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Question

Work out the following :
$$(i)\ \dfrac{2}{3} + \dfrac{3}{4}$$
$$(ii)\ \dfrac{5}{7} - \dfrac{4}{9} $$
$$(iii)\ \dfrac{1}{2} + \dfrac{3}{5} $$
$$(iv)\ 1 \dfrac{4}{9} + 3 \dfrac{5}{12}$$

Solution
Verified by Toppr

$$(i)\ \dfrac{2}{3} + \dfrac{3}{4}$$ $$(L.C.M$$ of $$7 \, $$and $$9 = 63)$$
$$= \dfrac{\left(2 \times 4\right) + \left(3 \times 3\right)}{12} = \dfrac{8 + 9}{12}$$
$$= \dfrac{17}{12} = 1 \dfrac{5}{12}$$

$$(ii)\ \dfrac{5}{7} - \dfrac{4}{9}$$ $$(L.C.M$$ of$$ \, 7\, $$and$$ \, 9 = 63)$$
$$= \dfrac{\left(5 \times 9 \right) -\left(7 \times 4\right)}{63} = \dfrac{45 - 28}{63} = \dfrac{17}{63}$$

$$(iii)\ \dfrac{1}{2} + \dfrac{3}{5}$$ $$(L.C.M.$$ of$$ \,2 $$ and$$ \, 5 = 10)$$
$$= \dfrac{\left(1 \times 5\right) + \left(3 \times 2 \right)}{10} = \dfrac{5 + 6}{10}=\dfrac{11}{10} = 1 \dfrac{1}{10}$$

$$(iv)\ 1 \dfrac{4}{9} + 3 \dfrac{5}{12}$$
$$\Rightarrow \dfrac{13}{9} + \dfrac{41}{12}$$ $$(L.C.M. \, $$of$$ \, 9 \,$$and$$ \, 12 = 36)$$
$$= \dfrac{\left(13 \times 4\right) + \left(41 \times 3\right)}{36} = \dfrac{52 + 123}{36}$$
$$= \dfrac{175}{36} = 4 \dfrac{31}{36}$$

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