Mister Exam
(x-2):(x+5)>0 inequation
The solution
Detail solution
Given the inequality:
$$\frac{x - 2}{x + 5} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x - 2}{x + 5} = 0$$
Solve:
Given the equation:
$$\frac{x - 2}{x + 5} = 0$$
Multiply the equation sides by the denominator 5 + x
we get:
$$x - 2 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 2$$
$$x_{1} = 2$$
$$x_{1} = 2$$
This roots
$$x_{1} = 2$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$\frac{19}{10}$$
substitute to the expression
$$\frac{x - 2}{x + 5} > 0$$
$$\frac{-2 + \frac{19}{10}}{\frac{19}{10} + 5} > 0$$
Then
$$x < 2$$
no execute
the solution of our inequality is:
$$x > 2$$
$$\frac{x - 2}{x + 5} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x - 2}{x + 5} = 0$$
Solve:
Given the equation:
$$\frac{x - 2}{x + 5} = 0$$
Multiply the equation sides by the denominator 5 + x
we get:
$$x - 2 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 2$$
$$x_{1} = 2$$
$$x_{1} = 2$$
This roots
$$x_{1} = 2$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$\frac{19}{10}$$
substitute to the expression
$$\frac{x - 2}{x + 5} > 0$$
$$\frac{-2 + \frac{19}{10}}{\frac{19}{10} + 5} > 0$$
-1/69 > 0
Then
$$x < 2$$
no execute
the solution of our inequality is:
$$x > 2$$
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x1
Rapid solution
[src]
Or(And(-oo < x, x < -5), And(2 < x, x < oo))
$$\left(-\infty < x \wedge x < -5\right) \vee \left(2 < x \wedge x < \infty\right)$$
((-oo < x)∧(x < -5))∨((2 < x)∧(x < oo))
Rapid solution 2
[src]
(-oo, -5) U (2, oo)
$$x\ in\ \left(-\infty, -5\right) \cup \left(2, \infty\right)$$
x in Union(Interval.open(-oo, -5), Interval.open(2, oo))