Question

1. What is the additive inverse of $\sf\frac{5}{13}$

2. Write the multiplicative inverse of $\sf\frac{-2}{3}$

3. Which number has no reciprocal?

4. Find the value :-

$\sf \:(i) \: \frac{2}{5} \:+ \: \frac{3}{4}$

$\sf (ii) \: \frac{-2}{7} \: + \: \frac{4}{3}$

$\sf (iii) \: \frac{4}{3} \:-\: \frac{2}{3} \: + \: \frac{3}{11}$

Answer 1

Required Solution:-

1. What's the Additive inverse of 5/13

Ans: Hence, You know that Result of any additive inverse is 0.

So, $\sf\dfrac{5}{13}+\dfrac{-5}{13} = 0$

∴ Solution is $\sf\dfrac{-5}{13}$

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2. Write Multiplicative inverse of -2/3?

Ans: Multiplicative inverse of $\sf\dfrac{ - 2}{3} \: is \: \dfrac{3}{ - 2}$

It is so, Because other name of multiplicative inverse is reciprocal. Means when they are Multiply . They will get cancel to each other.

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3. Which number has no reciprocal?

Ans: Zero doesn't have any Reciprocal .

Because if we take ( x × 0 ) it will be answered as 1. This clearly Identified that it is impossible that 0 have reciprocal of 0 itself.

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4. Find the value :-

$\sf \:(i) \: \dfrac{2}{5} \:+ \: \dfrac{3}{4}$

Ans:

Firstly, We have to find the LCM of 5 and 4 is 20

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$\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:4\:, \: 5 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:2\:, \: 5 \:\:}} \\\underline{\sf{5}}&\underline{\sf{\:\:1\:, \: 5 \:\:}}\\\underline{\sf{}}&{\sf{\:\:1\:, \: 1 \:\:}} \end{array}\end{gathered}\end{gathered}$

LCM of 5 and 4 = 2 × 2 × 5 = ${\boxed{\frak{\pink{20}}}}$

As we know that: To find the sum of Additions we will divide the LCM with Denominators and then we will multiply the outcome with Numerator.

⠀⠀⠀⠀

$: \implies \sf \dfrac{2}{5} + \dfrac{3}{4} \\ \\ \\ : \implies \sf \dfrac{8 + 15}{20} \\ \\ \\ : \implies\sf\dfrac{23}{20}$

∴ Solution is 23/20

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$\sf (ii) \: \dfrac{-2}{7} \: + \: \dfrac{4}{3}$

Ans: LCM of 7 and 3 is 21.

$: \implies \sf\dfrac{ - 2}{7} + \dfrac{4}{3} \\ \\ \\ : \implies \sf \dfrac{ - 6 + 28}{21} \\ \\ \\ : \implies \sf\dfrac{ - 22}{21}$

∴ Hence, Solution is -22/21

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$\sf (iii) \: \dfrac{4}{3} \:-\: \dfrac{2}{3} \: + \: \dfrac{3}{11}$

Ans: LCM of 3 and 11 is 33

$: \implies\sf \: \dfrac{4}{3} \:-\: \dfrac{2}{3} \: + \: \dfrac{3}{11} \\ \\ \\ : \implies \sf \dfrac{4 - 2}{3} + \dfrac{3}{11} \\ \\ \\ : \implies \sf \dfrac{2}{3} + \dfrac{3}{11} \\ \\ \\ \dashrightarrow \sf \: LCM \: of \: 3 \: and \: 11 \: = 33 \\ \\ \\ : \implies \sf \dfrac{22 + 9}{33} \\ \\ \\ : \implies \sf \dfrac{31}{33}$

∴ Solution is 31/33

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Answer 2

Answer:

Required Solution:-

1. What's the Additive inverse of 5/13

Ans: Hence, You know that Result of any additive inverse is 0.

So, 513+−513=0\sf\dfrac{5}{13}+\dfrac{-5}{13} = 0

13

5

+

13

−5

=0

∴ Solution is −513\sf\dfrac{-5}{13}

13

−5

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2. Write Multiplicative inverse of -2/3?

