Question

\(\left\{{}\begin{matrix}3x^2-5x=2\\4x+2y=-6\end{matrix}\right.\)

Answer 1

(1) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{3}\end{matrix}\right.\)

+ với x =2, thay vào (2) ta được: \(4\cdot2+2y=-6\Leftrightarrow y=-7\)

Vậy \(\left\{{}\begin{matrix}x=2\\y=-7\end{matrix}\right.\)

+ với \(x=-\dfrac{1}{3}\) thay vào (2), ta được: \(4\cdot\left(-\dfrac{1}{3}\right)+2y=-6\Leftrightarrow y=-\dfrac{7}{3}\)

Vậy \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=-\dfrac{7}{3}\end{matrix}\right.\)

Kl: \(\left\{{}\begin{matrix}x=2\\y=-7\end{matrix}\right.\), \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=-\dfrac{7}{3}\end{matrix}\right.\)