Given the inequality:
$$\frac{5 - 6 x}{3 x + 4} > 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{5 - 6 x}{3 x + 4} = 1$$
Solve:
Given the equation:
$$\frac{5 - 6 x}{3 x + 4} = 1$$
Multiply the equation sides by the denominator 4 + 3*x
we get:
$$5 - 6 x = 3 x + 4$$
Move free summands (without x)
from left part to right part, we given:
$$- 6 x = 3 x - 1$$
Move the summands with the unknown x
from the right part to the left part:
$$\left(-9\right) x = -1$$
Divide both parts of the equation by -9
x = -1 / (-9)
$$x_{1} = \frac{1}{9}$$
$$x_{1} = \frac{1}{9}$$
This roots
$$x_{1} = \frac{1}{9}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{9}$$
=
$$\frac{1}{90}$$
substitute to the expression
$$\frac{5 - 6 x}{3 x + 4} > 1$$
$$\frac{5 - \frac{6}{90}}{\frac{3}{90} + 4} > 1$$
148
--- > 1
121
the solution of our inequality is:
$$x < \frac{1}{9}$$
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