To solve:

$$
\frac{x-1}{x+1}>\frac{x}{x-1}
$$
I. Solution. We move every part of the inequality to the left side and bring it to a common denominator, thus we get:

$$
\begin{gathered}
\frac{(x-1)^{2}-x(x+1)}{x^{2}-1}>0 \quad \text { or } \quad-\frac{3 x-1}{(x+1)(x-1)}>0 \\
\text { or } \frac{3 x-1}{(x+1)(x-1)}<0 \text{, so } x \text{ cannot be between -1 and +1. The two conditions are } x<0 \text{, i.e., } x<0 \text{, and } -1<x<1.
$$

We can also write it in the form:

$$
y_{1}=1-3 x \quad \text { and } \quad y_{2}=x^{2}-1
$$

functions. The solutions to the inequality are all $x$ values where both function graphs lie below or above the $X$-axis.

The function $y_{1}=1-3 x$ corresponds to a straight line passing through the points $(0,1)$ and $\left(\frac{1}{3}, 0\right)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9e7fbf08b0c346405710g-1.jpg?height=735&width=637&top_left_y=1416&top_left_x=730)

The image of the function $y_{2}=x^{2}-1$ is a parabola with its vertex at the point $(0,-1)$ and this is a lower vertex.

As we can see, both graphs are above the $X$-axis at the points where $x<-1$, and below the $X$-axis at the points where $\frac{1}{3}<x<1$.

László Sebők (Bencés g. V. o. Győr.)