Question

Find the sum of $\frac{5}{6}$ and $\frac{11}{13}$, $\frac{4}{7}$ and $\frac{-3}{5}$, $\frac{-7}{4}$ and $\frac{-3}{7}$:

Answer 1

$\frac{5}{6}+\frac{11}{13}=\frac{(5\times 13)+(11\times 6)}{(6\times 13)}$
$=\frac{65+66}{78}$
$=\frac{131}{78}$
Similarly, the sum of $\frac{4}{7}$ and $\frac{-3}{5}$ is:
$\frac{4}{7}+\frac{-3}{5}=\frac{(4\times 5)+\left(\right(-3)\times 7)}{7\times 5}$
$=\frac{20+(-21)}{35}$
$=\frac{-1}{35}$.
The sum of $\frac{-7}{4}$ and $\frac{-3}{7}$ is:
$\frac{-7}{4}+\frac{-3}{7}=\frac{(-7)\times 7+(-3)\times 4}{4\times 7}$
$=\frac{(-49)+(-12)}{28}$
$=\frac{-61}{28}$.