\(\dfrac{2x}{3y}=-\dfrac{1}{3}\\ \Rightarrow3y=2x:-\dfrac{1}{3}=\dfrac{2x.3}{-1}=-6x\\ \Rightarrow y=-\dfrac{6x}{3}=-2x\)
Thế \(y=-2x\) vào \(2x+3y^2=\dfrac{161}{4}\) được:
\(2x+3.\left(-2x\right)^2=\dfrac{161}{4}\\ \Leftrightarrow2x+12x^2-\dfrac{161}{4}=0\\ \Leftrightarrow48x^2+8x-161=0\\ \Leftrightarrow\left(48x^2+92x\right)+\left(-84x-161\right)=0\\ \Leftrightarrow4x\left(12x+23\right)-7\left(12x+23\right)=0\\ \Leftrightarrow\left(4x-7\right)\left(12x+23\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{4}\Rightarrow y=-\dfrac{2.7}{4}=-\dfrac{7}{2}\\x=-\dfrac{23}{12}\Rightarrow y=-2.-\dfrac{23}{12}=\dfrac{23}{6}\end{matrix}\right.\)
Vậy phương trình có nghiệm \(\left\{x;y\right\}=\left\{\dfrac{7}{4};-\dfrac{7}{2}\right\}\) hoặc \(\left\{x;y\right\}=\left\{-\dfrac{23}{12};\dfrac{23}{6}\right\}\)
