Решебник по алгебре 9 класс Макарычев Задание 1080

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\[\left\{ \begin{matrix} x + y + z = 14 \\ x + yz = 19\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 14 - y - z\ \ \ \ \ \ \ \ \ \ \ \\ 14 - y - z + yz = 19 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 14 - y - z\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y(z - 1) - (z - 1) = 6 \\ \end{matrix} \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} x = 14 - y - z\ \ \ \ \ \ \ \ \\ (z - 1)(y - 1) = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow (z - 1)(y - 1) = 1 \cdot 6\ \ или\ \ \]

\[(z - 1)(y - 1) = 2 \cdot 3 \Longrightarrow\]

\[1)\ z - 1 = 1,\ \ y - 1 = 6,\]

\[\left\{ \begin{matrix} z = 2 \\ y = 7 \\ x = 5 \\ \end{matrix} \right.\ \]

\[2)\ z - 1 = 6,\ \ y - 1 = 1,\]

\[\left\{ \begin{matrix} z = 7 \\ y = 2 \\ x = 5 \\ \end{matrix} \right.\ \]

\[3)\ z - 1 = 2,\ \ y - 1 = 3,\]

\[\left\{ \begin{matrix} z = 3 \\ y = 4 \\ x = 7 \\ \end{matrix} \right.\ \]

\[4)\ z - 1 = 3,\ \ y - 1 = 2,\]

\[\left\{ \begin{matrix} z = 4 \\ y = 3 \\ x = 7. \\ \end{matrix} \right.\ \]