a: Ta có: 2x-3y=9
nên 2x=9+3y
hay \(x=\dfrac{3y+9}{2}\)
Vậy: \(\left\{{}\begin{matrix}y\in R\\x=\dfrac{3y+9}{2}\end{matrix}\right.\)
b: Ta có: 2x+0y=5
nên 2x=5
hay \(x=\dfrac{5}{2}\)
Vậy: \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y\in R\end{matrix}\right.\)