Question

Срочно даю сто балов решите систему методом интервалов{2x^2-3xy=-4 3x-y=5

Answer 1

$\left \{ {{2x^{2}-3xy=-4 } \atop {3x-y=5}} \right.\\\\\left \{ {{y=3x-5} \atop {2x^{2}-3x*(3x-5)+4=0 }} \right.\\\\\left \{ {{y=3x-5} \atop {2x^{2}-9x^{2}+15x+4=0}} \right. \\\\\left \{ {{y=3x-5} \atop {-7x^{2}+15x+4=0 }} \right.\\\\\left \{ {{y=3x-5} \atop {7x^{2}-15x-4=0}} \right.\\\\7x^{2}-15x-4=0\\\\D=(-15)^{2}-4*7*(-4)=225+112=337$

$x_{1}=\frac{15-\sqrt{337}}{14}\\\\x_{2} =\frac{15+\sqrt{337}}{14} \\\\y_{1}=3*\frac{15-\sqrt{337}}{14}-5=\frac{45-3\sqrt{337}-70}{14}=-\frac{25+3\sqrt{337} }{14}\\\\y_{2}=3*\frac{15+\sqrt{337}}{14}-5=\frac{45+3\sqrt{337}-70}{14}=-\frac{25-3\sqrt{337}}{14}\\\\Otvet:\boxed{(\frac{15-\sqrt{337}}{14};-\frac{25+3\sqrt{337}}{14}),(\frac{15+\sqrt{337}}{14};-\frac{25-3\sqrt{337}}{14})}$