Question

5x=2y,xy-10=0 

tiếp nha

Answer 1

Theo đề, ta có:  \(5x=2y\Leftrightarrow x=\dfrac{2y}{5}\left(1\right)\)

Thế vào ta có:  

\(xy-10=0\) 

\(\dfrac{2y}{5}.y-10=0\)

\(2y^2=50\)

\(y^2=25\)

\(\Rightarrow y=\pm5\)

Khi \(y=5\Rightarrow x=2\)

Khi \(y=-5\Rightarrow x=-2\)

Answer 2

Ta có : \(5x=2y\Leftrightarrow x=\dfrac{2y}{5}\left(1\right)\)

Thay \(\left(1\right)\) vào \(xy-10=0\)

\(\Leftrightarrow\dfrac{2y^2}{5}-10=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)

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