3a=5b=6c

3a=5b=6c,  then

$$\dfrac{(a+b)}{(b-c)}\ =\ ?$$  

$$\small{\text{ $ \begin{array}{lcl} &3a=5b=6c \\ (1) & 3a=5b\\\\ (2) & 3a=6c & \quad |\quad a=\frac{6}{3}c = 2c\\\\ (3) & 5b=6c & \quad |\quad b=\frac{6}{5}c \end{array} $ }}\\\\\\ \small{\text{ $ \dfrac{(a+b)}{(b-c)} = \dfrac{ 2c +\frac{6}{5}c }{\frac{6}{5}c -c } = \dfrac{ c\left(2 +\dfrac{6}{5}\right) }{ c\left(\dfrac{6}{5} -1\right) } =\dfrac{ 2 +\dfrac{6}{5} }{ \dfrac{6}{5} -1 } =\dfrac{ \dfrac{16}{5} }{ \dfrac{1}{5} } =\dfrac{16}{5} * \dfrac{5}{1} = 16 $ }}$$

3a=6c     so    a=2c

5b=6c       so   b=6c/5

$$\\\frac{a+b}{b-c}=(a+b)\div(b-c)\\\\ =\left(2c+\frac{6c}{5}\right)\div\left(\frac{6c}{5}-c}\right)\\\\ =\left(\frac{10c+6c}{5}\right)\div\left(\frac{6c-5c}{5}}\right)\\\\ =\frac{16c}{5}\div\frac{c}{5}}\\\\ =\frac{16c}{5}\times\frac{5}{c}}\\\\\ =16$$