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
- 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.
- The solution, therefore, is
x < -5 or x > 3
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