Ta có :3x+21=4y
\(3x⋮3,21⋮3\Rightarrow4y⋮3\)mà \(\left(4,3\right)=1\)
\(\Rightarrow y⋮3\left(1\right)\)
Lại có: \(4y=3x+21\ge3+21=24\Rightarrow y\ge6\left(2\right)\)
Mặt khác : y<10 kết hợp (1),(2)
\(\orbr{\begin{cases}y=6\\y=9\end{cases}\Leftrightarrow}\orbr{\begin{cases}3x=3\\3x=15\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}\left(TM\right)}}\)
Vậy \(\left(x,y\right)\in\left\{\left(1,6\right);\left(5,9\right)\right\}\)