Solve each of the following systems of equations in R.

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

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: $|4-x| +1 < 3$

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

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: $\frac{|x-2|}{x-2}$ $> 0$

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

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

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

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

$\\$

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

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

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: $|x-1|+|x-2|+|x-3| \geq 6$

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

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 $(0, 1] \cup ([3, 4)$

$\\$

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

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: $|x+1|+|x| > 3$

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: $1 \leq |x-2| \leq 3$

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: $|3-4x| \geq 9$

$\Rightarrow 3 - 4x \geq 9 \text{ or } 3 - 4x \leq -9$
$\Rightarrow 4x \leq -6 \text{ or } 12 \leq 4x$
$\Rightarrow x \leq$ $\frac{-3}{2}$ $\text{ or } x \geq 3$
Therefore the solution set will be $\Big(- \infty,$ $\frac{-3}{2}$ $\Big) \cup [3, \infty)$
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