Subtracting Whole Numbers with Addition in Parentheses

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Subtraction of whole numbers with addition in parentheses refers to a situation where we must perform the mathematical operation of subtraction on the sum of some terms that are in parentheses.
In this case, we must remember that the subtraction will be performed on each and every term separately.

The rule is as follows:

aβˆ’(b+c)=aβˆ’bβˆ’ca - (b +c) = a - b - c

a - (b +c) = a - b - c

Each of the terms has its own numerical value.

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\( 100-(5+55)= \)

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Practice example for the subtraction of whole numbers with addition in parentheses

Task:

12βˆ’(3+2)=12 - (3+2) =

We subtract each of the numbers in parentheses from 12 separately and get:

12βˆ’3βˆ’2=12 - 3 - 2 =

9βˆ’2=9 - 2 =

77


Another practice example for subtracting whole numbers with addition in parentheses

Task:

25βˆ’(10+5)=25βˆ’10βˆ’5=25 - (10+5)= 25 - 10 - 5 =

15βˆ’5=1015 - 5 = 10

Note:

You can solve subtraction exercises by following the order of operations.
We add the numbers in parentheses first, and then perform the subtraction.
The impressive part is that we will arrive at the same result, as we have done the same operation, only altering the order.


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Exercising subtraction of whole numbers with parentheses in which there are additions:

Exercise 1

Task:

βˆ’4βˆ’((βˆ’14)+(βˆ’23))= -4-\left(\left(-14\right)+\left(-23\right)\right)=

Solution:

First we address the expression in parentheses.

βˆ’4βˆ’(βˆ’14βˆ’23)= -4-\left(-14-23\right)=

We continue solving the expression in parentheses.

βˆ’4βˆ’(βˆ’37)= -4-\left(-37\right)=

We solve the expression according to the rules.

βˆ’4+37= -4+37=

We reorder the expression and solve.

37βˆ’4=33 37-4=33

Answer:

33 33


Exercise 2

Task:

48βˆ’(35+(βˆ’3))= 48-\left(35+\left(-3\right)\right)=

Solution:

We first address the expression in parentheses.

48βˆ’(35βˆ’3)= 48-\left(35-3\right)=

We continue with the parentheses and solve accordingly.

48βˆ’32=16 48-32=16

Answer:

16 16


Do you know what the answer is?

Exercise 3

Task:

βˆ’45βˆ’(8+10)= -45-\left(8+10\right)=

Solution:

First we solve the expression in parentheses, and then the rest.

βˆ’45βˆ’18=βˆ’63 -45-18=-63

Answer:

βˆ’63 -63


Exercise 4

Task:

13βˆ’(7+4)= 13-\left(7+4\right)=

Solution:

First we solve the exercise in parentheses and then the rest.

13βˆ’11=2 13-11=2

Answer:

22


Check your understanding

Exercise 5

Task:

7βˆ’(4+2)= 7-\left(4+2\right)=

Solution:

First we solve the expression in parentheses, and then the rest.

7βˆ’6=1 7-6=1

Answer:

1 1


Review questions

What is the subtraction of whole numbers with addition in parentheses?

The parentheses are a grouping sign. As the name implies, they helps us to group operations where certain mathematical operations need to be performed. These parentheses indicate that the operations must be performed from the inside out, that is, first we must solve the operations that are inside the parentheses, in this case the additions, and then use that number in our subtraction operation.


Do you think you will be able to solve it?

What is the law of signs?

In order to be able to solve operations with parentheses, the law of signs for multiplication must be applied. This law can be seen as follows:

+Γ—+=+ +\times+=+

+Γ—βˆ’=βˆ’ +\times-=-

βˆ’Γ—+=βˆ’ -\times+=-

βˆ’Γ—+=βˆ’ -\times+=-


How to solve with parentheses?

To solve an operation with parentheses, all we have to do is complete the operations from the inside out, that is, perform the operations that are inside the parentheses and then the operations outside the parentheses. In this case we will solve the subtractions that are inside the parentheses and then we use sign laws and perform the next operation.


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How to eliminate parentheses in addition and subtraction?

In order to eliminate the parentheses in an addition or a subtraction expression, we first perform the operations that are inside the parentheses, and in this way the parentheses are eliminated, let's see some examples:

Example 1

Task:

15βˆ’(6+2)= 15-\left(6+2\right)=

Solving the sum inside the parentheses we get:

15βˆ’(8)= 15-\left(8\right)=

Here we can remove the parentheses because all the operations have already been performed.

15βˆ’8=7 15-8=7


Example 2

Task:

βˆ’8βˆ’(9βˆ’3)= -8-\left(9-3\right)=

βˆ’8βˆ’(6)=βˆ’8βˆ’6 -8-\left(6\right)=-8-6

βˆ’8βˆ’6=βˆ’14 -8-6=-14


Example 3

Task:

19βˆ’(6+5)= 19-\left(6+5\right)=

We can perform the operation inside the parentheses or we can apply the sign law, as shown below.

19βˆ’6βˆ’5=13βˆ’5 19-6-5=13-5

13βˆ’5=8 13-5=8


Do you know what the answer is?

examples with solutions for subtracting whole numbers with addition in parentheses

Exercise #1

38βˆ’(18+20)= 38-(18+20)=

Video Solution

Step-by-Step Solution

According to the order of operations, first we solve the exercise within parentheses:

18+20=38 18+20=38

Now, the exercise obtained is:

38βˆ’38=0 38-38=0

Answer

0 0

Exercise #2

8βˆ’(2+1)= 8-(2+1)=

Video Solution

Step-by-Step Solution

According to the order of operations, we first solve the exercise within parentheses:

2+1=3 2+1=3

Now we solve the rest of the exercise:

8βˆ’3=5 8-3=5

Answer

5 5

Exercise #3

22βˆ’(28βˆ’3)= 22-(28-3)=

Video Solution

Step-by-Step Solution

According to the order of operations, we first solve the exercise within parentheses:

28βˆ’3=25 28-3=25

Now we obtain the exercise:

22βˆ’25=βˆ’3 22-25=-3

Answer

βˆ’3 -3

Exercise #4

12:(2Γ—2)= 12:(2\times2)=

Video Solution

Step-by-Step Solution

According to the order of operations, we first solve the exercise within parentheses:

2Γ—2=4 2\times2=4

Now we divide:

12:4=3 12:4=3

Answer

3 3

Exercise #5

100βˆ’(30βˆ’21)= 100-(30-21)=

Video Solution

Step-by-Step Solution

According to the order of operations, we first solve the exercise within parentheses:

30βˆ’21=9 30-21=9

Now we obtain:

100βˆ’9=91 100-9=91

Answer

91 91

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