If $$\sqrt{x-3}+\sqrt{y+1} = \dfrac{x+y}{2} $$
Solve for $$x+y$$

Question

If $$\sqrt{x-3}+\sqrt{y+1} = \dfrac{x+y}{2}      $$
Solve for $$x+y$$

Uniteasy Admin · Accepted Answer

$$\sqrt{x-3}+\sqrt{y+1} = \dfrac{x+y}{2}   $$n
Let 
$$a =\sqrt{x-3}$$n
$$b=\sqrt{y+1}$$n
Then
$$x = a^2+3$$n
$$y=b^2-1$$n
and
$$a+b=\dfrac{a^2+b^2+2}{2} $$n
$$2a+2b = a^2+b^2+2 $$
$$(a-1)^2+(b-1)^2+0$$
Therefore
$$a = 1$$ and $$b=1$$
Substituting to (4) and (5) gives
$$x = 4$$ and $$y = 0$$
Therefor
$$x+y = 4$$

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