Question

x^2+3xy-10y^2=0
x^2+2xy-y^2=28
_____________
4x^2-4xy+y^2=9
3x^2+2xy-y^2=36

Answer 1

$1)\; \; \left \{ {{x^2+3xy-10y^2=0} \atop {x^2+2xy-y^2=28}} \right.\\\\x^2+3xy-10y^2=0\; |:y^2\ne 0\; \; \; \Rightarrow\quad (\frac{x}{y})^2+3\cdot\frac{x}{y}-10=0\\\\t=\frac{x}{y}\; ,\; \; t^2+3t-10=0\; ,\; \; t_1=-5\; ,\; t_2=2\\\\a)\; \; \frac{x}{y}=-5\; ,\; \; x=-5y\; ,\; \; (-5y)^2+2\cdot (-5y)\cdot y-y^2=28\; ,\\\\25y^2-10y^2-y^2=28\; ,\; \; 14y^2=28\; ,\; \; y^2=2\; ,\; \; y_{1,2}=\pm \sqrt2\\\\x_1=-5\sqrt2\; ,\; \; x_2=5\sqrt2$

$b)\; \; \frac{x}{y}=2\; ,\; \; x=2y\; \; ,\; \; 4y^2+4y^2-y^2=28\; ,\; \; 7y^2=28\; ,\; \; y^2=4\\\\y=\pm 2\; ,\; \; x=\pm 4\\\\Otvet:\; \; (-5\sqrt2;\sqrt2)\; ,\; (5\sqrt2;\; -\sqrt2)\; ,\; (4,2)\; ,\; (-4,-2)\; .$

$2)\; \; \left \{ {{4x^2-4xy+y^2=9\, |\cdot (-4)} \atop {3x^2+2xy-y^2=36}} \right.\; \oplus \; \left \{ {4x^2-4xy+y^2=9} \atop {-13x^2+18xy-5y^2=0}} \right.\\\\13x^2-18xy+5y^2=0\, |:y^2\ne 0\; \; \; \Rightarrow \; \; \; 13\cdot (\frac{x}{y})^2 -18\cdot \frac{x}{y}+5=0\; ,\\\\t=\frac{x}{y}\; ,\; \; 13t^2-18t+5=0\; ,\; D/4=16\; ,\; \; t_1=\frac{5}{13}\; ,\; \; t_2=\frac{13}{13}=1\\\\a)\; \; \frac{x}{y}=\frac{5}{13}\; \; ,\; \; x=\frac{5y}{13}\; \; ,\; \; 4\cdot \frac{25}{169}y^2-4\cdot \frac{5}{13}y^2+y^2=9\; ,$

$\frac{9}{169}\, y^2=9\; ,\; \; y^2=169\; ,\; \; y=\pm 13\\\\x=\pm 5\\\\b)\; \; \frac{x}{y}=1\; ,\; \; x=y\; ,\; 4y^2-4y^2+y^2=9\; ,\; \; y^2=9\; ,\; \; y=\pm 3\\\\x=\pm 3\\\\Otvet:\; \; (5,13)\; ,\; (-5,-13)\; ,\; (3,3)\; ,\; (-3,-3)\; .$