How to solve the linear inequalities in one variable
Steps to solve the Linear inequalities in one variable.
Obtain the linear inequation
Pull all the terms having variable on one side and all the constant term on another side of the inequation
Simplify the equation in the form given above
$ ax> b $
or
$ax \geq b$
or
$ax< b $
or $ax \leq b$
Divide the coefficent of the variable on the both side.If the coefficent is positive,direction of the inequality does not changes,but if it is negative, direction of the inequation changed
Put the result of this equation on number line and get the solution set in interval form
Example:
$ x- 2 > 2x+15 $ Solution
$x-2 > 2x+15$
Subtract x from both the side
$x-2 -x > 2x+15 -x $
$-2 >x +15$
Subtract 15 from both the sides
$-17 > x $
Solution set
$(-\infty,-17)$
Some Problems to practice
$2x > 9 $
$x + 5 > 111 $
$3x < 4 $
$2(x + 3) < x+ 1$
Steps to solve the linear inequality of the fraction form
$ \frac {ax+b}{cx+d} > k $ or similar type
Take k on the LHS
Simplify LHS to obtain the inequation in the form
$ \frac {px+q}{ex+f} > 0 $
Make the coefficent positive if not
Find out the end points solving the equatoon $px+q=0 $ and $ex+f=0$
Plot these numbers on the Number line. This divide the number into three segment
Start from LHS side of the number and Substitute some value in the equation in all the three segments to find out which segments satisfy the equation
Write down the solution set in interval form
Or there is more method to solves these
Take k on the LHS
Simplify LHS to obtain the inequation in the form
$ \frac {px+q}{ex+f} > 0 $
Make the coefficent positive if not
For the equation to satisfy both the numerator and denominator must have the same sign
So taking both the part +, find out the variable x interval
So taking both the part -, find out the variable x interval
Lets find the end points of the equation
Here it is clearly
x=3 and x=-5
Now plots them on the Number line
Now lets start from left part of the most left number
i.e Case 1
$x < -5 $ ,Let takes x=-6 then $\frac {-6-3}{-6+5} > 0$
$3 > 0$
So it is good Case 2
Now take x=-5
as x+5 becomes zero and we cannot have zero in denominator,it is not the solution Case 3
Now x > -5 and x < 3, lets take x=1 then $\frac {1-3}{1+5} > 0$
$\frac {-1}{6} > 0 $
Which is not true Case 4
Now take x =3,then
0> 0 ,So this is also not true case 5
x> 3 ,Lets x=4
$\frac {4-3}{4+5} > 0 $
$\frac {1}{9} > 0 $
So this is good
So the solution is
x < -5 or x > 3
or
$(-\infty,-5)\cup (3,\infty)$
Method B
the numerator and denominator must have the same sign. Therefore, either
1) $x - 3 > 0$ and $x + 5 > 0,
or
2) $x - 3 < 0$ and $x + 5 < 0$
Now, 1) implies x > 3 and x > -5.
Which numbers are these that are both greater than 3 and greater than -5?
Clearly, any number greater than 3 will also be greater than -5. Therefore, 1) has the solution
x > 3.
Next, 2) implies
x < 3 and x < -5.
Which numbers are these that are both less than 3 and less than -5?
Clearly, any number less than -5 will also be less than 3. Therefore, 2) has the solution
x < -5.
