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Question

Find the difference
$$7-\left[4\left(\cfrac{2}{3}\right)\right]$$

Solution
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$$7-\left[4\left(\cfrac{2}{3}\right)\right]$$
We have to covert mixed fraction in improper fraction(a fraction in which numerator is greater than denominator).
$$\therefore 4\left(\cfrac{2}{3}\right)=\cfrac{4\times 3+2}{3}=\cfrac{14}{3}$$
$$\Rightarrow 7-\cfrac{14}{3}$$
$$7$$ can also be written as $$\cfrac{7}{1}$$
$$\Rightarrow \cfrac71-\cfrac{14}{3}$$
To subtract rational numbers their denominator should be same, so we can simply subtract their numerator and denominator will remain common.
$$LCM$$ of $$1$$ and $$3$$ is $$3$$
So, let's make numbers with same denominator equal to their $$LCM$$
$$\cfrac{7}1=\cfrac{(7\times (3))}{(1\times (3))}=\cfrac{21}{3}$$
So,
$$\cfrac71-\cfrac{14}{3}=\cfrac{21}{3}-\cfrac{14}{3}$$
$$=\cfrac{21-14}{3}$$
$$=\cfrac{7}{3}$$
$$=\cfrac{6+1}{3}$$
$$=2+\cfrac13$$
$$=2\left(\cfrac13\right)$$

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