Solve each of the following systems of equations in R.

Question 1: \Big| x+ \frac{1}{3} \Big| > \frac{8}{3}

Answer:

Given \Big| x+ \frac{1}{3} \Big| > \frac{8}{3}

Case I: x + \frac{1}{3} > \frac{8}{3} \Rightarrow x > \frac{-1}{3} + \frac{8}{3} \Rightarrow x > \frac{7}{3}

Case II: x + \frac{1}{3} < \frac{-8}{3} \Rightarrow  x < \frac{-1}{3} - \frac{8}{3} \Rightarrow  x < - 3

Therefore the solution set is \Big( \frac{7}{3} , \infty \Big) \cup ( -\infty, - 3)

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Question 2: |4-x| +1 < 3

Answer:

Given |4-x| +1 < 3

\Rightarrow -2 <4 - x < 2

\Rightarrow -6 < -x < - 2

\Rightarrow 2 < x < 6

Therefore the solution set is (2, 6)

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Question 3: \Big| \frac{3x-4}{2} \Big| \leq \frac{5}{12}

Answer:

Given \Big| \frac{3x-4}{2} \Big| \leq \frac{5}{12}

\Rightarrow - \frac{5}{12} \leq \frac{3x-4}{2} \leq \frac{5}{12}

\Rightarrow -5 \leq 18x-24 \leq 5

\Rightarrow 19 \leq 18x \leq 29

\Rightarrow \frac{19}{18} \leq x \leq \frac{29}{18}

Therefore the solution set is \Big[ \frac{19}{18} , \frac{29}{18} \Big ] 

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Question 4: \frac{|x-2|}{x-2} > 0

Answer:

Given \frac{|x-2|}{x-2} > 0

Case I: When x-2 \geq 0 \Rightarrow x \geq 2

\therefore \frac{|x-2|}{x-2} > 0   \Rightarrow \frac{x-2}{x-2} > 0   \Rightarrow x >2

\therefore x \in ( 2, \infty)

Case II: When x-2 < 0 \Rightarrow x < 2

\therefore \frac{|x-2|}{x-2} > 0   \Rightarrow \frac{-(x-2)}{x-2} > 0   \Rightarrow \frac{x-2} {x-2} < 0

There is no possible solution. Therefore x \in \phi

Combining Case I and Case II we get the solution set as ( 2, \infty)

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Question 5: \frac{1}{|x|-3} < \frac{1}{2}

Answer:

Given \frac{1}{|x|-3} < \frac{1}{2}   \Rightarrow \frac{1}{|x|-3} - \frac{1}{2} < 0   \Rightarrow \frac{2 - ( |x|-3)}{2 ( |x|-3)} < 0   \Rightarrow \frac{5 -  |x|}{2 |x|-6} < 0

Case I: When x \geq 0 \Rightarrow |x| = x

Therefore \frac{5 -  |x|}{2 |x|-6} < 0 \Rightarrow \frac{5 -  x}{2 x-6} < 0

Case I(a):  5-x < 0  \text{     and     } 2x - 6 > 0

\Rightarrow x > 5 \text{     and     } x > 3  \Rightarrow  x > 5

Case I(b):  5-x > 0  \text{     and     } 2x - 6 < 0

\Rightarrow x < 5 \text{     and     } x < 3  \Rightarrow  x < 3

Therefore the solution set in Case I is [0, 3) \cup ( 5, \infty)

Case II: When x < 0 \Rightarrow |x| = -x

Therefore \frac{5 -  |x|}{2 |x|-6} < 0 \Rightarrow \frac{5 -  (-x)}{2 (-x)-6} < 0 \Rightarrow \frac{5 x}{-2x-6} < 0

Case II(a):  5+x < 0  \text{     and     } -2x - 6 > 0

\Rightarrow x < -5 \text{     and     } x <-3  \Rightarrow  x < - 5

Case II(b): 5+x > 0  \text{     and     } -2x - 6 < 0

\Rightarrow x > -5 \text{     and     } x > -3  \Rightarrow  x > - 3

Therefore the solution set in Case II is (- \infty, -5) \cup ( -3, \infty)

