-(3+x):x>1 inequation

Given the inequality:
$$\frac{- x - 3}{x} > 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{- x - 3}{x} = 1$$
Solve:
Given the equation:
$$\frac{- x - 3}{x} = 1$$
Multiply the equation sides by the denominator x
we get:
$$- x - 3 = x$$
Move free summands (without x)
from left part to right part, we given:
$$- x = x + 3$$
Move the summands with the unknown x
from the right part to the left part:
$$\left(-2\right) x = 3$$
Divide both parts of the equation by -2

x = 3 / (-2)

$$x_{1} = - \frac{3}{2}$$
$$x_{1} = - \frac{3}{2}$$
This roots
$$x_{1} = - \frac{3}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{3}{2} + - \frac{1}{10}$$
=
$$- \frac{8}{5}$$
substitute to the expression
$$\frac{- x - 3}{x} > 1$$
$$\frac{-3 - - \frac{8}{5}}{- \frac{8}{5}} > 1$$

7/8 > 1

Then
$$x < - \frac{3}{2}$$
no execute
the solution of our inequality is:
$$x > - \frac{3}{2}$$

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