$\left\{ \begin{gathered} 3x - y = 5 \hfill \\ x + 2y = 4 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} 6x - 2y = 10 \hfill \\ x + 2y = 4 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} 7x = 14 \hfill \\ x + 2y = 4 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} x = 2 \hfill \\ y = 1 \hfill \\ \end{gathered} \right.$ Vậy $(x;y)=(2;1)$
b)
$\left\{ \begin{gathered}
4x + 7y = 16 \hfill \\
4x - 3y = - 24 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
10y = 40 \hfill \\
4x + 7y = - 24 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
x = 4 \hfill \\
y = - 3 \hfill \\
\end{gathered} \right.$
Vậy $(x;y)=(3;-4)$
