If $\theta$ lies in the first quadrant and $5\;\tan \theta=4$, then $\large\frac{5 \sin \theta -3 \cos \theta}{\sin \theta+ 2 \cos \theta}$ is equal to

Question

$$\begin {array} {1 1} (a)\;\frac{5}{14} & \quad (b)\;\frac{3}{14} \\ (c)\;\frac{1}{14} & \quad (d)\;0 \end {array}$$