Đáp án:$\left( {x;y} \right) = \left\{ {\left( { - 3; - 1} \right);\left( {\frac{{13}}{5};\frac{9}{5}} \right)} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
x - 2y + 1 = 0\\
{x^2} + {y^2} = 10
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2y - 1\\
{x^2} + {y^2} = 10
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2y - 1\\
{\left( {2y - 1} \right)^2} + {y^2} = 10
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2y - 1\\
5{y^2} - 4y + 1 - 10 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2y - 1\\
5{y^2} - 4y - 9 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2y - 1\\
\left[ \begin{array}{l}
y = - 1\\
y = \frac{9}{5}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
y = - 1;x = - 3\\
y = \frac{9}{5};x = \frac{{13}}{5}
\end{array} \right.
\end{array}$
