Add the following rational numbers :

(i) $\dfrac{-2}{3}$ and $\dfrac{3}{4}$

(ii) $\dfrac{-4}{9}$ and $\dfrac{5}{6}$

(iii) $\dfrac{-5}{18}$ and $\dfrac{11}{27}$

(iv) $\dfrac{-7}{12}$ and $\dfrac{-5}{24}$

(v) $\dfrac{-1}{18}$ and $\dfrac{-7}{27}$

(vi) $\dfrac{21}{-4}$ and $\dfrac{-11}{8}$

(i) $\dfrac{-2}{3}$ and $\dfrac{3}{4}$

We have:

$\dfrac{-2}{3}$ + $\dfrac{3}{4}$

Let us find L.C.M. of denominators 3 and 4

$\begin{array}{r|l} 2 & 3, 4 \ \hline 2 & 3, 2 \ \hline 3 & 3, 1 \ \hline & 1, 1 \end{array}$

L.C.M. = 2 x 2 x 3 = 12

Now, expressing each fraction with denominator 12:

$\dfrac{-2}{3} = \dfrac{-2 \times 4}{3 \times 4} = \dfrac{-8}{12} \\[1em] \dfrac{3}{4} = \dfrac{3 \times 3}{4 \times 3} = \dfrac{9}{12} \\[1em] \therefore \dfrac{-2}{3} + \dfrac{3}{4} = \dfrac{-8}{12} + \dfrac{9}{12} \\[1em] = \dfrac{(-8) + 9}{12} = \dfrac{1}{12}$

Hence, the answer is $\dfrac{1}{12}$

(ii) $\dfrac{-4}{9}$ and $\dfrac{5}{6}$

We have:

$\dfrac{-4}{9}$ + $\dfrac{5}{6}$

Let us find L.C.M. of denominators 9 and 6

$\begin{array}{r|l} 3 & 9, 6 \ \hline 3 & 3, 2 \ \hline 2 & 1, 2 \ \hline & 1, 1 \end{array}$

L.C.M. = 3 x 3 x 2 = 18

Now, expressing each fraction with denominator 18:

$\dfrac{-4}{9} = \dfrac{-4 \times 2}{9 \times 2} = \dfrac{-8}{18} \\[1em] \dfrac{5}{6} = \dfrac{5 \times 3}{6 \times 3} = \dfrac{15}{18} \\[1em] \therefore \dfrac{-4}{9} + \dfrac{5}{6} = \dfrac{-8}{18} + \dfrac{15}{18} \\[1em] = \dfrac{(-8) + 15}{18} = \dfrac{7}{18}$

Hence, the answer is $\dfrac{7}{18}$

(iii) $\dfrac{-5}{18}$ and $\dfrac{11}{27}$

We have:

$\dfrac{-5}{18}$ + $\dfrac{11}{27}$

Let us find LCM of denominators 18 and 27

$\begin{array}{r|l} 3 & 18, 27 \ \hline 3 & 6, 9 \ \hline 3 & 2, 3 \ \hline 2 & 2, 1 \ \hline & 1, 1 \end{array}$

L.C.M. = 3 x 3 x 3 x 2 = 54

Now, expressing each fraction with denominator 54:

$\dfrac{-5}{18} = \dfrac{-5 \times 3}{18 \times 3} = \dfrac{-15}{54} \\[1em] \dfrac{11}{27} = \dfrac{11 \times 2}{27 \times 2} = \dfrac{22}{54} \\[1em] \therefore \dfrac{-5}{18} + \dfrac{11}{27} = \dfrac{-15}{54} + \dfrac{22}{54} \\[1em] = \dfrac{(-15) + 22}{54} = \dfrac{7}{54}$

Hence, the answer is $\dfrac{7}{54}$

(iv) $\dfrac{-7}{12}$ and $\dfrac{-5}{24}$

We have:

$\dfrac{-7}{12}$ + $\dfrac{-5}{24}$

Let us find LCM of denominators 12 and 24

$\begin{array}{r|l} 2 & 12, 24 \ \hline 2 & 6, 12 \ \hline 3 & 3, 6 \ \hline 2 & 1, 2 \ \hline & 1, 1 \end{array}$

L.C.M. = 2 x 2 x 3 x 2 = 24

Now, expressing each fraction with denominator 24:

$\dfrac{-7}{12} = \dfrac{-7 \times 2}{12 \times 2} = \dfrac{-14}{24} \\[1em] \dfrac{-5}{24} = \dfrac{-5 \times 1}{24 \times 1} = \dfrac{-5}{24} \\[1em] \therefore \dfrac{-7}{12} + \dfrac{-5}{24} = \dfrac{-14}{24} + \dfrac{-5}{24} \\[1em] = \dfrac{(-14) + (-5)}{24} = \dfrac{-19}{24}$

Hence, the answer is $\dfrac{-19}{24}$

(v) $\dfrac{-1}{18}$ and $\dfrac{-7}{27}$

We have:

$\dfrac{-1}{18}$ + $\dfrac{-7}{27}$

LCM of denominators 18 and 27

$\begin{array}{r|l} 3 & 18, 27 \ \hline 3 & 6, 9 \ \hline 3 & 2, 3 \ \hline 2 & 2, 1 \ \hline & 1, 1 \end{array}$

L.C.M. = 3 x 3 x 3 x 2 = 54

Now, expressing each fraction with denominator 54

$\dfrac{-1}{18} = \dfrac{-1 \times 3}{18 \times 3} = \dfrac{-3}{54} \\[1em] \dfrac{-7}{27} = \dfrac{-7 \times 2}{27 \times 2} = \dfrac{-14}{54} \\[1em] \therefore \dfrac{-1}{18} + \dfrac{-7}{27} = \dfrac{-3}{54} + \dfrac{-14}{54} \\[1em] = \dfrac{(-3) + (-14)}{54} = \dfrac{-17}{54}$.

Hence, the answer is $\dfrac{-17}{54}$

(vi) $\dfrac{21}{-4}$ and $\dfrac{-11}{8}$

We have:

$\dfrac{21}{-4}$ + $\dfrac{-11}{8}$

First, multiply the numerator and denominator of $\dfrac{21}{-4}$ by -1 to make denominator positive:

$\dfrac{21 \times (-1)}{-4 \times (-1)} = \dfrac{-21}{4}$.

LCM of denominators 4 and 8

$\begin{array}{r|l} 2 & 4, 8 \ \hline 2 & 2, 4 \ \hline 2 & 1, 2 \ \hline & 1, 1 \end{array}$

L.C.M. = 2 x 2 x 2 = 8

Now, expressing each fraction with denominator 8:

$\dfrac{-21}{4} = \dfrac{-21 \times 2}{4 \times 2} = \dfrac{-42}{8} \\[1em] \dfrac{-11}{8} = \dfrac{-11 \times 1}{8 \times 1} = \dfrac{-11}{8} \\[1em] \therefore \dfrac{-21}{4} + \dfrac{-11}{8} = \dfrac{-42}{8} + \dfrac{-11}{8} \\[1em] = \dfrac{(-42) + (-11)}{8} = \dfrac{-53}{8}$

Hence, the answer is $\dfrac{-53}{8}$