Evaluate :

(i) $\dfrac{2}{-3} + \dfrac{-4}{9}$

(ii) $\dfrac{-1}{2} + \dfrac{-3}{4}$

(iii) $\dfrac{7}{-9} + \dfrac{-5}{6}$

(iv) $2 + \dfrac{-3}{4}$

(v) $3 + \dfrac{-5}{6}$

(vi) $-4 + \dfrac{2}{3}$

(i) $\dfrac{2}{-3} + \dfrac{-4}{9}$

First, express $\dfrac{2}{-3}$ with a positive denominator: $\dfrac{2 \times (-1)}{-3 \times (-1)} = \dfrac{-2}{3}$.

Let us find LCM of denominators 3 and 9

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

L.C.M. = 3 x 3 = 9

Now,

$\dfrac{-2}{3} = \dfrac{-2 \times 3}{3 \times 3} = \dfrac{-6}{9} \\[1em] \therefore \dfrac{-6}{9} + \dfrac{-4}{9} = \dfrac{(-6) + (-4)}{9} \\[1em] = \dfrac{-10}{9}$

Hence, the answer is $\dfrac{-10}{9}$

(ii) $\dfrac{-1}{2} + \dfrac{-3}{4}$

Let us find LCM of denominators 2 and 4.

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

L.C.M. = 2 x 2 = 4

Now,

$\dfrac{-1}{2} = \dfrac{-1 \times 2}{2 \times 2} = \dfrac{-2}{4} \\[1em] \therefore \dfrac{-2}{4} + \dfrac{-3}{4} = \dfrac{(-2) + (-3)}{4} \\[1em] = \dfrac{-5}{4}$

Hence, the answer is $\dfrac{-5}{4}$

(iii) $\dfrac{7}{-9} + \dfrac{-5}{6}$

First, express $\dfrac{7}{-9}$ with a positive denominator: $\dfrac{7 \times -1}{-9 \times -1} = \dfrac{-7}{9}$.

Let us find LCM 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:

$\dfrac{-7}{9} = \dfrac{-7 \times 2}{9 \times 2} = \dfrac{-14}{18} \\[1em] \dfrac{-5}{6} = \dfrac{-5 \times 3}{6 \times 3} = \dfrac{-15}{18} \\[1em] \therefore \dfrac{-14}{18} + \dfrac{-15}{18} \\[1em] = \dfrac{(-14) + (-15)}{18} \\[1em] = \dfrac{-29}{18}$

Hence, the answer is $\dfrac{-29}{18}$

(iv) $2 + \dfrac{-3}{4}$

Express 2 as $\dfrac{2}{1}$.

LCM of denominators 1 and 4 is 4.

Now,

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

Hence, the answer is $\dfrac{5}{4}$

(v) $3 + \dfrac{-5}{6}$

Express 3 as $\dfrac{3}{1}$.

LCM of denominators 1 and 6 is 6.

Now,

$\dfrac{3}{1} = \dfrac{3 \times 6}{1 \times 6} = \dfrac{18}{6} \\[1em] \therefore \dfrac{18}{6} + \dfrac{-5}{6} \\[1em] = \dfrac{18 + (-5)}{6} \\[1em] = \dfrac{13}{6}$

Hence, the answer is $\dfrac{13}{6}$

(vi) $-4 + \dfrac{2}{3}$

Express -4 as $\dfrac{-4}{1}$.

LCM of denominators 1 and 3 is 3.

Now,

$\dfrac{-4}{1} = \dfrac{-4 \times 3}{1 \times 3} = \dfrac{-12}{3} \\[1em] \therefore \dfrac{-12}{3} + \dfrac{2}{3} \\[1em] = \dfrac{(-12) + 2}{3} \\[1em] = \dfrac{-10}{3}$

Hence, the answer is $\dfrac{-10}{3}$