Question

Simplify:

(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) $\frac{4}{5}+\frac{2}{3}$ (f) $\frac{3}{4}-\frac{1}{3}$ (g) $\frac{5}{6}-\frac{1}{3}$

Answer 1

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

Now, LCM of $3$ and $7$ is $21$. [LCM of co-prime numbers is their product]

$=\frac{2\times 7}{3\times 7}+\frac{1\times 3}{7\times 3}=\frac{14}{21}+\frac{3}{21}$

$=\frac{17}{21}$

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

We will first find the LCM of the given denominators.

$\begin{array}{cc}2& 10,15\\ 3& 5,15\\ 5& 5,5\\ & 1\end{array}$

LCM of $10$ and $15$=$2\times 3\times 5=30$

$=\frac{3\times 15}{10\times 15}+\frac{7\times 10}{15\times 10}$

$=\frac{45}{150}+\frac{70}{150}$

$=\frac{45+70}{150}$

$=\frac{115}{150}$

$=\frac{23}{30}$

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

Since, $9$ and $7$ are co prime numbers, therefore, their LCM will be $9\times 7=63$

$=\frac{4\times 7}{9\times 7}+\frac{2\times 9}{7\times 9}$

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

$=\frac{46}{63}$

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

Since, $7$ and $3$ are co prime numbers, their LCM will be their product ie. $21$

$=\frac{5\times 3}{7\times 3}+\frac{1\times 7}{3\times 7}$

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

$=\frac{22}{21}$

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

Since, $5$ and $3$ are are co prime numbers their LCM will be their product ie. $15$

$=\frac{4\times 3}{5\times 3}+\frac{2\times 5}{3\times 5}$

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

$=\frac{22}{15}$

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

Since, $4$ and $3$ are co prime numbers their LCM will be their product ie. $12$.

$=\frac{3\times 3}{4\times 3}-\frac{1\times 4}{3\times 4}$

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

$=\frac{5}{12}$

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

Since, $6$ is the multiple of $3$ their LCM will be $6$

$=\frac{5}{6}-\frac{1\times 2}{3\times 2}$

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

$=\frac{3}{6}$

$=\frac{1}{2}$