Solving a System of Absolute Value Equations: When Does |x+2|=0?

$\begin{cases} |x+2|=|x-2| \\ \lvert x+2\rvert=0 \end{cases}$

To solve this problem, we'll follow these steps:

• Step 1: Solve the equation $|x+2| = 0$
  This implies $x+2 = 0$, hence $x = -2$.
• Step 2: Solve the equation $|x+2| = |x-2|$.
  This leads to two cases:
  - Case 1: $x+2 = x-2$ results in $2 = -2$, which is false.
  - Case 2: $x+2 = -(x-2)$ simplifies to $2x = 0$, hence $x = 0$.
• Step 3: Check for intersection of solutions.
  We have solutions $x = -2$ from the first equation and $x = 0$ from the absolute value equation.
• Step 4: Verify if any common $x$ satisfies both equations.

Substituting $x = -2$ into $|x+2| = |x-2|$, we find that it does not satisfy $|x-2| = |-4| = 4$. Therefore, $x = -2$ is not a solution to the system.

Substituting $x = 0$ into both equations does not satisfy $|x+2| = 0$. Therefore, there is no overlap of solutions.

The solution to the system is No solution.