a: 3x-2y=4
=>3x=4+2y
hay \(x=\dfrac{2y+4}{3}\)
Vậy: \(\left\{{}\begin{matrix}y\in R\\x=\dfrac{2y+4}{3}\end{matrix}\right.\)
b: 12x-9y=15
=>4x-3y=5
=>4x=3y+5
hay \(x=\dfrac{3y+5}{4}\)
Vậy: \(\left\{{}\begin{matrix}y\in R\\x=\dfrac{3y+5}{4}\end{matrix}\right.\)