$\begin{array}{l}
\dfrac{{\sin x - \cos x}}{{2\sin x + 6\cos x}}\\
= \dfrac{{\dfrac{{\sin x}}{{\cos x}} - 1}}{{2\dfrac{{\sin x}}{{\cos x}} + 6}} = \dfrac{{\tan x - 1}}{{2\tan x + 6}} = \dfrac{{5 - 1}}{{2.5 + 6}}\\
= \dfrac{4}{{16}} = \dfrac{1}{4}
\end{array}$