Ans: Multiplicative inverse of −23is3−2\sf\dfrac{ - 2}{3} \: is \: \dfrac{3}{ - 2}

3

−2

is

−2

3

It is so, Because other name of multiplicative inverse is reciprocal. Means when they are Multiply . They will get cancel to each other.

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3. Which number has no reciprocal?

Ans: Zero doesn't have any Reciprocal .

Because if we take ( x × 0 ) it will be answered as 1. This clearly Identified that it is impossible that 0 have reciprocal of 0 itself.

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4. Find the value :-

(i)25+34\sf \:(i) \: \dfrac{2}{5} \:+ \: \dfrac{3}{4}(i)

5

2

+

4

3

Ans:

Firstly, We have to find the LCM of 5 and 4 is 20

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2‾4,5‾2‾2,5‾5‾1,5‾‾1,1\begin{lgathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:4\:, \: 5 \:\:}}}\\ {\underline{\sf{2}}}& \underline{\sf{\:\:2\:, \: 5 \:\:}} \\\underline{\sf{5}}&\underline{\sf{\:\:1\:, \: 5 \:\:}}\\\underline{\sf{}}&{\sf{\:\:1\:, \: 1 \:\:}} \end{array}\end{gathered}\end{gathered}\end{lgathered}

2

2

5

4,5

2,5

1,5

1,1

LCM of 5 and 4 = 2 × 2 × 5 = 20{\boxed{\frak{\pink{20}}}}

20

As we know that: To find the sum of Additions we will divide the LCM with Denominators and then we will multiply the outcome with Numerator.

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:⟹25+34:⟹8+1520:⟹2320\begin{lgathered}: \implies \sf \dfrac{2}{5} + \dfrac{3}{4} \\ \\ \\ : \implies \sf \dfrac{8 + 15}{20} \\ \\ \\ : \implies\sf\dfrac{23}{20}\end{lgathered}

:⟹

5

2

+

4

3

:⟹

20

8+15

:⟹

20

23

∴ Solution is 23/20

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(ii)−27+43\sf (ii) \: \dfrac{-2}{7} \: + \: \dfrac{4}{3}(ii)

7

−2

+

3

4

Ans: LCM of 7 and 3 is 21.

:⟹−27+43:⟹−6+2821:⟹−2221\begin{lgathered}: \implies \sf\dfrac{ - 2}{7} + \dfrac{4}{3} \\ \\ \\ : \implies \sf \dfrac{ - 6 + 28}{21} \\ \\ \\ : \implies \sf\dfrac{ - 22}{21}\end{lgathered}

:⟹

7

−2

+

3

4

:⟹

21

−6+28

:⟹

21

−22

∴ Hence, Solution is -22/21

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(iii)43−23+311\sf (iii) \: \dfrac{4}{3} \:-\: \dfrac{2}{3} \: + \: \dfrac{3}{11}(iii)

3

4

−

3

2

+

11

3

Ans: LCM of 3 and 11 is 33

:⟹43−23+311:⟹4−23+311:⟹23+311⇢LCMof3and11=33:⟹22+933:⟹3133\begin{lgathered}: \implies\sf \: \dfrac{4}{3} \:-\: \dfrac{2}{3} \: + \: \dfrac{3}{11} \\ \\ \\ : \implies \sf \dfrac{4 - 2}{3} + \dfrac{3}{11} \\ \\ \\ : \implies \sf \dfrac{2}{3} + \dfrac{3}{11} \\ \\ \\ \dashrightarrow \sf \: LCM \: of \: 3 \: and \: 11 \: = 33 \\ \\ \\ : \implies \sf \dfrac{22 + 9}{33} \\ \\ \\ : \implies \sf \dfrac{31}{33}\end{lgathered}

:⟹

3

4

−

3

2

+

11

3

:⟹

3

4−2

+

11

3

:⟹

3

2

+

11

3

⇢LCMof3and11=33

:⟹

33

22+9

:⟹

33

31

∴ Solution is 31/33

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