Question

Помогите,пожалуйста решить систему!!
{log2(x+y)=2
{log√3x+log√3y=2

Answer 1

X+Y =4         X=4-Y
X*Y=3           (4-Y)Y=3

Y=3       Y=1
X=1       X=3

Answer 2

log₂(x+y) = 2                                    x + y = 4        x = 4 - y
log√3X + log√3Y = 2                        x * y = 3         (4-y)*y = 3

x = 4 - y               x = 4 - y                          x = 4 - y
4y - y² = 3            y² - 4y + 3 = 0                 y = 3  или y = 1

Если y = 3 ,   то     x = 4 - 3 = 1
Если y = 1,    то     x = 4 - 1 = 3
Ответ: (1 , 3), ( 3 ; 1)