Lineaire functies en vergelijkingen *: Eerstegraadsvergelijkingen oplossing *
Solve by reduction
Linear inequalities with one unknown can be solved by reduction.
Each inequality of the form #ax+b\ge0# with #a\ne0# can be rewritten as \[ \begin{cases}x\ge -\dfrac{b}{a}& \text{ als }a\gt 0\\ x\le -\dfrac{b}{a}& \text{ als }a\lt0\end {cases}\]
For this rules 4 and 5 of the Ordination of real numbers are used. We wil use the case #a\lt0# :
\[ \begin{array}{rclcl}ax+b\ge0&\Leftrightarrow&ax\ge -b &\phantom{x}&\color{blue}{\text{ substract }b \text{ on both sides}}\\&\Leftrightarrow&x\le -\dfrac{b}{a} &\phantom{x}&\color{blue}{\text{ divide by }a\text{ on both sides }}\\ \end {array}\]
Several inequalities can be put together using the operators' #\lor# 'and' #\land# ', which stand for 'or' and 'and'.
Using this we can reformulate the solution above as #ax+b\ge0#
\[\left(x\ge -\dfrac{b}{a}\land a\gt 0\right) \lor \left( x\le -\dfrac{b}{a}\land a\lt0\right)\tiny.\]
A set of inequalities is equivalent to another set if they have the exact same solutions. For example: #x\gt 5\wedge x\gt 2# is equivalent to #x\gt 5#.
Using #\Leftrightarrow# between two expressions, indicates that they are equivalent. For example:
#x\gt 5\wedge x\gt 2\Leftrightarrow x\gt 5#.
After All,
\[ \begin{array}{rclcl} \left(x\gt 9\land x\gt 10\right)\lor \left(x\gt 5\right) &\Leftrightarrow&x\gt 10\lor x>5&\phantom{x}&\color{blue}{\text{ }10=\max(9,10)}\\ &\Leftrightarrow&x\gt 5 &\phantom{x}& \color{blue}{\text{ }5=\min(10,5)} \\ \end {array}\]
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