Solve
(a) $ \frac{2}{3}+\frac{1}{7} $
(b) $ \frac{3}{10}+\frac{7}{15} $
(c) $ \frac{4}{9}+\frac{2}{7} $
(d) $ \frac{5}{7}+\frac{1}{3} $
(e) $ \frac{2}{5}+\frac{1}{6} $
(f) $ \frac{4}{5}+\frac{2}{3} $
(g) $ \frac{3}{4}-\frac{1}{3} $
(h) $ \frac{5}{6}-\frac{1}{3} $
(i) $ \frac{2}{3}+\frac{3}{4}+\frac{1}{2} $
(j) $ \frac{1}{2}+\frac{1}{3}+\frac{1}{6} $
(k) $ 1 \frac{1}{3}+3 \frac{2}{3} $
(l) $ 4 \frac{2}{3}+3 \frac{1}{4} $
(m) $ \frac{16}{5}-\frac{7}{5} $
(n) $ \frac{4}{3}-\frac{1}{2} $

To do :

We have to solve the given expressions. 

Solution :

(a) $\frac{2}{3} + \frac{1}{7}$

LCM of denominators 3 and 7 is 21.

This implies,

$\frac{2}{3} + \frac{1}{7}=\frac{2\times7+1\times3}{21}$

$= \frac{14+3}{21}$

$=\frac{17}{21}$

The value of $\frac{2}{3} + \frac{1}{7}$ is $\frac{17}{21}$.

(b) $\frac{3}{10} + \frac{7}{15}$

LCM of denominators 10 and 15 is 30.

This implies,

$\frac{3}{10} + \frac{7}{15}=\frac{3\times3+7\times2}{30}$

$= \frac{9+14}{30}$

$=\frac{23}{30}$

The value of $\frac{3}{10} + \frac{7}{15}$ is $\frac{23}{30}$.

(c) $\frac{4}{9}+\frac{2}{7}$

LCM of denominators 9 and 7 is 63.

This implies,

$\frac{4}{9} + \frac{2}{7}=\frac{4\times7+2\times9}{63}$

$= \frac{28+18}{63}$

$=\frac{46}{63}$

The value of $\frac{4}{9} + \frac{2}{7}$ is $\frac{46}{63}$.

(d) $\frac{5}{7}+\frac{1}{3}$

LCM of denominators 7 and 3 is 21.

This implies,

$\frac{5}{7} + \frac{1}{3}=\frac{5\times3+1\times7}{21}$

$= \frac{15+7}{21}$

$=\frac{22}{21}$

The value of $\frac{5}{7} + \frac{1}{3}$ is $\frac{22}{21}$.

(e) $\frac{2}{5}+\frac{1}{6}$

LCM of denominators 5 and 6 is 30.

This implies,

$\frac{2}{5} + \frac{1}{6}=\frac{2\times6+1\times5}{30}$

$= \frac{12+5}{30}$

$=\frac{17}{30}$

The value of $\frac{2}{5} + \frac{1}{6}$ is $\frac{17}{30}$.

(f) $\frac{4}{5}+\frac{2}{3}$

LCM of denominators 5 and 3 is 15.

This implies,

$\frac{4}{5} + \frac{2}{3}=\frac{4\times3+2\times5}{15}$

$= \frac{12+10}{15}$

$=\frac{22}{15}$

The value of $\frac{4}{5} + \frac{2}{3}$ is $\frac{22}{15}$.

(g) $\frac{3}{4}-\frac{1}{3}$

LCM of denominators 4 and 3 is 12.

This implies,

$\frac{3}{4} - \frac{1}{3}=\frac{3\times3-1\times4}{12}$

$= \frac{9-4}{12}$

$=\frac{5}{12}$

The value of $\frac{3}{4} - \frac{1}{3}$ is $\frac{5}{12}$.

(h) $\frac{5}{6}-\frac{1}{3}$

LCM of denominators 6 and 3 is 6.

This implies,

$\frac{5}{6} - \frac{1}{3}=\frac{5\times1-1\times2}{6}$

$= \frac{5-2}{6}$

$=\frac{3}{6}$

$=\frac{1}{2}$

The value of $\frac{5}{6} - \frac{1}{3}$ is $\frac{1}{2}$.

(i) \( \frac{2}{3}+\frac{3}{4}+\frac{1}{2} \)

LCM of denominators 3, 4 and 2 is 12.

This implies,

$\frac{2}{3}+\frac{3}{4}+\frac{1}{2}=\frac{2\times4+3\times3+1\times6}{12}$

$= \frac{8+9+6}{12}$

$=\frac{23}{12}$

The value of $\frac{2}{3}+\frac{3}{4}+\frac{1}{2}$ is $\frac{23}{12}$.

(j) \( \frac{1}{2}+\frac{1}{3}+\frac{1}{6} \)

LCM of denominators 2, 3 and 6 is 6.

This implies,

$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1\times3+1\times2+1\times1}{6}$

$= \frac{3+2+1}{6}$

$=\frac{6}{6}$

$=1$

The value of $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ is $1$.

(k) \( 1 \frac{1}{3}+3 \frac{2}{3} \)

$1\frac{1}{3}+3\frac{2}{3}=\frac{1\times3+1}{3}+\frac{3\times3+2}{3}$

$= \frac{3+1}{3}+\frac{9+2}{3}$

$=\frac{4}{3}+\frac{11}{3}$

$=\frac{4+11}{3}$

$=\frac{15}{3}$

$=5$

The value of $1\frac{1}{3}+3\frac{2}{3}$ is $5$.

(l) \( 4 \frac{2}{3}+3 \frac{1}{4} \)

$4\frac{2}{3}+3\frac{1}{4}=\frac{4\times3+2}{3}+\frac{3\times4+1}{4}$

$= \frac{12+2}{3}+\frac{12+1}{4}$

$=\frac{14}{3}+\frac{13}{4}$

LCM of denominators 3 and 4 is 12.

Therefore,

$\frac{14}{3}+\frac{13}{4}=\frac{14\times4+13\times3}{12}$

$=\frac{56+39}{12}$

$=\frac{95}{12}$

The value of $4\frac{2}{3}+3\frac{1}{4}$ is $\frac{95}{12}$.

(m) \( \frac{16}{5}-\frac{7}{5} \)

$\frac{16}{5}-\frac{7}{5}=\frac{16-7}{5}$

$= \frac{9}{5}$

Therefore,

The value of $\frac{16}{5}-\frac{7}{5}$ is $\frac{9}{5}$.

(n) \( \frac{4}{3}-\frac{1}{2} \)

LCM of 3 and 2 is 6.

$\frac{4}{3}-\frac{1}{2}=\frac{4\times2-1\times3}{6}$

$= \frac{8-3}{6}$

$=\frac{5}{6}$

Therefore,

The value of $\frac{4}{3}-\frac{1}{2}$ is $\frac{5}{6}$.