Evaluate :

(i) $\dfrac{-3}{8} + \dfrac{5}{8} + \dfrac{7}{8}$

(ii) $\dfrac{11}{3} + \dfrac{-5}{3} + \dfrac{-2}{3}$

(iii) $-1 + \dfrac{2}{-3} + \dfrac{5}{6}$

(iv) $\dfrac{7}{26} + \dfrac{-11}{13} + 2$

(v) $3 + \dfrac{-7}{8} + \dfrac{-3}{4}$

(vi) $\dfrac{-13}{8} + \dfrac{7}{16} + \dfrac{-3}{4}$

(i) $\dfrac{-3}{8} + \dfrac{5}{8} + \dfrac{7}{8}$

Since the denominators are already the same and positive, we simply add the numerators.

$\dfrac{-3 + 5 + 7}{8} \\[1em] = \dfrac{2 + 7}{8} \\[1em] = \dfrac{9}{8}$

Hence, the answer is $\dfrac{9}{8}$

(ii) $\dfrac{11}{3} + \dfrac{-5}{3} + \dfrac{-2}{3}$

Since the denominators are already the same and positive, we simply add the numerators.

$\dfrac{11 + (-5) + (-2)}{3} \\[1em] = \dfrac{6 + (-2)}{3} \\[1em] = \dfrac{4}{3}$

Hence, the answer is $\dfrac{4}{3}$

(iii) $-1 + \dfrac{2}{-3} + \dfrac{5}{6}$

Express numbers as positive denominators: $\dfrac{-1}{1} + \dfrac{-2}{3} + \dfrac{5}{6}$.

LCM of denominators = LCM (1, 3, 6):

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

LCM = 3 x 2 = 6.

Now, expressing each fraction with denominator 6:

$\dfrac{-1 \times 6}{1 \times 6} = \dfrac{-6}{6} \\[1em] \dfrac{-2 \times 2}{3 \times 2} = \dfrac{-4}{6} \\[1em] \dfrac{5 \times 1}{6 \times 1} = \dfrac{5}{6} \\[1em] \Rightarrow \dfrac{-6}{6} + \dfrac{-4}{6} + \dfrac{5}{6} \\[1em] \Rightarrow \dfrac{-6 + (-4) + 5}{6} \\[1em] \Rightarrow \dfrac{-10 + 5}{6} \\[1em] \Rightarrow \dfrac{-5}{6}$

Hence, the answer is $\dfrac{-5}{6}$

(iv) $\dfrac{7}{26} + \dfrac{-11}{13} + 2$

LCM of denominators = LCM (26, 13, 1):

$\begin{array}{r|l} 13 & 26, 13, 1 \ \hline 2 & 2, 1, 1 \ \hline & 1, 1, 1 \end{array}$

LCM = 13 x 2 = 26.

Now, expressing each fraction with denominator 26:

$\dfrac{7 \times 1}{26 \times 1} = \dfrac{7}{26} \\[1em] \dfrac{-11 \times 2}{13 \times 2} = \dfrac{-22}{26} \\[1em] \dfrac{2 \times 26}{1 \times 26} = \dfrac{52}{26} \\[1em] \Rightarrow \dfrac{7}{26} + \dfrac{-22}{26} + \dfrac{52}{26} \\[1em] \Rightarrow \dfrac{7 + (-22) + 52}{26} \\[1em] \Rightarrow \dfrac{-15 + 52}{26} \\[1em] \Rightarrow \dfrac{37}{26}$

Hence, the answer is $\dfrac{37}{26}$

(v) $3 + \dfrac{-7}{8} + \dfrac{-3}{4}$

LCM of denominators = LCM (1, 8, 4):

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

LCM = 2 x 2 x 2 = 8.

Now, expressing each fraction with denominator 8:

$\dfrac{3 \times 8}{1 \times 8} = \dfrac{24}{8} \\[1em] \dfrac{-7 \times 1}{8 \times 1} = \dfrac{-7}{8} \\[1em] \dfrac{-3 \times 2}{4 \times 2} = \dfrac{-6}{8} \\[1em] \Rightarrow \dfrac{24}{8} + \dfrac{-7}{8} + \dfrac{-6}{8} \\[1em] \Rightarrow \dfrac{24 + (-7) + (-6)}{8} \\[1em] \Rightarrow \dfrac{17 + (-6)}{8} \\[1em] \Rightarrow \dfrac{11}{8}$

Hence, the answer is $\dfrac{11}{8}$

(vi) $\dfrac{-13}{8} + \dfrac{7}{16} + \dfrac{-3}{4}$

LCM of denominators = LCM (8, 16, 4):

$\begin{array}{r|l} 2 & 8, 16, 4 \ \hline 2 & 4, 8, 2 \ \hline 2 & 2, 4, 1 \ \hline 2 & 1, 2, 1 \ \hline & 1, 1, 1 \end{array}$

LCM = 2 x 2 x 2 x 2 = 16.

Now, expressing each fraction with denominator 16:

$\dfrac{-13 \times 2}{8 \times 2} = \dfrac{-26}{16} \\[1em] \dfrac{7 \times 1}{16 \times 1} = \dfrac{7}{16} \\[1em] \dfrac{-3 \times 4}{4 \times 4} = \dfrac{-12}{16} \\[1em] \Rightarrow \dfrac{-26}{16} + \dfrac{7}{16} + \dfrac{-12}{16} \\[1em] \Rightarrow \dfrac{-26 + 7 + (-12)}{16} \\[1em] \Rightarrow \dfrac{-19 + (-12)}{16} \\[1em] \Rightarrow \dfrac{-31}{16}$.

Hence, the answer is $\dfrac{-31}{16}$