解$:(3)$方程组可变形为$\begin{cases}{\dfrac 35a_1x+\dfrac 25b_1y=c_1}\\{\dfrac 35a_2x+\dfrac 25b_2y=c_2}\end{cases}$
∵方程组$\begin{cases}{a_1x+b_1y=c_1}\\{a_2x+b_2y=c_2}\end{cases}$的解为$\begin{cases}{x=3}\\{y=4}\end{cases}$
∴所求方程组的解满足$\begin{cases}{\dfrac 35x=3}\\{\dfrac 25y=4}\end{cases}$
∴方程组的解为$\begin{cases}{x=5}\\{y=10}\end{cases}$