Question

помогите пожалуйста оч надо

Answer 1

Решить систему уравнений.

$\begin{cases}\dfrac{2x+7y}{4}+\dfrac{3x-2y}{3} = \dfrac{2}{3},\\\\\dfrac{3x+2y}{2}-\dfrac{4x-6y}{7} = \dfrac{39}{14};\end{cases}\Longleftrightarrow\begin{cases}\dfrac{3(2x+7y) + 4(3x-2y)}{12} = \dfrac{2\cdot4}{12},\\\\\dfrac{7(3x+2y) - 2(4x-6y)}{14} = \dfrac{39}{14};\end{cases}\Longleftrightarrow\\\\\begin{cases}6x+21y + 12x-8y = 8,\\21x+14y - 8x+12y = 39;\end{cases}\Longleftrightarrow\begin{cases}18x+13y = 8,\\13x+26y = 39; |:13\end{cases}\Longleftrightarrow$

$\begin{cases}18x+13y = 8,\\x+2y = 3;\end{cases}\Longleftrightarrow\begin{cases}18x+13y = 8,\\x = 3 - 2y;\end{cases}\\\\18(3 - 2y) + 13y = 8,\\54 - 36y + 13y = 8,\\54 - 8 = 23y,\\46 = 23y,\\y = \dfrac{46}{23},\\y = 2;\\x = 3 - 2y = 3 - 2\cdot2 = 3 - 4 = -1.$

Ответ: $\textbf{(x; y) = (-1; 2)}.$