`20xy+8x-15y=0`
`⇔5y(4x-3)+8x=0`
`⇔5y(4x-3)+8x-6=-6`
`⇔5y(4x-3)+2(4x-3)=-6`
`⇔(4x-3)(5y+2)=-6`
\(⇔\left[ \begin{array}{l}\left\{ \begin{array}{l}4x-3=-6\\ 5y+2=1 \end{array} \right.\\\left\{ \begin{array}{l} 4x-3=6\\ 5y+2=-1 \end{array} \right.\end{array} \right.\)
\(⇔\left[ \begin{array}{l}\left\{ \begin{array}{l} x=\dfrac{-3}{4}(L)\\y= \dfrac{-1}{5}(L)\end{array} \right.\\\left\{ \begin{array}{l} x=\dfrac{9}{4}(TM)\\y= \dfrac{-3}{5}(L)\end{array} \right.\end{array} \right.\)
Vậy pt vô nghiệm
