Question

Помогите пожалуйста!!

Answer 1

$\begin{equation*}\begin{cases}16x^2 + 8xy + y^2 = 36\\3x - y = 8\end{cases}\end{equation*}\ \ \ \Leftrightarrow\ \begin{equation*}\begin{cases}(4x + y)^2 = 36\\3x - y = 8\end{cases}\end{equation*}$

Выразим в нижнем уравнении $y$ через $x$:

$3x - y = 8\\\\-y = 8-3x\ \ \ \ \ \Big| :(-1)\\\\y = 3x - 8$

А теперь рассмотрим верхнее уравнение:

$(4x+y)^2 = 36\\\\\left[\begin{gathered}4x + y = 6\\4x + y = -6\end{gathered}$

Значит, наша система будет иметь две пары решений.

$\left[\begin{gathered}\begin{equation*}\begin{cases}4x + y = 6\\y = 3x -8\end{cases}\\\begin{equation*}\begin{cases}4x + y = -6\\y = 3x - 8\end{cases}\end{equation*}\\\end{gathered}\ \ \ \Leftrightarrow\ \left[\begin{gathered}\begin{equation*}\begin{cases}4x + 3x - 8 = 6\\y = 3x -8\end{cases}\\\begin{equation*}\begin{cases}4x + 3x - 8 = -6\\y = 3x - 8\end{cases}\end{equation*}\\\end{gathered}\ \ \ \Leftrightarrow$    $\left[\begin{gathered}\begin{equation*}\begin{cases}7x = 14\\y = 3x -8\end{cases}\\\begin{equation*}\begin{cases}7x = 2\\y = 3x - 8\end{cases}\end{equation*}\\\end{gathered}\ \ \ \Leftrightarrow$

$\Leftrightarrow\ \ \left[\begin{gathered}\begin{equation*}\begin{cases}x = 2\\y = 3x -8\end{cases}\\\begin{equation*}\begin{cases}x = \dfrac{2}{7}\\y = 3x - 8\end{cases}\end{equation*}\\\end{gathered}\ \ \ \Leftrightarrow\ \ \left[\begin{gathered}\begin{equation*}\begin{cases}x = 2\\y = 6 -8\end{cases}\\\begin{equation*}\begin{cases}x = \dfrac{2}{7}\\\\y = \dfrac{6}{7} - 8\end{cases}\end{equation*}\\\end{gathered}\ \ \ \Leftrightarrow$    $\left[\begin{gathered}\begin{equation*}\begin{cases}x = 2\\y = -2\end{cases}\\\begin{equation*}\begin{cases}x = \dfrac{2}{7}\\y = -7\dfrac{1}{7}\end{cases}\end{equation*}\\\end{gathered}$

Ответ: $(2,\ -2);\ \left(\dfrac{2}{7}\ ,\ -7\dfrac{1}{7}\right)$ .