Question

ПОМОЩЬПОМОЩЬПОМОЩЬПОМОЩЬ

Answer 1

Решение во вложении смотрим

Answer 2

Сделаем замену :

$\frac{x}{y}=m\Rightarrow \frac{y}{x}=\frac{1}{m}\\\\m+\frac{1}{m}=\frac{17}{4}\\\\m+\frac{1}{m}-\frac{17}{4}=0\\\\\frac{4m^{2}-17m+4 }{4m}=0,m\neq0\\\\4m^{2}-17m+4=0\\\\m_{1}=4\\\\m_{2}=\frac{1}{4}\\\\1)\left \{ {{\frac{x}{y}=4} \atop {x+y=3}} \right.\\\\\left \{ {{x=4y} \atop {4y+y=3}} \right.\\\\\left \{ {{x=4y} \atop {5y=3}} \right. \\\\\left \{ {{x=4*0,6} \atop {y=0,6}} \right.\\\\\left \{ {{x=\frac{12}{5} } \atop {y=\frac{3}{5} }} \right.$

$2)\left \{ {{\frac{x}{y}=\frac{1}{4}} \atop {x+y=3}} \right.\\\\\left \{ {{y=4x} \atop {x+4x=3}} \right.\\\\\left \{ {{y=4x} \atop {5x=3}} \right.\\\\\left \{ {{x=\frac{3}{5}} \atop {y=\frac{12}{5} }} \right.\\\\Otvet:\left \{ {{x_{1}=\frac{3}{5}} \atop {y_{1}=\frac{12}{5}}} \right.;\left \{ {{x_{2}=\frac{12}{5}} \atop {y_{2=\frac{3}{5}} } \right.$