Mister Exam
(3/(x+2))-(2/(x-3))≥0 inequation
The solution
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3 2 ----- - ----- >= 0 x + 2 x - 3
$$\frac{3}{x + 2} - \frac{2}{x - 3} \geq 0$$
3/(x + 2) - 2/(x - 3) >= 0
Detail solution
Given the inequality:
$$\frac{3}{x + 2} - \frac{2}{x - 3} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{3}{x + 2} - \frac{2}{x - 3} = 0$$
Solve:
Given the equation:
$$\frac{3}{x + 2} - \frac{2}{x - 3} = 0$$
Use proportions rule:
From a1/b1 = a2/b2 should a1*b2 = a2*b1,
In this case
so we get the equation
$$3 \left(x - 3\right) = 2 \left(x + 2\right)$$
$$3 x - 9 = 2 x + 4$$
Move free summands (without x)
from left part to right part, we given:
$$3 x = 2 x + 13$$
Move the summands with the unknown x
from the right part to the left part:
$$x = 13$$
We get the answer: x = 13
$$x_{1} = 13$$
$$x_{1} = 13$$
This roots
$$x_{1} = 13$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 13$$
=
$$\frac{129}{10}$$
substitute to the expression
$$\frac{3}{x + 2} - \frac{2}{x - 3} \geq 0$$
$$- \frac{2}{-3 + \frac{129}{10}} + \frac{3}{2 + \frac{129}{10}} \geq 0$$
but
Then
$$x \leq 13$$
no execute
the solution of our inequality is:
$$x \geq 13$$
$$\frac{3}{x + 2} - \frac{2}{x - 3} \geq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{3}{x + 2} - \frac{2}{x - 3} = 0$$
Solve:
Given the equation:
$$\frac{3}{x + 2} - \frac{2}{x - 3} = 0$$
Use proportions rule:
From a1/b1 = a2/b2 should a1*b2 = a2*b1,
In this case
a1 = 3
b1 = 2 + x
a2 = 2
b2 = -3 + x
so we get the equation
$$3 \left(x - 3\right) = 2 \left(x + 2\right)$$
$$3 x - 9 = 2 x + 4$$
Move free summands (without x)
from left part to right part, we given:
$$3 x = 2 x + 13$$
Move the summands with the unknown x
from the right part to the left part:
$$x = 13$$
We get the answer: x = 13
$$x_{1} = 13$$
$$x_{1} = 13$$
This roots
$$x_{1} = 13$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 13$$
=
$$\frac{129}{10}$$
substitute to the expression
$$\frac{3}{x + 2} - \frac{2}{x - 3} \geq 0$$
$$- \frac{2}{-3 + \frac{129}{10}} + \frac{3}{2 + \frac{129}{10}} \geq 0$$
-10 ----- >= 0 14751
but
-10 ----- < 0 14751
Then
$$x \leq 13$$
no execute
the solution of our inequality is:
$$x \geq 13$$
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Solving inequality on a graph
Rapid solution
[src]
Or(And(-2 < x, x < 3), 13 <= x)
$$\left(-2 < x \wedge x < 3\right) \vee 13 \leq x$$
(13 <= x)∨((-2 < x)∧(x < 3))
Rapid solution 2
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(-2, 3) U [13, oo)
$$x\ in\ \left(-2, 3\right) \cup \left[13, \infty\right)$$
x in Union(Interval.open(-2, 3), Interval(13, oo))