-6(x+1)-(2+x)_>0 inequation

Given the inequality:
$$\left(- x - 2\right) - 6 \left(x + 1\right) \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- x - 2\right) - 6 \left(x + 1\right) = 0$$
Solve:
Given the linear equation:

-6*(x+1)-(2+x) = 0

Expand brackets in the left part

-6*x-6*1-2-x = 0

Looking for similar summands in the left part:

-8 - 7*x = 0

Move free summands (without x)
from left part to right part, we given:
$$- 7 x = 8$$
Divide both parts of the equation by -7

x = 8 / (-7)

$$x_{1} = - \frac{8}{7}$$
$$x_{1} = - \frac{8}{7}$$
This roots
$$x_{1} = - \frac{8}{7}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{8}{7} + - \frac{1}{10}$$
=
$$- \frac{87}{70}$$
substitute to the expression
$$\left(- x - 2\right) - 6 \left(x + 1\right) \geq 0$$
$$\left(-2 - - \frac{87}{70}\right) - 6 \left(- \frac{87}{70} + 1\right) \geq 0$$

7/10 >= 0

the solution of our inequality is:
$$x \leq - \frac{8}{7}$$

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