Mister Exam
(x+5):(x-3)>0 inequation
The solution
Detail solution
Given the inequality:
$$\frac{x + 5}{x - 3} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x + 5}{x - 3} = 0$$
Solve:
Given the equation:
$$\frac{x + 5}{x - 3} = 0$$
Multiply the equation sides by the denominator -3 + x
we get:
$$x + 5 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -5$$
$$x_{1} = -5$$
$$x_{1} = -5$$
This roots
$$x_{1} = -5$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-5 + - \frac{1}{10}$$
=
$$- \frac{51}{10}$$
substitute to the expression
$$\frac{x + 5}{x - 3} > 0$$
$$\frac{- \frac{51}{10} + 5}{- \frac{51}{10} - 3} > 0$$
the solution of our inequality is:
$$x < -5$$
$$\frac{x + 5}{x - 3} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x + 5}{x - 3} = 0$$
Solve:
Given the equation:
$$\frac{x + 5}{x - 3} = 0$$
Multiply the equation sides by the denominator -3 + x
we get:
$$x + 5 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = -5$$
$$x_{1} = -5$$
$$x_{1} = -5$$
This roots
$$x_{1} = -5$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-5 + - \frac{1}{10}$$
=
$$- \frac{51}{10}$$
substitute to the expression
$$\frac{x + 5}{x - 3} > 0$$
$$\frac{- \frac{51}{10} + 5}{- \frac{51}{10} - 3} > 0$$
1/81 > 0
the solution of our inequality is:
$$x < -5$$
_____
\
-------ο-------
x1
Solving inequality on a graph
Rapid solution
[src]
Or(And(-oo < x, x < -5), And(3 < x, x < oo))
$$\left(-\infty < x \wedge x < -5\right) \vee \left(3 < x \wedge x < \infty\right)$$
((-oo < x)∧(x < -5))∨((3 < x)∧(x < oo))
Rapid solution 2
[src]
(-oo, -5) U (3, oo)
$$x\ in\ \left(-\infty, -5\right) \cup \left(3, \infty\right)$$
x in Union(Interval.open(-oo, -5), Interval.open(3, oo))