Mister Exam
|0,5x+1|<=0 inequation
The solution
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|x | |- + 1| <= 0 |2 |
$$\left|{\frac{x}{2} + 1}\right| \leq 0$$
|x/2 + 1| <= 0
Detail solution
Given the inequality:
$$\left|{\frac{x}{2} + 1}\right| \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{\frac{x}{2} + 1}\right| = 0$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$\frac{x}{2} + 1 \geq 0$$
or
$$-2 \leq x \wedge x < \infty$$
we get the equation
$$\frac{x}{2} + 1 = 0$$
after simplifying we get
$$\frac{x}{2} + 1 = 0$$
the solution in this interval:
$$x_{1} = -2$$
2.
$$\frac{x}{2} + 1 < 0$$
or
$$-\infty < x \wedge x < -2$$
we get the equation
$$- \frac{x}{2} - 1 = 0$$
after simplifying we get
$$- \frac{x}{2} - 1 = 0$$
the solution in this interval:
$$x_{2} = -2$$
but x2 not in the inequality interval
$$x_{1} = -2$$
$$x_{1} = -2$$
This roots
$$x_{1} = -2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-2 + - \frac{1}{10}$$
=
$$- \frac{21}{10}$$
substitute to the expression
$$\left|{\frac{x}{2} + 1}\right| \leq 0$$
$$\left|{\frac{-21}{2 \cdot 10} + 1}\right| \leq 0$$
but
Then
$$x \leq -2$$
no execute
the solution of our inequality is:
$$x \geq -2$$
$$\left|{\frac{x}{2} + 1}\right| \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{\frac{x}{2} + 1}\right| = 0$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$\frac{x}{2} + 1 \geq 0$$
or
$$-2 \leq x \wedge x < \infty$$
we get the equation
$$\frac{x}{2} + 1 = 0$$
after simplifying we get
$$\frac{x}{2} + 1 = 0$$
the solution in this interval:
$$x_{1} = -2$$
2.
$$\frac{x}{2} + 1 < 0$$
or
$$-\infty < x \wedge x < -2$$
we get the equation
$$- \frac{x}{2} - 1 = 0$$
after simplifying we get
$$- \frac{x}{2} - 1 = 0$$
the solution in this interval:
$$x_{2} = -2$$
but x2 not in the inequality interval
$$x_{1} = -2$$
$$x_{1} = -2$$
This roots
$$x_{1} = -2$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-2 + - \frac{1}{10}$$
=
$$- \frac{21}{10}$$
substitute to the expression
$$\left|{\frac{x}{2} + 1}\right| \leq 0$$
$$\left|{\frac{-21}{2 \cdot 10} + 1}\right| \leq 0$$
1/20 <= 0
but
1/20 >= 0
Then
$$x \leq -2$$
no execute
the solution of our inequality is:
$$x \geq -2$$
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