Solve the following inequality:

$$
\frac{2 x}{|x-3|-5}+\frac{1}{x+2} \geq 1
$$
$$
\frac{2 x}{|x-3|-5}+\frac{1}{x+2} \geq 1
$$

Due to the appearance of the $|x-3|$ term, we need to distinguish two cases.

I. If $x-3 \geq 0$, that is, $x \geq 3$, then the first fraction in (1) can also be written as:

$$
\frac{2 x}{x-3-5}=2+\frac{16}{x-8}
$$

and with this, (1) can be transformed into:

$$
\frac{16}{x-8} \geq-1-\frac{1}{x+2}
$$

The denominator on the right side is $x+2 \geq 5$, so in the entire interval

$$
-1>-1-\frac{1}{x+2} \geq-1-\frac{1}{5}=-1.2
$$

However, the zero point of the left side divides the interval into two parts. If

$$
x>8,
$$

then the left side is positive, which according to (2) and (3) is a solution to the inequality. If, on the other hand, $3 \leq x < 8$, then the left side is negative, and the inequality does not hold.

II. If $x-3 < 0$, that is, $x < 3$, then the first fraction in (1) can be written as:

$$
\frac{2 x}{-(x-3)-5}=\frac{2 x}{-x-2}=-2+\frac{4}{x+2}
$$

and with this, (1) can be transformed into:

$$
-2+\frac{4}{x+2}+\frac{1}{x+2} \geq 1
$$

Simplifying, we get:

$$
\frac{4}{x+2}+\frac{1}{x+2} \geq 3
$$

$$
\frac{5}{x+2} \geq 3
$$

$$
5 \geq 3(x+2)
$$

$$
5 \geq 3x + 6
$$

$$
-1 \geq 3x
$$

$$
x \leq -\frac{1}{3}
$$

Combining the results from both cases, the solutions to the inequality (1) are:

$$
x \leq -\frac{1}{3} \quad \text{or} \quad x > 8
$$

Remark. The graph shows the variation of the left side of (1). The curve is continuous at the point $x=3$ because $x-3$ is continuous there, but there is a cusp at that point.

![](https://cdn.mathpix.com/cropped/2024_05_02_1fc58b59470ad360b79ag-1.jpg?height=639&width=1152&top_left_y=1941&top_left_x=476)