Помогите решить систему log0,5х+log0,5у=-1 х-2у=3

ОДЗ x > 0, y > 0
$\left \{ {{log0,5_x + log0,5_y = -1} \atop {x-2y=3}} \right. =\ \textgreater \ \left \{ {{log0,5_(x*y) = log0,5_2} \atop {x-2y=3}} \right. =\ \textgreater \ \left \{ {{x*y=2} \atop {x-2y=3}} \right.$

$\left \{ {{(3+2y)*y=2} \atop {x=3+2y}} \right. =\ \textgreater \ \left \{ {{3y+2y^2-2=0} \atop {x=3+2y}} \right. =\ \textgreater \ \left \{ {{2y^2+3y-2=0} \atop {x=3+2y}} \right.$

2y^2+3y-2=0
D=25
y1 = -2
y2 = 0,5

x1 = 3 + 2* (-2) = -1
x2 = 3 + 2*(0,5)= 4

(-1;-2) не подходят по одз

Ответ: (4;0,5)