Removing Brackets

It is vital to always try and simplify the terms inside brackets () first. If they will not simplify, remove the brackets.

To remove the brackets, you must apply the sign rules.

i.e. A positive +  sign before a bracket does not change the signs of the terms inside the brackets.

For example;

\( \scriptsize 3a \: + \: \left(5a \; – \: 2a\right) \)

The sign before the opening bracket is positive ( + ), so if we remove the bracket the sign will not change.

We therefore have;

\( \scriptsize 3a \: + \: 5a \; – \: 2a\)

We can go ahead and solve the equation.

\( \scriptsize 8a \: – \: 2a\)

= \( \scriptsize 6a\)   

Also, a negative ( – ) sign before an opening bracket will change all the positive ( + ) signs to negative ( – )  and also change all the negative ( – ) signs inside the bracket to positive ( + ), when the bracket is removed.

For example,

\( \scriptsize 20x \; – \: \left(8x \: + \: 5x \right) \)

First, we solve the equation in the brackets (Remember – BODMAS)

\( \scriptsize 20x \; – \: \left(13x \right) \)

When we remove the brackets + 13x will change to -13x because of the negative sign in front of the opening bracket. 

= \( \scriptsize 20x \; – \: 13x \) 

= \( \scriptsize 7x \) 

If there is no sign in front of a number it typically means that the number is positive. For example, +13 is positive, −13 is negative, and 13 is positive.

Example 3.2.1:

Simplify the following by removing the bracket: 

i. \( \scriptsize9 \; – \: \left(5 \: + \: 2 \right) \)
ii. \( \scriptsize a \: + \: \left(a \; – \: b \right) \)
iii. \( \scriptsize – \: \left(b \; – \: 2c \right) \; – \: 5a \)
iv. \( \scriptsize \left(x \; – \:3y \right) \: + \: \left(x \; – \:3y \right) \)
v. \( \scriptsize 3a \; – \: \left( 5b \; – \: 2c \right) \: + \: 4d \)

Solution

i. \( \scriptsize 9 \; – \: \left(5 \: + \: 2 \right) \)

Remove the brackets. There is a negative sign in front of the opening bracket so 5 and 2 will change signs. The sign in front of 5 is + and the sign in front of 2 is +

The negative sign in front of the opening bracket will change both signs to negative when we remove the brackets as discussed earlier.

The equation now becomes;

\( \scriptsize 9 \; – \: 5 \; – \: 2 \)

\(\scriptsize 9 \; – \: 7 \)

or

\( \scriptsize 9 \; – \: \left(5 \: + \: 2 \right) \)

\( \scriptsize 9 \; – \: \left(7\right) \)

\( \scriptsize 9 \; – \: 7 \)

= 2

ii. \( \scriptsize a \: + \: \left(a \; – \: b \right) \)

A positive + sign before an opening bracket does not change the signs of the terms inside the brackets.

Therefore,

\( \scriptsize a \: + \: a \; – \: b \)

Solve the like terms

= \( \scriptsize 2a \; – \: b \)

iii. \( \scriptsize – \: \left(b \; – \: 2c \right) \; – \: 5a \)

The negative sign changes the signs of the terms in the brackets. When we remove the brackets. +b becomes -b, and -2c becomes +2c

Therefore,

\( \scriptsize – b \: + \: 2c \; – \: 5a \)

There are no like terms to solve so we can simply rearrange the equation in alphabetical order.

= \( \scriptsize -5a \; – \: b \: + \: 2c \)

iv. \( \scriptsize \left(x \; – \:3y \right) \: + \: \left(x \; – \:3y \right) \)

A positive + sign before a bracket does not change the signs of the terms inside the brackets.

Therefore if we remove the brackets we now have;

\( \scriptsize x \; – \:3y \: + \: x \; – \:3y \)

Collect like terms

\( \scriptsize x \; + \: x \; – \:3y \; – \: 3y \)

= \( \scriptsize 2x \; – \: 6y \)

v. \( \scriptsize 3a \; – \: \left( 5b \; – \: 2c \right) \: + \: 4d \)

The negative sign changes the signs of the terms in the brackets. When we remove the brackets. +5b becomes -5b, and -2c becomes +2c

= \( \scriptsize 3a \: + \:5b \: + \: 2c \: + \: 4d \)