Combining both Case I and Case II, we get the complete solution set as

[0, 3) \cup ( 5, \infty) \ \cup (- \infty, -5) \cup ( -3, \infty)

\Rightarrow ( - \infty, -5) \cup ( -3, 3) \cup ( 5, \infty)

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Question 6: \frac{|x+2|-x}{x} < 2

Answer:

Given \frac{|x+2|-x}{x} < 2   \Rightarrow \frac{|x+2|-x}{x} - 2 <0   \Rightarrow  \frac{|x+2|-3x}{x} < 0

Case I: When x + 2 \geq 0 \Rightarrow x \geq -2

\Rightarrow |x+2| = ( x+2)

\therefore   \frac{|x+2|-3x}{x} < 0   \Rightarrow   \frac{x+2-3x}{x} < 0   \Rightarrow   \frac{2-2x}{x} < 0

Case I(a):  2-2x < 0  \text{     and     } x > 0

\Rightarrow x > 1 \text{     and     } x > 0  \Rightarrow  x > 1

Case I(b): 2-2x > 0  \text{     and     } x < 0

\Rightarrow x < 1 \text{     and     } x < 0  \Rightarrow  x < 0

Therefore the solution set in Case I and x \geq -2 is [-2, 0) \cup ( 1, \infty)

Case II: When x + 2 < 0 \Rightarrow x < -2

\Rightarrow |x+2| = -( x+2)

\therefore   \frac{|x+2|-3x}{x} < 0   \Rightarrow   \frac{-(x+2)-3x}{x} < 0   \Rightarrow   \frac{-4x-2}{x} < 0   \Rightarrow   \frac{4x+2}{x} > 0

Case II(a):  4x+2 > 0  \text{     and     } x > 0

\Rightarrow x >   \frac{-1}{2} \text{     and     } x > 0  \Rightarrow  x > 0

Case II(b): 4x+2 < 0  \text{     and     } x < 0

\Rightarrow x <   \frac{-1}{2} \text{     and     } x < 0  \Rightarrow  x >   \frac{-1}{2}

Therefore the solution set in Case I and x < -2 is (- \infty, -2)  \cup ( 0, \infty)

Combining both Case I and Case II,  and we get the complete solution set as [-2, 0) \cup ( 1, \infty) \  \cup \  (- \infty, -2)  \cup ( 0, \infty)

\Rightarrow ( - \infty, 0) \cup ( 1, \infty)

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Question 7: \Big| \frac{2x-1}{x-1} \Big| > 2

Answer:

Given \Big| \frac{2x-1}{x-1} \Big| > 2

Case I:  \frac{2x-1}{x-1} >2 \Rightarrow \frac{2x-1}{x-1} -2 > 0 \Rightarrow \frac{2x-1-2x+2}{x-1} > 0 \Rightarrow \frac{1}{x-1} >0

\Rightarrow x - 1 > 0 \Rightarrow x > 0

Therefore the solution set for Case I is ( 1, \infty)

Case II:  \frac{2x-1}{x-1} < -2 \Rightarrow \frac{2x-1}{x-1} +2 < 0 \Rightarrow \frac{2x-1+2x-2}{x-1} < 0 \Rightarrow \frac{4x-3}{x-1} < 0

Case II(a):  4x-3 > 0  \text{     and     } x-1 < 0

\Rightarrow x >   \frac{3}{4} \text{     and     } x < 1  \Rightarrow  \frac{3}{4} < x < 1

Case II(b): 4x-3 < 0  \text{     and     } x-1 > 0

\Rightarrow x <   \frac{3}{4} \text{     and     } x > 1  \Rightarrow  \text{ No possible solution }

Combining both Case I and Case II,  and we get the complete solution set as (1, \infty) \cup (   \frac{3}{4} , 1)

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Question 8: |x-1|+|x-2|+|x-3| \geq 6

Answer:

