Question

Cho hệ phương trình: \(\left\{{}\begin{matrix}x-3y=0\\\left(a-1\right)x-3y=2\end{matrix}\right.\) (với a là tham số).

Tìm giá trị của a để hệ có nghiệm (x ; y) với x, y > 0.

Answer 1

Ta có : \(\left\{{}\begin{matrix}x-3y=0\\\left(a-1\right)x-3y=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=3y\\\left(a-1\right)x-2=3y\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left(a-1\right)x-2=x\\3y\left(a-1\right)-2=3y\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left(a-1\right)x-x=2\\3y\left(a-1\right)-3y=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left(a-2\right)x=2\\3y\left(a-2\right)=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{2}{a-2}\\y=\frac{2}{3\left(a-2\right)}\end{matrix}\right.\)

- Ta có : \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\frac{2}{a-2}>0\\\frac{2}{3\left(a-2\right)}>0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a-2>0\\3\left(a-2\right)>0\end{matrix}\right.\)

=> a > 2

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