For the system to have infinitely many solutions, the second equation must be a constant multiple of the first. Comparing the coefficients of $x$, the second equation must be $\frac{2}{5}$ times the first equation. Therefore, $q+r=\left(\frac{2}{5}\right)\times (7)+\left(\frac{2}{5}\right)\times(8)=6$. 
14. Consider the following system of equations: $$\begin{cases} 5 x+7 y=8 \textbackslash\textbackslash{} 2 x+q y=r \end{cases}$$, where $q$ and $r$ are constants. If $q$ and $r$ are chosen such that this system has infinitely many solutions $(x, y)$, find $q+r=$~\uline{~~~~~~~~~~}~. 
Options:
A: $$6$$ 
B: $$8$$ 
C: $$9$$ 
D: $$12$$ 