So the required rational number is $$x$$
It can be written as
$$ = \dfrac{-3}{5}+x=2 $$
$$x=2+\dfrac{3}{5}$$
LCM of $$1$$ and $$5$$ is $$5$$
$$ = \dfrac{2 \times 5}{1 \times 5} + \dfrac{3 \times 1}{5 \times 1} $$
By further calculation
$$ = \dfrac{10 + 3}{5} $$
So we get
$$ = \dfrac{13}{5} $$
$$ = 2 \dfrac{3}{5} $$