Answer

\sqrt{x-3}+\sqrt{y+1} = \dfrac{x+y}{2}
(1)

Let

a =\sqrt{x-3}
(2)
b=\sqrt{y+1}
(3)

Then

x = a^2+3
(4)
y=b^2-1
(5)

and

a+b=\dfrac{a^2+b^2+2}{2}
(6)

2a+2b = a^2+b^2+2

(a-1)^2+(b-1)^2+0

Therefore

a = 1 and b=1

Substituting to (4) and (5) gives

x = 4 and y = 0

Therefor

x+y = 4


Steven Zheng posted 23 hours ago

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