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|2x-5|>=2 inequation

A inequation with variable

The solution

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|2*x - 5| >= 2
$$\left|{2 x - 5}\right| \geq 2$$
|2*x - 5| >= 2
Detail solution
Given the inequality:
$$\left|{2 x - 5}\right| \geq 2$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{2 x - 5}\right| = 2$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.

1.
$$2 x - 5 \geq 0$$
or
$$\frac{5}{2} \leq x \wedge x < \infty$$
we get the equation
$$\left(2 x - 5\right) - 2 = 0$$
after simplifying we get
$$2 x - 7 = 0$$
the solution in this interval:
$$x_{1} = \frac{7}{2}$$

2.
$$2 x - 5 < 0$$
or
$$-\infty < x \wedge x < \frac{5}{2}$$
we get the equation
$$\left(5 - 2 x\right) - 2 = 0$$
after simplifying we get
$$3 - 2 x = 0$$
the solution in this interval:
$$x_{2} = \frac{3}{2}$$


$$x_{1} = \frac{7}{2}$$
$$x_{2} = \frac{3}{2}$$
$$x_{1} = \frac{7}{2}$$
$$x_{2} = \frac{3}{2}$$
This roots
$$x_{2} = \frac{3}{2}$$
$$x_{1} = \frac{7}{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{3}{2}$$
=
$$\frac{7}{5}$$
substitute to the expression
$$\left|{2 x - 5}\right| \geq 2$$
$$\left|{-5 + \frac{2 \cdot 7}{5}}\right| \geq 2$$
11/5 >= 2

one of the solutions of our inequality is:
$$x \leq \frac{3}{2}$$
 _____           _____          
      \         /
-------•-------•-------
       x2      x1

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq \frac{3}{2}$$
$$x \geq \frac{7}{2}$$
Solving inequality on a graph
Rapid solution [src]
Or(And(7/2 <= x, x < oo), And(x <= 3/2, -oo < x))
$$\left(\frac{7}{2} \leq x \wedge x < \infty\right) \vee \left(x \leq \frac{3}{2} \wedge -\infty < x\right)$$
((7/2 <= x)∧(x < oo))∨((x <= 3/2)∧(-oo < x))
Rapid solution 2 [src]
(-oo, 3/2] U [7/2, oo)
$$x\ in\ \left(-\infty, \frac{3}{2}\right] \cup \left[\frac{7}{2}, \infty\right)$$
x in Union(Interval(-oo, 3/2), Interval(7/2, oo))