Given: |x-1|+|x-2|+|x-3| \geq 6  

For |x-1|,

when x-1 \geq 0 \Rightarrow x \geq 1

when (x-1) < 0 \Rightarrow x < 0

For |x-2|,

when x-2 \geq 0 \Rightarrow x \geq 2

when (x-2) < 0 \Rightarrow x < 2

For |x-3|,

when x-3 \geq 0 \Rightarrow x \geq 3

when (x-3) < 0 \Rightarrow x < 3

Case I: When x < 1

(1-x)+ ( 2- x) + ( 3-x ) \geq 6 \Rightarrow 6 - 3x \geq 6 \Rightarrow x \leq 0

Hence the solutions set is ( - \infty, 0]

Case II: When 1 \leq x < 2

(x-1)+ ( 2- x) + ( 3-x ) \geq 6 \Rightarrow 4 - x \geq 6 \Rightarrow x \leq -2

Hence the solutions set is null.

Case III: When 2 \leq x < 3

(x-1)+ ( x-2) + ( 3-x ) \geq 6 \Rightarrow  x \geq 6

Hence the solutions set is null.

Case IV: When x \geq 3

(x-1)+ ( x-2) + ( x-3 ) \geq 6 \Rightarrow  3x-6 \geq 6 \Rightarrow x \geq 4

Hence the solutions set is [4, \infty )

From all the four cases the solution set is ( - \infty, 0] \cup [4, \infty )

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Question 9: \frac{|x-2|-1}{|x-2|-2} \leq 0

Answer:

Given: \frac{|x-2|-1}{|x-2|-2} \leq 0

Case I:  When x - 2 \geq 0 \Rightarrow x \geq 2

Therefore \frac{x-2-1}{x-2-2} \leq 0 \Rightarrow \frac{x-3}{x-4} \leq 0

Case I(a):  x-3 \leq 0  \text{     and     } x-4 > 0

\Rightarrow x \leq 3 \text{     and     } x > 4  

Therefore the solution set is null set.

Case I(b): x-3 \geq 0  \text{     and     } x-4 < 0

\Rightarrow x \geq 3 \text{     and     } x < 4  

Therefore the solution set is [3, 4) .

Therefore the solution set in Case I and x \geq 2 is [3, 4)

Case II:  When x - 2 \leq 0 \Rightarrow x \leq 2

Therefore \frac{2-x-1}{2-x-2} \leq 0 \Rightarrow \frac{1-x}{-x} \leq 0 \Rightarrow \frac{x-1}{x} \geq 0

Case II(a)x-1 \leq 0  \text{     and     } x > 0

\Rightarrow x \leq 1 \text{     and     } x > 0  

Therefore the solution set is (0, 1] .

Case II(b)x-1 \geq 0  \text{     and     } x < 0

\Rightarrow x \geq 1 \text{     and     } x < 0  

Therefore the solution set is null.

Therefore the solution set in Case I and x \leq 2 is (0, 1]

Combining both Case I and Case II,  and we get the complete solution set as (0, 1] \cup ([3, 4)

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Question 10: \frac{1}{|x|-3} \leq \frac{1}{2}

Answer:

Given: \frac{1}{|x|-3} \leq \frac{1}{2}

\Rightarrow \frac{1}{|x|-3} - \frac{1}{2} \leq 0   \Rightarrow \frac{2 - |x| + 3}{2 ( |x|-3)} \leq 0   \Rightarrow \frac{5 - |x|}{2 |x| - 6} \leq 0

Case I: When x \geq 0 \Rightarrow |x| = x

\therefore \frac{5 - |x|}{2 |x| - 6} \leq 0 \Rightarrow \frac{5-x}{2x-6} \leq 0

Case I(a):  5-x \leq 0  \text{     and     } 2x-6 > 0

\Rightarrow x \geq 5 \text{     and     } x > 3  

Therefore the solution set is [5, \infty).

Case I(b): 5-x \geq 0  \text{     and     } 2x-6 < 0

\Rightarrow x \leq 5 \text{     and     } x < 3  

Therefore the solution set is along with x \geq 0 is [0, 3) .

