Incorrect General Statement for Modulus Inequalities $$|x + 1| = x$$
This, quite evidently, has no solution.
Through solving many inequalities, I came to the conclusion that,

If,
$$|f(x)| = g(x)$$
Then,
$$f(x) = ±g(x)$$

And this was quite successful in solving many inequalities. However, applying the above to this particular inequality:
$$|x + 1| = x$$
$$x + 1 = ±x$$
When ± is +, there is no solution. However, when ± is -:
$$x + 1 = -x$$
$$2x = -1$$
$$x = -\frac{1}{2}$$
Which does not make sense. What is wrong with the supposedly general statement that seemed to work flawlessly?
 A: The definition of the absolute value is
$$
  \lvert x \rvert = 
  \begin{cases}
    x & \text{if $x\geqslant 0$} \\
    -x & \text{if $x < 0$},
  \end{cases}
$$
so
$$
  \lvert f(x) \rvert = g(x)
  \iff
  \begin{cases}
     f(x) = g(x) \\
     f(x) \geqslant 0
  \end{cases}
  \qquad\text{or}\qquad
  \begin{cases}
     -f(x) = g(x) \\ 
     f(x) < 0.
  \end{cases}
$$
In particular,
$$
\begin{align*}
  \lvert x+1 \rvert = x
  &\iff
  \begin{cases}
     x + 1 = x \\
     x + 1 \geqslant 0
  \end{cases}
  \qquad\text{or}\qquad
  \begin{cases}
    - (x + 1) = x \\
    x + 1 < 0
  \end{cases} \\
  &\iff
  \begin{cases}
    1 = 0 \\
    x \geqslant -1
  \end{cases}
  \qquad\text{or}\qquad
  \begin{cases}
     x = -1/2 \\
     x < - 1.
  \end{cases}
\end{align*}
$$
As $1 = 0$ and $-1/2 < -1$ are not true, we conclude that the equation $\lvert x+1 \rvert = x$ has no real solutions.
A: solving ∣x+1∣=x amounts to finding A={x:∣x+1∣=x}. Now A⊂{x:x+1=x}∪{x:x+1=-x}=φ∪{1/2}={1/2}. Thus A⊂{1/2},A=φ. This is possible.
