\(\left\{\begin{matrix}x^2+4xy=18\left(1\right)\\8y+2x=10\left(2\right)\end{matrix}\right.\)
Từ \(\left(2\right)\Rightarrow8y=10-2x\Rightarrow y=\frac{10-2x}{8}\) thay vào (1) ta có:
\(x^2+4x\cdot\frac{10-2x}{8}=18\)
\(\Rightarrow x^2+5x-x^2=18\)
\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)