Question

Решите уравнение на фотографии. Буду благодарен)

Answer 1

$+\left \{ {{x^{2}+7xy=-6 } \atop {9y^{2}-xy=10 }} \right. \\-------\\x^{2}+6xy+9y^{2}=4\\\\(x+3y)^{2} =4\\\\\left[\begin{array}{ccc}x+3y=2\\x+3y=-2\end{array}\right\\\\1)\left \{ {{x+3y=2} \atop {x^{2} +7xy=-6}} \right.\\\\\left \{ {{x=2-3y} \atop {(2-3y)^{2}+7y\cdot(2-3y)+6=0 }} \right. \\\\\left \{ {{x=2-3y} \atop {4-12y+9y^{2}+14y-21y^{2}+6=0}} \right. \\\\\left \{ {{x=2-3y} \atop {12y^{2}-2y-10=0 }} \right. \\\\\left \{ {{x=2-3y} \atop {6y^{2}-y-5=0 }} \right.$

$6y^{2}-y-5=0\\\\y_{1} =1 \ \Rightarrow \ x_{1}=2-3*1=-1\\\\y_{2}=-\frac{5}{6} \ \Rightarrow \ x_{2}=2-3*(-\frac{5}{6} )=2+2,5=4,5\\\\\\2)\left \{ {{x+3y=-2} \atop {9y^{2}-xy=10 }} \right.\\\\\left \{ {{x=-3y-2} \atop {9y^{2}-y\cdot(-3y-2)=10 }} \right.\\\\\left \{ {{x=-3y-2} \atop {9y^{2}+3y^{2}+2y-10=0}} \right.\\\\\left \{ {{x=-3y-2} \atop {12y^{2}+2y-10=0 }} \right.\\\\\left \{ {{x=-3y-2} \atop {6y^{2} +y-5=0}} \right.$

$y_{3} =-1 \ \Rightarrow \ x_{3}=-3\cdot(-1)-2=1\\\\y_{4}=\frac{5}{6} \ \Rightarrow \ x_{4}=-3*\frac{5}{6} -2=-2,5-2=-4,5$

$Otvet:\boxed{(-1;1) \ , \ (4,5;-\frac{5}{6}) \ ,(1;-1) \ ,(-4,5;\frac{5}{6} )}$