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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

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