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Answer 1

$\large\underline{\bold{Question - }}$

$\sf \:What \: is \: the \: additive \: inverse \: of \: \dfrac{5}{13}$

$\large\underline{\bold{Solution-}}$

Basic Concept :- An additive inverse of a number is defined as the value, which on adding with the given number results in zero value. It is the value we add to a number to give resultant zero.

Let, a is any real number, then its additive inverse will be - a so that, a + (-a) = a – a = 0.

$\sf \: Hence \: additive \: inverse \: of \: \dfrac{5}{13} = \bf \: - \: \dfrac{5}{13}$

$\sf \: Write \: the \: multiplicative \: inverse \: of \: - \: \dfrac{2}{3}$

$\large\underline{\bold{Solution-}}$

• Basic concept :- The multiplicative inverse of a number x (should be non - zero) is given by 1/x, such that when it is multiplied by given number, it results in value equal to 1. It is also called Reciprocal of a given real number.

• If p/q is a fraction, then the multiplicative inverse of p/q should be such that, when it is multiplied to the fraction, then the result should be 1. Hence, q/p is the multiplicative inverse of fraction p/q.

$\sf \: Hence \:multiplicative \: inverse \: of \: - \: \dfrac{2}{3} = \bf \: - \: \dfrac{3}{2}$

$\bf \: Find \: the \: value \: of \:$

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

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

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

So, LCM of 5 and 4 is 20

• As we know that,

• To find the operation of additions or subtraction in rational numbers, we will divide the LCM by denominator and then we will multiply the outcome with numerator.

So,

$= \: \sf \: \dfrac{2 \times 4 + 3 \times 5}{20}$

$= \: \sf \: \dfrac{8 + 15}{20}$

$= \: \sf \: \dfrac{23}{20}$

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

Firstly, We have to find the LCM of 7 and 3

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{3}}}&{\underline{\sf{\:\:7\:, \: 3 \:\:}}}\\ {\underline{\sf{7}}}& \underline{\sf{\:\:7\:, \: 1 \:\:}} \\{\sf{}}&\underline{\sf{\:\:1\:, \: 1 \:\:}}\end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

So, LCM of 3 and 7 is 21

As we know that,

To find the operations of addition or subtraction of rational numbers, we will divide the LCM by enominator and then we will multiply the outcome with numerator.

So,

$= \: \sf \: \dfrac{ - 2 \times 3 + 4 \times 7}{21}$

$= \: \sf \: \dfrac{ - 6 + 28}{21}$

$= \: \sf \: \dfrac{22}{21}$

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

Firstly, We have to find the LCM of 3 and 11.

$★\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{3}}}&{\underline{\sf{\:\:11\:, \: 3 \:\:}}}\\ {\underline{\sf{11}}}& \underline{\sf{\:\:11\:, \: 1 \:\:}} \\{\sf{}}&\underline{\sf{\:\:1\:, \: 1 \:\:}}\end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

• So, LCM of 11 and 3 is 33.

• As we know that,

To find the operations of addition or subtraction of rational numbers, we will divide the LCM by enominator and then we will multiply the outcome with numerator.

$= \: \sf \: \dfrac{4 \times 11 - 2 \times 11 + 3 \times 3}{33}$

$= \: \sf \: \dfrac{44 - 22 + 9}{33}$

$= \: \sf \: \dfrac{22 + 9}{33}$

$= \: \sf \: \dfrac{31}{33}$