Hey how would I solve this? (it's absolute values if the writing doesn't make it clear!)
|(x-1)/(x+1)| <1
\[ \text{The solution is equivalent to that of}\\ |x-1| < |x+1| \]
(However, note that \(x \neq -1\) because otherwise in the original equation we have dividing by 0 issues.)
Recall
The definition of the absolute value states that
\[ |x| = \begin{cases} x & \text{if }x\geq 0\\ -x & \text{if }x < 0\end{cases} \]
Therefore we have
\begin{align*} |x-1| &= \begin{cases} x-1 & \text{if }x-1 \geq 0\\ -(x-1) & \text{if }x-1 < 0\end{cases} \\ &= \begin{cases} x-1 & \text{if }x \geq 1\\ -x+1 &\text{if }x < 1\end{cases} \end{align*}
And similarly
\[ |x+1| = \begin{cases} x+1 &\text{ if }x \geq -1\\ -x-1 &\text{if }x < -1\end{cases} \]
\[ \text{For this inequality, noting the above}\\ \text{we can accordingly split the following cases.} \]
\[ \text{Case 1: } x < -1.\\ \text{For this case the inequality becomes}\\ \begin{align*} -x+1 &< -x-1\\ 1 &< -1\end{align*}\\ \text{This obviously has no solution}\\ \text{so no values of }x\text{ in this case belong in the solution.} \]
\[ \text{Case 2: }-1\leq x < 1.\\ \text{For this case the inequality becomes}\\ \begin{align*}-x+1 &< x+1\\ 0 &< 2x\\ x &> 0 \end{align*}\]
\[ \text{So overlapping }x > 0\text{ with }-1\leq x < 1\\ \text{we see that }\boxed{0 < x < 1}\text{ is a part of the solution.}\]
\[ \text{Case 3: }x\geq 1.\\ \text{For this case the inequality becomes}\\ \begin{align*}x-1 &< x+1\\ -1 &< 1 \end{align*}\]
\[ \text{This is obviously true for every value of }x.\\ \text{So all values of }x\text{ in this case, i.e. }\\\boxed{ x\geq 1}\text{, is a part of the solution.}\]
\[ \text{Combining all parts, our solution is thus}\\ \boxed{x > 0}.\]