Solve each of the following systems of equations in R. 

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

Answer:

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

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

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

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

\displaystyle \\

\displaystyle \text{Question 2: } |4-x| +1 < 3

Answer:

\displaystyle \text{Given } |4-x| +1 < 3

\displaystyle \Rightarrow -2 <4 - x < 2

\displaystyle \Rightarrow -6 < -x < - 2

\displaystyle \Rightarrow 2 < x < 6

\displaystyle \text{Therefore the solution set is } (2, 6)

\displaystyle \\

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

Answer:

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

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

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

\displaystyle \Rightarrow 19 \leq 18x \leq 29

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

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

\displaystyle \\

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

Answer:

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

\displaystyle \text{Case I: When }  x-2 \geq 0 \Rightarrow x \geq 2

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

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

\displaystyle \text{Case II: When }  x-2 < 0 \Rightarrow x < 2

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

\displaystyle \text{There is no possible solution. Therefore  }  x \in \phi

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

\displaystyle \\

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

Answer:

\displaystyle \text{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

\displaystyle \text{Case I: When }  x \geq 0 \Rightarrow |x| = x

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

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

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

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

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

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

\displaystyle \text{Case II: When }  x < 0 \Rightarrow |x| = -x

\displaystyle \text{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): \displaystyle 5+x < 0 \text{ and } -2x - 6 > 0

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

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

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

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

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

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

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

\displaystyle \\

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

Answer:

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

\displaystyle \text{Case I: When }  x + 2 \geq 0 \Rightarrow x \geq -2

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

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

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

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

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

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

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

\displaystyle \text{Case II: When }  x + 2 < 0 \Rightarrow x < -2

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

\displaystyle \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): \displaystyle 4x+2 > 0 \text{ and } x > 0

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

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

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

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

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

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

\displaystyle \\

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

Answer:

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

\displaystyle \text{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

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

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

\displaystyle \text{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): \displaystyle 4x-3 > 0 \text{ and } x-1 < 0

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

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

\displaystyle \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 \displaystyle (1, \infty) \cup ( \frac{3}{4} , 1)

\displaystyle \\

\displaystyle \text{Question 8: } |x-1|+|x-2|+|x-3| \geq 6

Answer:

\displaystyle \text{Given:  }  |x-1|+|x-2|+|x-3| \geq 6  

\displaystyle \text{For  }  |x-1|,

\displaystyle \text{When  }  x-1 \geq 0 \Rightarrow x \geq 1

\displaystyle \text{When  }  (x-1) < 0 \Rightarrow x < 0

\displaystyle \text{For  }  |x-2|,

\displaystyle \text{When  }  x-2 \geq 0 \Rightarrow x \geq 2

\displaystyle \text{When  }  (x-2) < 0 \Rightarrow x < 2

\displaystyle \text{For  }  |x-3|,

\displaystyle \text{When  }  x-3 \geq 0 \Rightarrow x \geq 3

\displaystyle \text{When  }  (x-3) < 0 \Rightarrow x < 3

\displaystyle \text{Case I: When }  x < 1

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

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

\displaystyle \text{Case II: When }  1 \leq x < 2

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

Hence the solutions set is null.

\displaystyle \text{Case III:  }  \text{When  }  2 \leq x < 3

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

Hence the solutions set is null.

Case IV: \displaystyle \text{When  }  x \geq 3

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

Hence the solutions set is \displaystyle [4, \infty )

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

\displaystyle \\

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

Answer:

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

\displaystyle \text{Case I: When }  x - 2 \geq 0 \Rightarrow x \geq 2

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

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

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

Therefore the solution set is null set.

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

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

\displaystyle \text{Therefore the solution set is } [3, 4) .

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

\displaystyle \text{Case II: When }  x - 2 \leq 0 \Rightarrow x \leq 2

\displaystyle \text{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): \displaystyle x-1 \leq 0 \text{ and } x > 0

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

\displaystyle \text{Therefore the solution set is } (0, 1] .

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

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

Therefore the solution set is null.

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

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

\displaystyle \\

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

Answer:

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

\displaystyle \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

\displaystyle \text{Case I: When }  x \geq 0 \Rightarrow |x| = x

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

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

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

\displaystyle \text{Therefore the solution set is } [5, \infty).

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

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

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

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

\displaystyle \text{Case II: When }  x < 0 \Rightarrow |x| = -x

\displaystyle \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): \displaystyle 5+x \geq 0 \text{ and } 2x+6 > 0

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

\displaystyle \Rightarrow x > -3

\displaystyle \text{Therefore the solution set is } (-3, \infty) .

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

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

\displaystyle \Rightarrow x \leq -5

\displaystyle \text{Therefore the solution set is } (-\infty, -5) .

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

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

\displaystyle \\

\displaystyle \text{Question 11: } |x+1|+|x| > 3

Answer:

\displaystyle \text{Given:  }  |x+1|+|x| > 3

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

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

Similarly, 

\displaystyle \text{When  }  x \geq 0 \text{ then } |x| = x

\displaystyle \text{When  }  x < 0 \text{ then } |x| = -x

\displaystyle \text{Case I: When }  x < -1

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

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

\displaystyle \text{Case II: When }  -1 \leq x < 0

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

This is not possible. \displaystyle \text{Therefore  }  x \in \phi

\displaystyle \text{Case III:  }  \text{When  }  x \geq 0

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

\displaystyle \text{Therefore  }  x \in (1, \infty)

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

\displaystyle \\

\displaystyle \text{Question 12: } 1 \leq |x-2| \leq 3

Answer:

\displaystyle \text{Given } 1 \leq |x-2| \leq 3

\displaystyle \text{For  }  |x-2| \geq 1

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

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

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

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

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

\displaystyle \text{For  }  |x-2| \leq 3

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

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

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

\displaystyle \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 \displaystyle (- \infty, 5] \cup [- 1, \infty) \Rightarrow [-1, 5]

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

\displaystyle \\

\displaystyle \text{Question 13: } |3-4x| \geq 9

Answer:

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

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

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

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