Step 1 - Combine the left-hand side into one fraction


\(\frac{1}{x-1}-\frac{1}{x+1}=\frac{(x+1)-(x-1)}{(x-1)(x+1)}=\frac{2}{x^2-1}\)



Step 2 - Set equal to \(\frac{2}{3}\)


\(\frac{2}{x^2-1}=\frac{2}{3}\)



Step 3 - Solve for x

Cancel the 2's:


\(\frac{1}{x^2-1}=\frac{1}{3}\)



Since the numerators are equal, equate denominators:


\(x^2-1=3 \quad \Rightarrow \quad x^2=4 \quad \Rightarrow \quad x= \pm 2\)



Step 4 - Check for extraneous solutions

- \(x=2 \rightarrow\) denominators: 1 and \(3 \rightarrow\) valid
- \(x=-2 \rightarrow\) denominators: -3 and \(-1 \rightarrow\) valid

Both are valid.

Step 5 - Match to answer choices

The options are: -4,-2,-1,0,1
Only -2 appears in the list.

\(-2\)