\(3x=4y+2\Rightarrow x=\frac{4y+2}{3}\)
Thay vào pt trên:
\(\left(\frac{4y+2}{3}\right)^2+y^2+4\left(\frac{4y+2}{3}\right)+4y-1=0\)
\(\Leftrightarrow16y^2+16y+4+9y^2+48y+24+36y-9=0\)
\(\Leftrightarrow25y^2+100y+19=0\Leftrightarrow\left[{}\begin{matrix}y=-\frac{1}{5}\Rightarrow x=\frac{2}{5}\\y=-\frac{19}{5}\Rightarrow x=-\frac{22}{5}\end{matrix}\right.\)