Question

# Divide the sum of $\dfrac{-4}{9}$ and $\dfrac{1}{3}$ by $\dfrac{5}{6}$.

Hint: In this question we have to divide the sum $\dfrac{-4}{9}$ and $\dfrac{1}{3}$ by $\dfrac{5}{6}$, for this we have to know that how to add two fractions.

We are going to add $\dfrac{-4}{9}$ and $\dfrac{1}{3}$, and to add these fractions we need to have same denominator, so for this we are going to find the L.C.M(least common multiple) of denominator,
i.e L.C.M(9,3) = 9.

Now we are going to add these fractions,
$\dfrac{-4}{9} +\dfrac{1}{3}$
Since, L.C.M is 9 so we have make 9 in the denominator of both fractions
= $\dfrac{-4}{9} +\dfrac{1\times 3}{3\times 3}$
= $\dfrac{-4}{9} +\dfrac{3}{9}$
= $\dfrac{\left( -4+3\right) }{9}$ [ since denominator is same so we can add the numerators easily]
= $\dfrac{\left( 3-4\right) }{9}$
= $\dfrac{-1}{9}$

Now we have to divide the above result i.e $\dfrac{-1}{9}$ by $\dfrac{5}{6}$.
So we can write,
$\dfrac{-1}{9} \div \dfrac{5}{6}$
Since, as we know that $\dfrac{a}{b} \div \dfrac{c}{d} =\dfrac{a}{b} \times \dfrac{d}{c}$
So by this we can write, $\dfrac{-1}{9} \div \dfrac{5}{6}$ as
$\dfrac{-1}{9} \div \dfrac{5}{6} =\dfrac{-1}{9} \times \dfrac{6}{5}$=$\dfrac{-6}{45}$
Dividing numerator and denominator by 3 we get,
= $\dfrac{-2}{15}$
So $\dfrac{-2}{25}$ is our required solution.

Note: In this type of question er generally forgot to take L.C.M of denominators while adding so keep it in mind and also the formula that we have used earlier i.e $\dfrac{a}{b} \div \dfrac{c}{d} =\dfrac{a}{b} \times \dfrac{d}{c}$.