Question

[Thầy Tùng Dương]

Giải hệ phương trình $\left\{\begin{array}{l}x-3 y=5 \\ 2 x+3 y=1\end{array}\right.$.

Answer 1

[Nguyễn Thị Thương Hoài]

  \(\left\{{}\begin{matrix}x-3y=5\\2x+3y=1\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}x-3y+2x+3y=5+1\\2x+3y=1\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}3x=6\\2x+3y=1\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}x=6:3\\2x+3y=1\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}x=2\\2x+3y=1\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}x=2\\2.2+3y=1\end{matrix}\right.\)

⇒\(\left\{{}\begin{matrix}x=2\\3y=1-4\end{matrix}\right.\)

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

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

Vậy (\(x\);y) =(2; -1)