Question

Помогите по алгебре!!!

Answer 1

Ответ:

$\left\{\begin{array}{l}(4x+y)(x+3y)=\dfrac{306}{25}\\\dfrac{4x+y}{x+3y}=\dfrac{17}{18}\end{array}\right\ \ \left\{\begin{array}{l}p=4x+y\\t=x+4y\end{array}\right\ \ \left\{\begin{array}{l}\ \ p\, t=\dfrac{306}{25}\\\dfrac{p}{t}=\dfrac{17}{18}\end{array}\right$

$\left\{\begin{array}{l}\ \ \ p=\dfrac{306}{25\, t}\\p=\dfrac{17\, t}{18}\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{306}{25\, t}=\dfrac{17\, t}{18}\\p=\dfrac{17\, t}{18}\end{array}\right\ \ \left\{\begin{array}{l}17\cdot 25\, t^2=306\cdot 18\\p=\dfrac{17\, t}{18}\end{array}\right\ \ \left\{\begin{array}{l}t^2=12,96\\p=\dfrac{17\, t}{18}\end{array}\right$

$\left\{\begin{array}{l}t_1=-3,6\ ,\ t_2=3,6\\p_1=-3,4\ ,\ p_2=3,4\end{array}\right\\\\\\1)\ \left\{\begin{array}{l}x+3y=-3,6\ |\cdot (-4)\\4x+y=-3,4\end{array}\right\ \oplus \ \left\{\begin{array}{l}x=-3,6-3y\\-11y=11\end{array}\right\ \ \left\{\begin{array}{l}x_1=-0,6\\y_1=-1\end{array}\right$

$2)\ \left\{\begin{array}{l}x+3y=-3,6\ |\cdot (-4)\\4x+y=3,4\end{array}\right\ \oplus \ \left\{\begin{array}{l}x=-3,6-3y\\-11y=17,8\end{array}\right\ \ \left\{\begin{array}{l}x_2=-\dfrac{69}{55}\\y_2=-\frac{89}{55}\end{array}\right$

$3)\ \left\{\begin{array}{l}x+3y=3,6\ |\cdot (-4)\\4x+y=-3,4\end{array}\right\ \oplus \ \left\{\begin{array}{l}x=3,6-3y\\-11y=-17,8\end{array}\right\ \ \left\{\begin{array}{l}x_3=-\dfrac{69}{55}\\y_3=\frac{89}{55}\end{array}\right$

$4)\ \left\{\begin{array}{l}x+3y=3,6\ |\cdot (-4)\\4x+y=3,4\end{array}\right\ \oplus \ \left\{\begin{array}{l}x=-3,6-3y\\-11y=-11\end{array}\right\ \ \left\{\begin{array}{l}x_4=-6,6\\y_4=1\end{array}\right$