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## JSS1: MATHEMATICS - 2ND TERM

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Lesson 3, Topic 1
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# Order of Operations in Algebra

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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
S – Subtraction

### Example 1

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

Solution

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

$$\scriptsize 12 \:\: – \: 8 \: \div \: 2$$

Solution

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 $$\scriptsize 3x \: + \: 12x$$

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

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 $$\scriptsize 27x \; – \: 18x$$

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, $$\scriptsize 48a \; – \: 10a$$

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;

$$\scriptsize 30y \; – \: 12y \; – \: 8y$$

$$\scriptsize 30y \; – \: 20y$$

$$\scriptsize 30y \; – \: 20y$$