3x/4+1/3x≤-9/2 inequation

Given the inequality:
$$\frac{x}{3} + \frac{3 x}{4} \leq - \frac{9}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x}{3} + \frac{3 x}{4} = - \frac{9}{2}$$
Solve:
Given the linear equation:

3*x/4+1/3*x = -9/2

Looking for similar summands in the left part:

13*x/12 = -9/2

Divide both parts of the equation by 13/12

x = -9/2 / (13/12)

$$x_{1} = - \frac{54}{13}$$
$$x_{1} = - \frac{54}{13}$$
This roots
$$x_{1} = - \frac{54}{13}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{54}{13} + - \frac{1}{10}$$
=
$$- \frac{553}{130}$$
substitute to the expression
$$\frac{x}{3} + \frac{3 x}{4} \leq - \frac{9}{2}$$
$$\frac{\left(- \frac{553}{130}\right) 3}{4} + \frac{-553}{3 \cdot 130} \leq - \frac{9}{2}$$

-553         
----- <= -9/2
 120         

the solution of our inequality is:
$$x \leq - \frac{54}{13}$$

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