Subtracting Whole Numbers with Subtraction in Parentheses

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Subtraction of whole numbers with subtractions in parentheses refers to a situation where we perform the mathematical operation of subtraction on the difference of some terms that are in parentheses.

For example:

12βˆ’(3βˆ’2)=12 - (3-2) =

One way to solve this exercise will be to distribute the parentheses. To do this, we must remember that according to the law of signs of addition/ subtraction, after removing parentheses, the expressions that were inside them change their sign.

C - Subtracting Whole Numbers with Subtraction in Parentheses

That is, in our example:

12βˆ’(3βˆ’2)=12 - (3-2) =

12βˆ’3+2=12 - 3 + 2 =

9+2=119 + 2 = 11

When distributing the parentheses, we will place a βˆ’ - in front of the number 3 3 and a + + before the 2 2 .
As you can see, in both cases the sign that was inside the parentheses has switched to the opposite sign.

Another way to solve this exercise is to use the order of operations, that is to say:

12βˆ’(3βˆ’2)=12 - (3-2) =

We will start by solving the expression in parentheses by using the order of operations and we will get:

12βˆ’1=1112 - 1 = 11


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

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Exercises for subtracting whole numbers with subtraction in parentheses

Exercise 1

Task:

Solve the following exercise:

βˆ’30βˆ’((βˆ’41)βˆ’((βˆ’4)βˆ’(βˆ’8)))=-30-\left(\left(-41\right)-\left(\left(-4\right)-\left(-8\right)\right)\right)=

Solution:

First we solve the innermost parentheses.

βˆ’30βˆ’((βˆ’41)βˆ’((βˆ’4)+8))=-30-\left(\left(-41\right)-\left(\left(-4\right)+8\right)\right)=

Order the expression in the innermost parentheses and solve it.

βˆ’30βˆ’((βˆ’41)βˆ’(8βˆ’4))= -30-\left(\left(-41\right)-\left(8-4\right)\right)=

βˆ’30βˆ’((βˆ’41)βˆ’4)= -30-\left(\left(-41\right)-4\right)=

We break 41 41 down into 2 2 numbers to make the calculation easier.

βˆ’30βˆ’(βˆ’40βˆ’1βˆ’4)= -30-\left(-40-1-4\right)=

βˆ’30βˆ’(βˆ’40βˆ’1βˆ’4)= -30-\left(-40-1-4\right)=

βˆ’30βˆ’(βˆ’45)= -30-\left(-45\right)=

We solve according to the rules.

βˆ’30+45= -30+45=

45βˆ’30=15 45-30=15

Answer:

1515


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Exercise 2

Task:

17βˆ’(3βˆ’(βˆ’7βˆ’4))= 17-\left(3-\left(-7-4\right)\right)=

Solution:

First, we solve the innermost parentheses and break down4 4 into 2 2 terms to make the calculation easier.

17βˆ’(3βˆ’(βˆ’7βˆ’3βˆ’1))= 17-\left(3-\left(-7-3-1\right)\right)=

17βˆ’(3βˆ’(βˆ’10βˆ’1))= 17-\left(3-\left(-10-1\right)\right)=

17βˆ’(3βˆ’(βˆ’11))= 17-\left(3-\left(-11\right)\right)=

After solving the inner parentheses we continue solving the exercise that remains in parentheses.

17βˆ’(3+11)= 17-\left(3+11\right)=

17βˆ’14= 17-14=

We break down the expression to make the calculation easier.

10+7βˆ’10βˆ’4= 10+7-10-4=

We solve the exercise in two parts.

10βˆ’10=0 10-10=0

7βˆ’4=3 7-4=3

Answer:

3 3


Exercise 3

Task:

βˆ’58βˆ’((βˆ’7)βˆ’(βˆ’12))= -58-\left(\left(-7\right)-\left(-12\right)\right)=

Solution:

We solve the expression in parentheses, first reordering the minus and plus signs.

βˆ’58βˆ’(βˆ’7+12)= -58-\left(-7+12\right)=

Solve the expression in parentheses accordingly.

βˆ’58βˆ’(12βˆ’7)= -58-\left(12-7\right)=

Solve according to the rules.

βˆ’58βˆ’5= -58-5=

We break down 5 5 into 2 2 terms to make the calculation easier.

βˆ’58βˆ’2βˆ’3= -58-2-3=

βˆ’60βˆ’3=βˆ’63 -60-3=-63

Answer:

βˆ’63 -63


Do you know what the answer is?

Exercise 4

Task:

49βˆ’(53βˆ’18)= 49-\left(53-18\right)=

Solution:

First we start with the expression in parentheses and break down the 53 53 into 2 2 numbers to make the calculation easier.

49βˆ’(58βˆ’5βˆ’18)= 49-\left(58-5-18\right)=

We solve the expression in parentheses and then solve accordingly.

49βˆ’(58βˆ’18βˆ’5)= 49-\left(58-18-5\right)=

49βˆ’(40βˆ’5)= 49-\left(40-5\right)=

49βˆ’35= 49-35=

We break down the 49 49 into 2 2 numbers to make the calculation easier.

45+4βˆ’35= 45+4-35=

We reorder the operations accordingly and solve.

45βˆ’35+4= 45-35+4=

10+4=14 10+4=14

Answer:

1414


Exercise 5

Task:

37βˆ’(4βˆ’7)= 37-\left(4-7\right)=

Solution:

First we tackle the expression in parentheses and solve.

37βˆ’(βˆ’3)= 37-\left(-3\right)=

We reorder the minus and plus signs accordingly, and solve.

37+3=40 37+3=40

Answer:

40 40


Check your understanding

Review questions

What is the subtraction of whole numbers with subtraction 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 subtraction, and then use that number in our subtraction operation.


What is the law of signs?

In order to be able to solve operations with parenthese, 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+=-


Do you think you will be able to solve it?

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.


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:

52βˆ’(25βˆ’11)= 52-\left(25-11\right)=

Solution:

We solve the subtraction inside the parentheses.

52βˆ’(14)= 52-\left(14\right)=

We remove the parentheses by applying the law of signs.

52βˆ’14= 52-14=

We break down the 14 14 into two terms to make the calculations easier.

52βˆ’12βˆ’2=40βˆ’2=38 52-12-2=40-2=38

Answer:

38 38


Example 2

Task:

36βˆ’((βˆ’3)βˆ’(βˆ’8))= 36-\left(\left(-3\right)-\left(-8\right)\right)=

Solution:

Eliminate the innermost parentheses using sign laws.

36βˆ’(βˆ’3+8)= 36-\left(-3+8\right)=

Solve the operations inside the parentheses.

36βˆ’(8βˆ’3)=36βˆ’(5) 36-\left(8-3\right)=36-\left(5\right)

Again we eliminate parentheses.

36βˆ’5=31 36-5=31

Answer:

31 31


Test your knowledge

examples with solutions for subtracting whole numbers with subtraction 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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