Could xᵢ = x̅ ???

Question

I know it's a weird question.
But this thing is confusing me.
(x̅) : average μ
    

①
∵ $\frac{1}{n}\sum\limits_{i=1}^n(x_i) = \bar{x}$
∴ $\sum\limits_{i=1}^n(x_i) = {n}\bar{x}$
    

②
∵ $\sum\limits_{i=1}^n(C) = {n}{C}$    | C = constant
∴ $\sum\limits_{i=1}^n(\bar{x}) = {n}{\bar{x}}$     

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From ①, ②
$(x_i) = (\bar{x})$ ???

How could this be true?? Am I missing something?

Answer 1

$$1+3=4$$ $$2+2=4$$

We can't conclude that $1=2$.

Answer 2

$\bar x$ is the average, so it's just some number that you can factor out of the summand:

$$\sum_{i=1}^n\bar x=\bar x\sum_{i=1}^n1=\bar x\left(\underbrace{1+1+\cdots+1}_{n\text{ times}}\right)=\bar x n$$