: 2a = \normalsize \frac{12\not{a}}{2\not{a}} \\ \scriptsize = 6 \)

The equation now becomes;

\( \scriptsize 8a \: \times \: 6 \; – \: 10a \)

3rd Step: Multiply, \( \scriptsize 8a \: \times \: 6 = 48a \)

4th Step: Subtract, \( \scriptsize 48a \; – \: 10a \)

Answer = 38a

iv) \(\scriptsize 5 \: \times \: 6y \; – \: 3y \: \times \: 4 \; – \: 4y \: \times \: 2 \)

Using BODMAS rule, we Multiply first, then subtract.

1st Step: Multiply

\( \scriptsize 5 \: \times \: 6y = 30y \)

\( \scriptsize 3y \: \times \: 4 = 12y \)

\( \scriptsize 4y \: \times \: 2 = 8y \)

What we now have is;

\( \scriptsize 30y \; – \: 12y \; – \: 8y \)

2nd Step: Subtract;

\( \scriptsize 30y \; – \: 12y \; – \: 8y \)

\( \scriptsize 30y \; – \: 20y \)

\( \scriptsize 30y \; – \: 20y \)

Answer = 10y

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