Question

Помогиие пожалуйста​

Answer 1

$\displaysty \left \{ {{2x^2-xy=33} \atop {4x-y=17} \right. \\\\y=4x-17\\\\2x^2-x(4x-17)=33\\\\2x^2-4x^2+17x-33=0\\\\-2x^2+17x-33=0\\\\D=289-264=25\\\\x_{1.2}=\frac{-17\pm 5}{-4}\\\\x_1=5.5; x_2=3\\\\y_1=4*5.5-17=5;y_2=4*3-17=-5\\\\(5.5;5)(3;-5)$

Answer 2

$\displaystyle\left \{ {{2x^{2} -xy=33} \atop {4x-y=17}} \right. \\\left \{ {{x(2x-y)=33} \atop {x=\frac{17+y}{4} }} \right. \\\left \{ {{\frac{17+y}{4}(2*\frac{17+y}{4}-y)=33} \atop {x=\frac{17+y}{4} }} \right. \\\left \{ {{\frac{17+y}{4}(\frac{17+y}{2}-y)=33} \atop {x=\frac{17+y}{4}}} \right. \\\left \{ {{\frac{17+y}{4}(\frac{17-y}{2} )=33} \atop {x=\frac{17+y}{4} }} \right.$

$\displaystyle\left \{ {{\frac{(17+y)(17-y)}{8} =33} \atop {x=\frac{17+y}{4} }} \right. \\\left \{ {{\frac{289-y^{2}}{8} =33 } \atop {x=\frac{17+y}{4} }} \right.\\\left \{ {{289-y^{2}=264} \atop {x=\frac{17+y}{4} }} \right. \\\left \{ {{y^{2}-25=0 } \atop {x=\frac{17+y}{4}}} \right.\\ \left \{ {{ \left[\begin{array}{ccc}y_{1}=5 \\y_{2}=-5\end{array}\right } \atop {\left[\begin{array}{ccc}x_{1}=5,5 \\x_{2}=3\end{array}\right }}} \right.$

$Otvet: x_{1} =5,5, y_{1} =5\ ili\ x_{2} =3, y_{2} =-5$