Linear inequality in one variable
A linear inequality in one variable is of the form:$ax+b\lt c$
$ax+b\le c$
$ax+b\gt c$
or$ax+b\ge c$
where $x$ is a variable, $a$,$b$ and $c$ are constants, i.e., some numbers.
Addition and subtraction properties of inequality
If $a\lt b$ then$a+c\lt b+c$
and$a-c\lt b-c$
Multiplication and division properties of inequality
For any real numbers $a, b$ and $c$,If $a\lt b$ then
$ac\lt bc$
and$\dfrac{a}{c}\lt \dfrac{b}{c}$ if $c$ is positive
$\dfrac{a}{c}\gt \dfrac{b}{c}$ if $c$ is negative
Compound inequalities
If we have two or more sets, we can combine them with union and intersection operations.If $A$ and $B$ are two sets, then the union of $A$ and $B$ is the set of elements which are in $A$ or in $B$ or in both $A$ and $B$.
Union of $A$ and $B$ or $A$ union $B$ is denoted as $A\cup B$
If $A$ and $B$ are two sets, then the intersection of $A$ and $B$ is the set of elements which are common to both $A$ and $B$.
Intersection of $A$ and $B$ or $A$ intersection $B$ is denoted as $A\cap B$
$A\cup B$
Absolute value equation
Absolute value of a number is the distance of the number from zero on the number line. Since distance is always positive, absolute value is positive. Absolute value of a number, $A$ is written as $|A|$.Solving absolute value equation
If $|x|=a$ then $x=a$ or $x=-b$ if $a$ is positive or zero.If $|x|=a$ then there is no solution if $a$ is negative.
Solving absolute value inequalities
For any positive real number, $a$, i.e., $a>0$,If $|x|\gt a$, then $x\lt -a\:$ or $\:x\gt a$
If $|x|\lt a$, then $-a\lt x\lt a$
Solving inequalities- test point method
There are four steps in this method.- Find the boundary points of the inequality. To find that replace the inequality sign with the equal sign and find the solutions. The solutions are the boundary points.
- Plot the boundary points on a number line. You will have different intervals in the number line.
- Pick a test point from each interval and substitute it into the inequality. If the test point makes the inequality true, then