 # 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
• Write down the solution set in interval form
Lets take one example to clarify the points
Question
$\frac{x - 3}{x + 5} > 0$
Solution
Method A
1. Lets find the end points of the equation
Here it is clearly
x=3 and x=-5
2. Now plots them on the Number line
3. 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
4. So the solution is
x < -5 or x > 3
or
$(-\infty,-5)\cup (3,\infty)$
Method B
1. 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\$
2. 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.
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.
4. The solution, therefore, is
x < -5 or x > 3