Question

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

Answer 1

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

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