$\begin{cases}\dfrac{2}{3}x+5(x+y)=3\\x-\dfrac{3}{4}y=2(y-2x-5)\end{cases}$
$⇔\begin{cases}\dfrac{2}{3}x+5x+5y=3\\x-\dfrac{3}{4}y=2y-4x-10\end{cases}$
$⇔\begin{cases}\dfrac{17}{3}x+5y=3\\x-\dfrac{3}{4}y-2y+4x=-10\end{cases}$
$⇔\begin{cases}\dfrac{17}{3}x+5y=3\\5x-\dfrac{11}{4}y=-10\end{cases}$
$⇔\begin{cases}\dfrac{85}{3}x+25y=15\\\dfrac{85}{3}x-\dfrac{187}{12}y=\dfrac{-170}{3}\end{cases}$
$⇔\begin{cases}\dfrac{487}{12}y=\dfrac{215}{3}\\\dfrac{85}{3}x+25y=15\end{cases}$
$⇔\begin{cases}y=\dfrac{860}{487}\\\dfrac{85}{3}x+25\cdot\dfrac{860}{487}=15\end{cases}$
$⇔\begin{cases}y=\dfrac{860}{487}\\\dfrac{85}{3}x=\dfrac{-14195}{487}\end{cases}$
$⇔\begin{cases}y=\dfrac{860}{487}\\x=\dfrac{-501}{487}\end{cases}$
Vậy hệ phương trình có nghiệm duy nhất là $(x;y)=(\dfrac{-501}{487};\dfrac{860}{487})$.
