Решить систему 1/у-1/х=1/3 х-2у=2

$\left \{ {{\frac{1}{y}-\frac{1}{x}=\frac{1}{3}} \atop {x-2y=2}} \right.$

$\left \{ {{\frac{1}{y}-\frac{1}{x}=\frac{1}{3}} \atop {x=2+2y}} \right.$

1) ${\frac{1}{y}-\frac{1}{2+2y}-\frac{1}{3}=0$

${\frac{3(2+2y)}{3y(2+2y)}-\frac{3y}{3y(2+2y)}-\frac{y(2+2y)}{3y(2+2y)}=0$

${\frac{3(2+2y)-3y-y(2+2y)}{3y(2+2y)}=0$

ОДЗ:

3y(2+2y)≠0

2+2y≠0 или y≠0

2y≠-2

y≠-1

$3(2+2y)-3y-y(2+2y)}=0$

$6+6y-3y-2y-2y^{2} =0$

$6+y-2y^{2}=0$ | ×(-1)

$2y^{2} -y-6=0$

$D=b^{2} -4ac=1+4*2*6=49$

$\sqrt{D} =7$

$y_{1} =\frac{-b+\sqrt{D}}{2a} =\frac{1+7}{2*2}=2$

$y_{2} =\frac{-b-\sqrt{D}}{2a} =\frac{1-7}{4}=-\frac{3}{2}=-1,5$

2) x₁=2+2*2=2+4=6

x₂=2+2*(-1,5)=2-3=-1

Ответ: x₁ = 6, y₁ = 2; x₂ = -1, y₂ = -1,5.