Question

ПОМОГИТЕ ПОЖАЛУЙСТААА!!!!!!!!
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Answer 1

$\left\{\begin{array}{l}(5x+y)(x+3y)=\dfrac{264}{49}\\\dfrac{5x+y}{x+3y}=\dfrac{6}{11}\end{array}\right\ \ \left\{\begin{array}{l}(5x+y)(x+3y)=\dfrac{264}{49}\\\\55x+11y=6x+18y\end{array}\right\\\\\\\left\{\begin{array}{l}(5x+y)(x+3y)=\dfrac{264}{49}\\49x=7y\end{array}\right\ \ \left\{\begin{array}{l}(5x+y)(x+3y)=\dfrac{264}{49}\\y=7x\end{array}\right$

$\left\{\begin{array}{l}(5x+7x)(x+21x)=\dfrac{264}{49}\\y=7x\end{array}\right\ \ \left\{\begin{array}{l}264x^2=\dfrac{264}{49}\\y=7x\end{array}\right\ \ \left\{\begin{array}{l}x^2=\dfrac{1}{49}\\y=7x\end{array}\right$

$\left\{\begin{array}{l}x=\pm \dfrac{1}{7}\\y=7x\end{array}\right\ \ \left\{\begin{array}{l}x_1=\dfrac{1}{7}\ ,\ x_2=-\dfrac{1}{7}\\y_1=1\ ,\ y_2=-1\end{array}\right\\\\\\Otvet:\ \ \Big(\ \dfrac{1}{7}\ ;\ 1\, \Big)\ \ ,\ \ \Big(-\dfrac{1}{7}\ ;\ -1\, \Big)\ .$