Mister Exam
3/|x-1|<1 inequation
The solution
Detail solution
Given the inequality:
$$\frac{3}{\left|{x - 1}\right|} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{3}{\left|{x - 1}\right|} = 1$$
Solve:
$$x_{1} = -2$$
$$x_{2} = 4$$
$$x_{1} = -2$$
$$x_{2} = 4$$
This roots
$$x_{1} = -2$$
$$x_{2} = 4$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-2 + - \frac{1}{10}$$
=
$$-2.1$$
substitute to the expression
$$\frac{3}{\left|{x - 1}\right|} < 1$$
$$\frac{3}{\left|{-2.1 - 1}\right|} < 1$$
one of the solutions of our inequality is:
$$x < -2$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -2$$
$$x > 4$$
$$\frac{3}{\left|{x - 1}\right|} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{3}{\left|{x - 1}\right|} = 1$$
Solve:
$$x_{1} = -2$$
$$x_{2} = 4$$
$$x_{1} = -2$$
$$x_{2} = 4$$
This roots
$$x_{1} = -2$$
$$x_{2} = 4$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-2 + - \frac{1}{10}$$
=
$$-2.1$$
substitute to the expression
$$\frac{3}{\left|{x - 1}\right|} < 1$$
$$\frac{3}{\left|{-2.1 - 1}\right|} < 1$$
0.967741935483871 < 1
one of the solutions of our inequality is:
$$x < -2$$
_____ _____
\ /
-------ο-------ο-------
x1 x2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < -2$$
$$x > 4$$
Solving inequality on a graph
Rapid solution 2
[src]
(-oo, -2) U (4, oo)
$$x\ in\ \left(-\infty, -2\right) \cup \left(4, \infty\right)$$
x in Union(Interval.open(-oo, -2), Interval.open(4, oo))
Rapid solution
[src]
Or(And(-oo < x, x < -2), And(4 < x, x < oo))
$$\left(-\infty < x \wedge x < -2\right) \vee \left(4 < x \wedge x < \infty\right)$$
((-oo < x)∧(x < -2))∨((4 < x)∧(x < oo))