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(3/(x+2))-(2/(x-3))≥0 inequation

You have entered [src]

  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

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

         _____  
        /
-------•-------
       x1

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 [src]

(-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))