Question

Помогите решить, пожалуйста ​

Answer 1

1)

$\left \{ {{2x-y=1} \atop {\frac{1}{x}+\frac{1}{y}=\frac{5}{6}}} \right. => \left \{ {{y=-1+2x} \atop {\frac{1}{x}+\frac{1}{y}=\frac{5}{6}}} \right.\\\\\frac{1}{x}+\frac{1}{-1+2x}=\frac{5}{6}\\\\\frac{5(-1+2x)+6x-5x(-1+2x)}{6x(-1+2x)}\\\frac{-6+23x-10x^2}{6x(-1+2x)}=0\\-6x+23x-10x^2=0\\10x^2-3x-20x+6=0\\x(10x-3)-2(10x-3)=0\\(10x-3)(x-2)=0\\10x-3=0\\x-2=0\\x_1=\frac{3}{10} = 0,3\\x_2=2\\\\y=-1+2*0,3 \\y=-1+2*2\\y_1=-0,4\\y_2=3$

2)

$\left \{ {{2x-y=10} \atop {x^2-3xy+y^2=4}} \right. => \left \{ {{y=-10+2x} \atop {x^2-3xy+y^2=4}} \right. \\\\x^2-3x*(-10+2x)+(-10+2x)^2=4\\x^2+30x-6x^2+4x^2-40x+100=4\\-x^2-10x+100=4\\x^2+10x-96=0\\x_1= -16\\x_2=6\\\\y=-10+2*(-16)\\y=-10+2*6\\y_1=-42\\y_2=2$