Modulus inequality

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We have learnt how to solve modulus equality,
now, we are going to learn how to solve modulus inequality.

Consider,

|x| \le 5

i) for x \le 0

- x \le 5

or, x \ge - 5

therefore,  - 5 \le x \le 0  ( because we take x \le 0 )   ......................(i)

ii) for,, x \ge 0

x \le 5

therefore, 0 \le x \le 5...........................(ii)

combining (i) and (ii)
x \in [ - 5,0]\bigcup {[0,5]}
or, x \in [ - 5,5], \forall x ( \forall  means "for every")

you can directly solve this inequality, first to put -x in place of |x| then x in place of |x|
|x| \le 5
or,  - x \le 5  \Rightarrow x \ge - 5

now, x \le 5
therefore, x \in [ - 5,5]

from graph,
|x| \le 5, 
we have two curves , y = |x| and y = 5, and we have to find those values of 'x' for which y = |x| is less than or equal to y = 5

clearly from graph, between -5 and 5 curve |x| is less than to 5

so immediately the answer will be , x \in [ - 5,5]

2. Suppose, question  is |x| \ge 5
clearly from graph, for x \le - 5, curve of |x| is greater than or equal to 5 
and for x \ge 5 , |x| is also greater than 5

so, x \in ( - \infty , - 5]\bigcup {[5,\infty )}

3. |x - 1| < 2

|x-1| consists of two lines 1-x and x-1 
1 - x < 2
- x < 1
or, x > - 1

and, x - 1 < 2
or, x < 3
therefore,  - 1 < x < 3
or, x \in ( - 1,3)

4.  3|x - 1| - |2x - 5| \le 2|x - 3|

There are three critical points 1 ( for |x - 1|), 5/2 (for |2x - 5|) and 3 (for |x - 3|) 
follw the post of solving the modulus equality
http://iitmathseasy.blogspot.in/2013/03/modulus-equality-solving-concept.html

 Step by step process, take from left


i) x \le 1
ii) 1 \le x \le \frac{5}{2}
iii)\frac{5}{2} \le x \le 3
iv) x \ge 3

 i) x \le 1,

 3(1 - x) - (5 - 2x) \le 2(3 - x)

 3 - 3x - 5 + 2x \le 6 - 2x

 or, x \le 8

 since we take x \le 1,
therefore, we take those values of x \le 8 which are lie in  x \le 1, i.e. intersection of both the intervals 
( - \infty ,8] and ( - \infty ,1]

 or, ( - \infty ,8]\bigcap {( - \infty ,1]}

  \Rightarrow  x \le 1 or x \in ( - \infty ,1]...............................................................(a)

 ii) 1 \le x \le \frac{5}{2}

 3(x - 1) - (5 - 2x) \le 2(3 - x)

3x - 3 - 5 + 2x \le 6 - 2x

 7x \le 14

 or, x \le 2

 we have to take intersection of [1,\frac{5}{2}] and ( - \infty ,2] i.e. only those values which lies on 1 \le x \le \frac{5}{2}
or, [1,\frac{5}{2}]\bigcap {( - \infty ,2]}

therefore,  1 \le x \le 2 or x \in [1,2]  ................................................... (b)

 iii) \frac{5}{2} \le x \le 3

 3(x - 1) - (2x - 5) \le 2(3 - x)

 3x - 3 - 2x + 5 \le 6 - 2x

 3x \le 4

 or, x \le \frac{4}{3}

 therefore, solution x \in ( - \infty ,\frac{4}{3}]\bigcap {\left[ {\frac{5}{2},3} \right]}
or,  x \in \emptyset   , therefore no solution in this interval............................................................................ (c)

 iv)x \ge 3

 3(x - 1) - (2x - 5) \le 2(x - 3)

 3x - 3 - 2x + 5 \le 2x - 6

 - x \le - 8

 x \ge 8

 therefore solution, x \in [8,\infty )\bigcap {[3,\infty )}
or, x \in [8,\infty )  ......................................................................(d)
since (a) , (b), (c) and (d) are solutions of inequality indifferent intervals ,
therefore, considering all solution is
x \in ( - \infty ,1]\bigcup {[1,2]} \bigcup {[8,\infty ]}

or x \in ( - \infty ,2]\bigcup {[8,\infty ]}   ( since  ( - \infty ,1]\bigcup {[1,2]} = ( - \infty ,2] )




















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