Therefore the solution set in Case I  is [0, 3) \cup [5, \infty)

Case II: When x < 0 \Rightarrow |x| = -x

\therefore \frac{5 - |x|}{2 |x| - 6} \leq 0 \Rightarrow \frac{5-(-x)}{2(-x)-6} \leq 0 \Rightarrow \frac{5+x}{-2x-6} \leq 0   \Rightarrow \frac{5+x}{2x+6} \geq 0

Case II(a):  5+x \geq 0  \text{     and     } 2x+6 > 0

\Rightarrow x \geq -5 \text{     and     } x > -3  

\Rightarrow x > -3

Therefore the solution set is (-3, \infty) .

Case II(b): 5+x \leq 0  \text{     and     } 2x+6 < 0

\Rightarrow x \leq -5 \text{     and     } x < -3  

\Rightarrow x \leq -5

Therefore the solution set is (-\infty, -5) .

Therefore the solution set in Case I  is (-3, \infty) \  \cup \ (-\infty, -5)

Combining both Case I and Case II,  and we get the complete solution set as [0, 3) \cup [5, \infty) \ \cup \  (-3, \infty) \  \cup \ (-\infty, -5) \Rightarrow ( - \infty, -5) \cup (-3, 3) \cup ([5, \infty)

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Question 11: |x+1|+|x| > 3

Answer:

Given: |x+1|+|x| > 3

When (x+1) \geq 0 \Rightarrow x \geq -1 \text{ then } |x+1| = (x+1)

When (x+1) < 0 \Rightarrow x < -1 \text{ then } |x+1| = -(x+1)

Similarly, 

When x \geq 0  \text{ then } |x| = x

When x < 0  \text{ then } |x| = -x

Case I: When x < -1

\Rightarrow -(x+1) - x > 3   \Rightarrow -2x-1 > 3   \Rightarrow -2x > 4 \Rightarrow x<-2

\therefore x \in ( - \infty , -2)

Case II: When -1 \leq x < 0

\Rightarrow (x+1) - x > 3   \Rightarrow 1 > 3

This is not possible. Therefore x \in \phi

Case III: When x \geq 0

\Rightarrow (x+1) + x > 3 \Rightarrow 2x+1 > 3   \Rightarrow x>1

Therefore x \in (1, \infty)

Combining all the three cases, the solution set is ( - \infty, -2) \cup ( 1,\infty)

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Question 12: 1 \leq |x-2| \leq 3

Answer:

Given 1 \leq |x-2| \leq 3

For  |x-2| \geq 1

When x - 2 \geq 0 \Rightarrow x \geq 2 \text{ then } |x-2|= (x-2)

\therefore  x-2 \geq 1 \Rightarrow x \geq 3  \text{ i.e } x \in [ 3, \infty)

When x - 2 < 0 \Rightarrow x < 2 \text{ then } |x-2| = -(x-2)

\therefore -(x-2) \geq 1 \Rightarrow x < 1 \text{ i.e } x \in ( - \infty, 1)

Hence the solution set for the first equation is ( - \infty, 1) \cup [ 3, \infty)

For |x-2| \leq 3

When x - 2 \geq 0 \Rightarrow x \geq 2 \text{ then } |x-2|= (x-2)

\therefore  x-2 \leq 3 \Rightarrow x \leq 5  \text{ i.e } x \in (- \infty, 5]

When x - 2 < 0 \Rightarrow x < 2 \text{ then } |x-2| = -(x-2)

\therefore -(x-2) \leq 3 \Rightarrow x \geq -1 \text{ i.e } x \in [- 1, \infty)

Hence the solution set for the first equation is (- \infty, 5] \cup [- 1, \infty) \Rightarrow [-1, 5]

Combining the two cases, we get the solution set as ( - \infty, 1) \cup [ 3, \infty) \cup [-1, 5] \Rightarrow  [-1, 1] \cup [3, 5]

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Question 13: |3-4x| \geq 9

Answer:

\Rightarrow 3 - 4x \geq 9  \text{ or } 3 - 4x \leq -9

\Rightarrow 4x \leq -6  \text{ or } 12 \leq 4x

\Rightarrow x \leq \frac{-3}{2} \text{ or } x \geq 3

Therefore the solution set will be \Big(- \infty, \frac{-3}{2} \Big)  \cup [3, \infty)

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