Question

Subtract the sum of $\frac{-36}{11}$ and $\frac{49}{22}$ from the sum of $\frac{33}{8}$ and $\frac{-19}{4}$.

Answer 1

$\begin{array}{l}\\ \text{First we will find the sum of}\frac{-36}{11}\text{and}\frac{49}{22}\text{.}\\ \\ \frac{-36}{11}+\frac{49}{22}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}11\mathrm{and}22\mathrm{is}22.\\ =\frac{-72+49}{22}\\ \\ =\frac{-23}{22}\\ \\ \text{Now. we have to find the sum of}\frac{33}{8}\mathrm{and}\frac{-19}{4}.\\ \frac{33}{8}+\frac{(-19)}{4}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}8\mathrm{and}4\mathrm{is}8.\\ \begin{array}{l}=\frac{33-38}{8}\\ =\frac{-5}{8}\end{array}\\ \\ \mathrm{Now},\frac{(-5)}{8}-\frac{(-23)}{22}\\ =\frac{-5}{8}+\frac{23}{22}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}8\mathrm{and}22\mathrm{is}88.\\ \begin{array}{l}=\frac{-55+92}{88}\\ =\frac{37}{88}\end{array}\end{array}$