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(x+1)*(7-x)/(8+x)*(x-5)<0 inequation

A inequation with variable

The solution

You have entered [src]
(x + 1)*(7 - x)            
---------------*(x - 5) < 0
     8 + x                 
$$\frac{\left(7 - x\right) \left(x + 1\right)}{x + 8} \left(x - 5\right) < 0$$
(((7 - x)*(x + 1))/(x + 8))*(x - 5) < 0
Detail solution
Given the inequality:
$$\frac{\left(7 - x\right) \left(x + 1\right)}{x + 8} \left(x - 5\right) < 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\left(7 - x\right) \left(x + 1\right)}{x + 8} \left(x - 5\right) = 0$$
Solve:
$$x_{1} = -1$$
$$x_{2} = 5$$
$$x_{3} = 7$$
$$x_{1} = -1$$
$$x_{2} = 5$$
$$x_{3} = 7$$
This roots
$$x_{1} = -1$$
$$x_{2} = 5$$
$$x_{3} = 7$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$\frac{\left(7 - x\right) \left(x + 1\right)}{x + 8} \left(x - 5\right) < 0$$
$$\frac{\left(- \frac{11}{10} + 1\right) \left(7 - - \frac{11}{10}\right)}{- \frac{11}{10} + 8} \left(-5 + - \frac{11}{10}\right) < 0$$
1647    
---- < 0
2300    

but
1647    
---- > 0
2300    

Then
$$x < -1$$
no execute
one of the solutions of our inequality is:
$$x > -1 \wedge x < 5$$
         _____           _____  
        /     \         /
-------ο-------ο-------ο-------
       x1      x2      x3

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x > -1 \wedge x < 5$$
$$x > 7$$
Solving inequality on a graph
Rapid solution [src]
Or(And(-oo < x, x < -8), And(-1 < x, x < 5), And(7 < x, x < oo))
$$\left(-\infty < x \wedge x < -8\right) \vee \left(-1 < x \wedge x < 5\right) \vee \left(7 < x \wedge x < \infty\right)$$
((-oo < x)∧(x < -8))∨((-1 < x)∧(x < 5))∨((7 < x)∧(x < oo))
Rapid solution 2 [src]
(-oo, -8) U (-1, 5) U (7, oo)
$$x\ in\ \left(-\infty, -8\right) \cup \left(-1, 5\right) \cup \left(7, \infty\right)$$
x in Union(Interval.open(-oo, -8), Interval.open(-1, 5), Interval.open(7, oo))