Elimination of Parentheses in Real Numbers

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In previous articles, we have studied real numbers and the grouping of terms, as well as the order of mathematical operations with parentheses. In this article, we move forward and combine these topics in order to understand when and how we can eliminate parentheses in real numbers.

What does the elimination of parentheses in real numbers mean?

When we perform grouping of like terms ("addition and subtraction") with real numbers, we confine the real number within parentheses.

Parentheses can be removed but when eliminating them, the following rules must be remembered:

  • ++=+++ = +
    1+(+3)=1+31+(+3) = 1+3
    Β 
  • βˆ’βˆ’=+-- = +
    1βˆ’(βˆ’3)=1+31-(-3) = 1+3
    Β 
  • +βˆ’=βˆ’+- = -
    1+(βˆ’3)=1βˆ’31+(-3) = 1-3
    Β 
  • βˆ’+=βˆ’-+ = -
    1βˆ’(+3)=1βˆ’31-(+3) = 1-3
Elimination of Parentheses in Real Numbers

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What is the opposite number of \( 5 \)

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The logic in this case is that the subtraction sign allows us to obtain the number opposite to the one given to us.
Consequently:

  • "Minus minus six" is equal to "plus six", that is, "six".
  • Similarly, "minus plus six" is equal to "minus six".

However, the plus sign does not indicate a modification in the number.
Therefore,

  • "plus minus six" is equal to "minus six"
  • and "plus plus six" is equal to "plus six", that is, "six".

Examples:

  • (+50)+(βˆ’20)=(+30)(+50)+(-20) = (+30)
  • (βˆ’8)βˆ’(+2)=(βˆ’10)(-8)-(+2) = (-10)
  • (βˆ’3)βˆ’(βˆ’4)=(+1)(-3)-(-4) = (+1)

Let's look again at the three previously solved exercises, now we will write them without parentheses.

  • (+50)+(βˆ’20)=(+30)(+50)+(-20) = (+30)
    50βˆ’20=3050-20 = 30
    Β 
  • (βˆ’8)βˆ’(+2)=(βˆ’10)(-8)-(+2) = (-10)
    βˆ’8βˆ’2=βˆ’10-8-2 = -10
    Β 
  • (βˆ’3)βˆ’(βˆ’4)=(+1)(-3)-(-4)=(+1)
    βˆ’3+4=1-3+4 = 1

As we surely remember from the class on "real numbers", when a number has no sign, we understand it to be positive.
Therefore,

  • in the first exercise, we can write "50" and "30" instead of "+50" and "+30".
  • However, we cannot remove the plus sign in the third exercise: in "+4".

Remember: We can only omit the plus sign if the number is the first in the sequence.

When solving exercises with real numbers, in the first phase we will remove the parentheses according to mathematical rules.

Example:

(+58)βˆ’(βˆ’34)+(+9)βˆ’(+5)+(βˆ’2)=(+58)-(-34)+(+9)-(+5)+(-2) =
58+34+9βˆ’5βˆ’2=9458+34+9-5-2 = 94


Exercises on Eliminating Parentheses in Real Numbers

Exercise 1

Complete:

  • –(βˆ’10)=–(-10) = __
  • +(+8)=+(+8) = __
  • __(βˆ’9)=9(-9) = 9
  • __(βˆ’9)=βˆ’9(-9) = -9
  • __+(βˆ’5)=βˆ’5+(-5) = -5
  • ((__3)=βˆ’33) = -3
  • -((__)=20) = 20

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Exercise 2

Solve the following exercises, first of all, remove the parentheses:

  • (βˆ’5)+(+35)βˆ’(βˆ’22)=(-5)+(+35)-(-22) =
  • (βˆ’9)βˆ’(+2)+(+10)=(-9)-(+2)+(+10) =
  • (+56)+(βˆ’43)βˆ’(βˆ’4)βˆ’(+5)=(+56)+(-43)-(-4)-(+5) =
  • (βˆ’12.8)βˆ’(βˆ’3.7)βˆ’(+5)=(-12.8)-(-3.7)-(+5) =
  • (βˆ’90)+(+4.7)βˆ’(βˆ’2.2)=(-90)+(+4.7)-(-2.2) =

