Order of Operations in Algebra

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 3.1.1:

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

Solution

Using the 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 3.1.2:

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

Solution

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

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

Similarly in algebra 

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

Example 3.1.3:

Simplify the following expressions using the 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 the 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, \( \scriptsize 3x \: + \: 12x \)

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 the 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