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

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