Exercise 3

Assignment

Mark the correct answer

[(3βˆ’2+4)2βˆ’22]:(9β‹…7)3= [(3-2+4)^2-2^2]:\frac{(\sqrt{9}\cdot7)}{3}=

Solution

We solve the expressions inside the parentheses according to the order of arithmetic operations

[(1+4)2βˆ’22]:(3β‹…7)3= [(1+4)^2-2^2]:\frac{(3\cdot7)}{3}=

We continue solving the expressions inside the parentheses accordingly.

[52βˆ’22]:213= [5^2-2^2]:\frac{21}{3}=

[25βˆ’4]:213= [25-4]:\frac{21}{3}=

21:7=3 21:7=3

Answer

3 3


Do you know what the answer is?

Exercise 4

Assignment

(7+2+3)(7+6)(12βˆ’3βˆ’4)=? (7+2+3)(7+6)(12-3-4)=\text{?}

Solution

First, we solve the expressions within the parentheses according to the laws of addition and subtraction

(9+3)(7+6)(9βˆ’4)=? \left(9+3\right)\left(7+6\right)\left(9-4\right)=?

12Γ—13Γ—5=? 12\times13\times5=\text{?}

We arrange the multiplication exercise we obtained to make it easier to solve.

12Γ—5Γ—13=? 12\times5\times13=\text{?}

We solve the exercise from left to right

12Γ—5=60 12\times5=60

60Γ—13=780 60\times13=780

Answer

780 780


Exercise 5

Assignment

(9+7+3)(4+5+3)(7βˆ’3βˆ’4)\left(9+7+3\right)\left(4+5+3\right)\left(7-3-4\right)

Solution

First, we solve the expressions within the parentheses according to the laws of addition and subtraction

(9+10)(9+3)(4βˆ’4)= \left(9+10\right)\left(9+3\right)\left(4-4\right)=

19Γ—12Γ—0= 19\times12\times0=

Note that we obtained a multiplication exercise with the number 0 0 and we solve it first to simplify the calculation.

12Γ—0=0 12\times0=0

19Γ—0=0 19\times0=0

Answer

0 0


Check your understanding

Exercise 6

Assignment

(8βˆ’3βˆ’1)Γ—4Γ—3= (8-3-1)\times4\times3=

Solution

First, we solve the operations inside the parentheses according to the rules of addition and subtraction

(5βˆ’1)Γ—4Γ—3= (5-1)\times4\times3=

4Γ—4Γ—3= 4\times4\times3=

We solve the operation from left to right

4Γ—4=16 4\times4=16

16Γ—3=48 16\times3=48

Answer

48 48


Exercise 7

Assignment

(7+2)Γ—(3+8)= (7+2)\times(3+8)=

Solution

We multiply the first element inside the parentheses by the elements of the second parentheses

Then we multiply the second element inside the primary parentheses by the elements of the second parentheses

7Γ—3+7Γ—8+2Γ—3+2Γ—8= 7\times3+7\times8+2\times3+2\times8=

We solve all the multiplication exercises from left to right

21+56+6+16= 21+56+6+16=

Now we add from left to right

21+56=77 21+56=77

77+6=83 77+6=83

83+16=99 83+16=99

Answer

99 99


Do you think you will be able to solve it?

examples with solutions for elimination of parentheses in real numbers

Exercise #1

What is the opposite number of 5 5

Video Solution

Answer

βˆ’5 -5

Exercise #2

What is the opposite number of 87 87

Video Solution

Answer

βˆ’87 -87

Exercise #3

What is the opposite number of 0.7 0.7

Video Solution

Answer

βˆ’0.7 -0.7

Exercise #4

What is the opposite number of βˆ’7 -7

Video Solution

Answer

7 7

Exercise #5

What is the opposite number of βˆ’0.25 -0.25

Video Solution

Answer

0.25 0.25

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