Question

# Number of real ordered pair $$(x, y)$$ satisfying $$x^{2}+1=y$$ and $$y^{2}+1=x$$ is

A
0
B
1
C
2
D
4

Solution

## The correct option is A $$0$$$$x^{2}+1=y$$ and $$y^{2}+1=x$$Subtracting $$x^{2}-y^{2}=y-x$$$$\Rightarrow \left ( x-y \right )\left ( x+y+1 \right )=0$$$$\Rightarrow x=y$$ or $$x+y+1 =0$$If $$x=y$$ then $$x^{2}+1=y \Rightarrow x^{2}-x+1=0$$ which has no real roots.If $$x + y + 1 = 0$$ then $$x + y = -1$$Adding given equations, $$x^{2}+y^{2}+2=x+y$$$$\Rightarrow x^{2}+y^{2}+2=-1 \Rightarrow x^{2}+y^{2}+3=0$$ which is not possible.Mathematics

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