sqrt(6x-11)<7 不等式

Given the inequality:
$$\sqrt{6 x - 11} < 7$$
To solve this inequality, we must first solve the corresponding equation:
$$\sqrt{6 x - 11} = 7$$
Solve:
Given the equation
$$\sqrt{6 x - 11} = 7$$
Because equation degree is equal to = 1/2 - does not contain even numbers in the numerator, then
the equation has single real root.
We raise the equation sides to 2-th degree:
We get:
$$\left(\sqrt{6 x - 11}\right)^{2} = 7^{2}$$
or
$$6 x - 11 = 49$$
Move free summands (without x)
from left part to right part, we given:
$$6 x = 60$$
Divide both parts of the equation by 6

x = 60 / (6)

We get the answer: x = 10

$$x_{1} = 10$$
$$x_{1} = 10$$
This roots
$$x_{1} = 10$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 10$$
=
$$\frac{99}{10}$$
substitute to the expression
$$\sqrt{6 x - 11} < 7$$
$$\sqrt{-11 + \frac{6 \cdot 99}{10}} < 7$$

     ____    
11*\/ 10     
--------- < 7
    5        
    

the solution of our inequality is:
$$x < 10$$

 _____          
      \    
-------ο-------
       x1