Consider $$x$$ as the required rational number
Other number $$ = \dfrac{-7}{12} $$
Sum of two numbers $$ = \dfrac{3}{8} $$
Using the question
$$ \dfrac{-7}{12} + x = \dfrac{3}{8} $$
So we get
$$ x = \dfrac{3}{8} - \left (-\dfrac{7}{12} \right ) $$
LCM of $$8$$ and $$12$$ is $$24$$
$$ x = \dfrac{3 \times 3}{8 \times 3} + \dfrac{7 \times 2}{12 \times 2} $$
By further calculation
$$ = \dfrac{9}{24} + \dfrac{14}{24} $$
So we get
$$ = \dfrac{9 + 14}{24} = \dfrac{23}{24} $$