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Solve:
(a) $ \frac{1}{18}+\frac{1}{18} $
(b) $ \frac{8}{15}+\frac{3}{15} $
(c) $ \frac{7}{7}-\frac{5}{7} $
(d) $ \frac{1}{22}+\frac{21}{22} $
(e) $ \frac{12}{15}-\frac{7}{15} $
(f) $ \frac{5}{8}+\frac{3}{8} $
(g) $ 1-\frac{2}{3}\left(1=\frac{3}{3}\right) $
(h) $ \frac{1}{4}+\frac{0}{4} $
(i) $ 3-\frac{12}{5} $
To do :
We have to solve the given expressions.
Solution :
(a) $\frac{1}{18} + \frac{1}{18} =\frac{1+1}{18}$
$= \frac{2}{18}$
$=\frac{1}{9}$
The value of $\frac{1}{18} + \frac{1}{18}$ is $\frac{1}{9}$
(b) $\frac{8}{15} + \frac{3}{15} =\frac{8+3}{15}$
$= \frac{11}{15}$
The value of $\frac{8}{15} + \frac{3}{15}$ is $\frac{11}{15}$.
(c) $\frac{7}{7} - \frac{5}{7} =\frac{7-5}{7}$
$= \frac{2}{7}$
The value of $\frac{7}{7} - \frac{5}{7}$ is $\frac{2}{7}$.
(d) $\frac{1}{22} + \frac{21}{22} =\frac{1+21}{22}$
$= \frac{22}{22}$
$=1$
The value of $\frac{1}{22} + \frac{21}{22}$ is $1$.
(e) $\frac{12}{15} - \frac{7}{15} =\frac{12-7}{15}$
$= \frac{5}{15}$
$=\frac{1}{3}$
The value of $\frac{12}{15} - \frac{7}{15}$ is $\frac{1}{3}$.
(f) $\frac{5}{8} + \frac{3}{8} =\frac{5+3}{8}$
$= \frac{8}{8}$
$=1$
The value of $\frac{5}{8} + \frac{3}{8}$ is $1$.
(g) We can write $1$ as,
$1=\frac{3}{3}$
Therefore,
$1- \frac{2}{3}=\frac{3}{3}-\frac{2}{3}$
$=\frac{3-2}{3}$
$= \frac{1}{3}$
The value of $1-\frac{2}{3}$ is $\frac{1}{3}$.
(h) $\frac{1}{4} + \frac{0}{4} =\frac{1+0}{4}$
$= \frac{1}{4}$
The value of $\frac{1}{4} + \frac{0}{4}$ is $\frac{1}{4}$.
(i) We can write 3 as,
$3=\frac{5\times3}{5}$
$=\frac{15}{5}$
Therefore,
$3- \frac{12}{5} =\frac{15}{5}-\frac{12}{5}$
$=\frac{15-12}{5}$
$= \frac{3}{5}$
The value of $3- \frac{12}{5}$ is $\frac{3}{5}$.