We are asked to simplify the expression $y = \frac{|x+1| - |x-1|}{2}$.

1. Problem Description

We are asked to simplify the expression $y = \frac{|x+1| - |x-1|}{2}$.

2. Solution Steps

We need to consider different cases based on the values of $x$.

Case 1: $x < -1$

In this case, $x+1 < 0$ and $x-1 < 0$, so $|x+1| = -(x+1)$ and $|x-1| = -(x-1)$.

Therefore,

$y = \frac{-(x+1) - (-(x-1))}{2} = \frac{-x-1 + x - 1}{2} = \frac{-2}{2} = -1$.

Case 2: $-1 \le x < 1$

In this case, $x+1 \ge 0$ and $x-1 < 0$, so $|x+1| = x+1$ and $|x-1| = -(x-1)$.

Therefore,

$y = \frac{(x+1) - (-(x-1))}{2} = \frac{x+1 + x - 1}{2} = \frac{2x}{2} = x$.

Case 3: $x \ge 1$

In this case, $x+1 > 0$ and $x-1 \ge 0$, so $|x+1| = x+1$ and $|x-1| = x-1$.

Therefore,

$y = \frac{(x+1) - (x-1)}{2} = \frac{x+1 - x + 1}{2} = \frac{2}{2} = 1$.

In summary:

$y = -1$ if $x < -1$

$y = x$ if $-1 \le x < 1$

$y = 1$ if $x \ge 1$

We can rewrite the solution as:

$y = \begin{cases} -1, & x < -1 \\ x, & -1 \le x < 1 \\ 1, & x \ge 1 \end{cases}$

3. Final Answer

$y = \begin{cases} -1, & x < -1 \\ x, & -1 \le x < 1 \\ 1, & x \ge 1 \end{cases}$