Помогите решить 2, 3 и 5. Заранее спасибо.

Question 1

Question 2

Решите задачу:

$2)\; \; \left \{ {{x^2=11x+6y} \atop {y^2=6x+11y}} \right. \; \Rightarrow \; \; x^2-y^2=5x-5y\\\\(x-y)(x+y)-5(x-y)=0\\\\(x-y)(x+y-5)=0\; \; \Rightarrow \; \; x-y=0\; ili\; \; x+y-5=0\\\\a)\; \; x=y\; \to \; \; \left \{ {{x^2=11x+6y} \atop {x=y}} \right. \; \left \{ {{x^2=11x+6x} \atop {x=y}} \right. \; \left \{ {{x^2-17x=0} \atop {x=y}} \right. \; \left \{ {{x(x-17)=0} \atop {x=y}} \right. \\\\ \left \{ {{x_1=0\; ili\; x_2=17} \atop {y_1=0\; ili\; y_2=17}} \right.$

$b)\; \; x=5-y\; \to \; \; \left \{ {{x=5-y} \atop {y^2=6(5-y)+11}} \right. \; \left \{ {{x=5-y} \atop {y^2=30-6y+11y}} \right. \; \left \{ {{x=5-y} \atop {y^2-5y-30=0}} \right. \\\\y^2-5y-30=0\; ,\; \; D=25+4\cdot 30=145\; ,\; \; y_{1,2}=\frac{5\pm \sqrt{145}}{2}\\\\x_1=5-\frac{5-\sqrt{145}}{2}=\frac{5+\sqrt{145}}{2} \; ,\; \; x_2=5-\frac{5+\sqrt{145}}{2}=\frac{5-\sqrt{145}}{2}\\\\Otvet:\; \; (0;0)\; ,\; (17;17)\; ,\; (\frac{5-\sqrt{145}}{2};\frac{5+\sqrt{145}}{2})\; ,\; (\frac{5+\sqrt{145}}{2};\frac{5-\sqrt{145}}{2})\; .$

$3)\; \; \left \{ {{4x^2+7xy-2y^2=0} \atop {3x^2-5xy+y^2=-1}} \right. \\\\4x^2+7xy-2y^2=0\, |:y^2\ne 0\; \; ,\; \; 4\cdot (\frac{x}{y})^2+7\cdot \frac{x}{y}-2=0\\\\t= \frac{x}{y}\; ,\; \; 4t^2+7t-2=0\; ,\; D=49+32=81\; ,\\\\t_1=\frac{-7-9}{8}=-2\; ,\; t_2=\frac{-7+9}{8}=\frac{1}{4}\\\\a)\; \; \frac{x}{y}=-2\; ,\; \; x=-2y\; ,\; \; 3\cdot 4y^2-5y\cdot (-2y)+y^2=-1\\\\23y^2=-1\; ,\; \; y^2=-\frac{1}{23}\; \Rightarrow \; \; y\in \varnothing \; \; (t.k.\; \; y^2 \geq 0)$

$b)\; \; \frac{x}{y}=\frac{1}{4}\; ,\; \; y=4x\; ,\; \; 3x^2-5x\cdot 4x+16x^2=-1\\\\-x^2=-1\; ,\; \; x^2=1\; \Rightarrow \; \; x_{1}=-1\; ,\; \; x_2=1\\\\y_1=-4\; ,\; x_2=4\\\\Otvet:\; \; (-1;-4)\; ,\; \; (1;4)\; .\\\\5)\; \; \left \{ {{x^2+y^2=13,} \atop {x+y+xy=11;}} \right. \; \; \; \; \; (x+y)^2=x^2+y^2+2xy\; \Rightarrow \\\\x^2+y^2=(x+y)^2-2xy\; ,\quad \left \{ {{(x+y)^2-2xy=13,} \atop {x+y+xy=11;}} \right. \; ,\; \; u=x+y\; ,\; v=xy \\\\\left \{ {{u^2-2v=13} \atop {u+v=11}} \right. \; \left \{ {{u^2-2(11-u)=13} \atop {v=11-u}} \right. \; \left \{ {{u^2+2u-35=0} \atop {v=11-u}} \right.$

$u^2+2u-35=0\; ,\; \; u_1=-7\; ,\; \; u_2=5\; \; (teorema\; Vieta)\\\\v_1=11-(-7)=18\; ,\; \; v_2=11-5=6\\\\a)\; \; \left \{ {{x+y=-7} \atop {xy=18}} \right. \; \left \{ {{y=-7-x} \atop {x(-7-x)=18}} \right. \; \left \{ {{y=-7-x} \atop {x^2+7x+18=0}} \right. \\\\x^2+7x+18=0\; ,\; \; D=-23\ \textless \ 0\; \Rightarrow \; \; x\in \varnothing \\\\b)\; \; \left \{ {{x+y=5} \atop {xy=6}} \right. \; \left \{ {{y=5-x} \atop {x(5-x)=6}} \right. \; \left \{ {{y=5-x} \atop {x^2-5x+6=0}} \right. \; \left \{ {{y_1=3\; ,\; y_2=2} \atop {x_1=2\; ,\; x_2=3}} \right. \\\\Otvet:\; \; (2;3)\; ,\; \; (3;2)\; .$