# 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 $\frac{3}{5}$.

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Updated on: 10-Oct-2022

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