Вопрос:

Решите систему уравнений: 2x/3-2+y/2; 2x/3+y=8.

Ответ:

\[\left\{ \begin{matrix} \frac{2x}{3} = 2 + \frac{y}{2}\ \ \ \ | \cdot 6 \\ \frac{2x}{3} + y = 8\ \ \ \ | \cdot 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2 \cdot 2x = 6 \cdot 2 + 3y \\ 2x + 3y = 3 \cdot 8\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 4x - 3y = 12\ \ \ (1) \\ 2x + 3y = 24\ \ \ (2) \\ \end{matrix} \right.\ \]

\[6x = 36\ \ \]

\[x = 6\]

\[\left\{ \begin{matrix} x = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4x - 3y = 12 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3y = 4x - 12 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y = \frac{4}{3} \cdot 6 - 4\ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} x = 6 \\ y = 4 \\ \end{matrix} \right.\ \]

\[Ответ:(6;4).\]


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