Solve the following linear inequations in $R$

Question 1: Solve: $12x < 50$ when

i) $x \in R$     ii) $x \in Z$     iii) $x \in N$

Given $12x < 50 \Rightarrow x <$ $\frac{25}{6}$

i)        If $\Rightarrow x \in R$

Therefore the solution of the given inequation is $\Big( -\infty,$ $\frac{25}{6}$ $\Big)$

ii)      If $x \in Z$

Therefore the solution of the given inequation is $\{ -\infty, \ldots , -3, -2, -1, 0 , 1, 2, 3, 4 \}$

iii)     If $x \in N$

Therefore the solution of the given inequation is $\{ 1, 2, 3, 4 \}$

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Question 2: Solve: $-4x > 30$ when

i) $x \in R$     ii) $x \in Z$     iii) $x \in N$

Given $-4x > 30 \Rightarrow x <$ $- \frac{15}{2}$

i)        If $\Rightarrow x \in R$

Therefore the solution of the given inequation is $\Big( -\infty,$ $\frac{-15}{2}$ $\Big)$

ii)      If $x \in Z$

Therefore the solution of the given inequation is $\{ -\infty, \ldots , -9, -8 \}$

iii)     If $x \in N$

Therefore the solution of the given inequation is $\phi$ This is null set.

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Question 3: Solve: $4x-2 < 8$ when

i) $x \in R$     ii) $x \in Z$     iii) $x \in N$

Given $4x-2 < 8 \Rightarrow 4x < 10 \Rightarrow x <$ $\frac{5}{2}$

i)        If $\Rightarrow x \in R$

Therefore the solution of the given inequation is $\Big( -\infty,$ $\frac{5}{2}$ $\Big)$

ii)      If $x \in Z$

Therefore the solution of the given inequation is $\{ -\infty, \ldots , -3, -2, -1, 0 , 1, 2 \}$

iii)     If $x \in N$

Therefore the solution of the given inequation is $\{ 1, 2 \}$

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Question 4: $3x-7 > x + 1$

Given $3x-7 > x + 1 \Rightarrow 2x > 8 \Rightarrow x > 4$

Therefore the solution of the given inequation is $(4, \infty)$

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Question 5: $x+5 > 4x-10$

Given $x+5 > 4x-10 \Rightarrow 3x < 15 \Rightarrow x < 5$

Therefore the solution of the given inequation is $(-\infty , 5)$

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Question 6: $3x+9 \geq -x+19$

Given $3x+9 \geq -x+19 \Rightarrow 4x \geq 10 \Rightarrow x \geq$ $\frac{5}{2}$

Therefore the solution of the given inequation is $\Big[$ $\frac{5}{2}$ $, -\infty \Big)$

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Question 7: $2( 3 - x) \geq$ $\frac{x}{5}$ $+4$

$2( 3 - x) \geq$ $\frac{x}{5}$ $+4$

$\Rightarrow 6-2x \geq$ $\frac{x}{5}$ $+ 4$

$\Rightarrow$ $\frac{x}{5}$ $+ 2x \leq 2$

$\Rightarrow x \leq$ $\frac{10}{11}$

Therefore the solution of the given inequation is $\Big( -\infty,$ $\frac{10}{11}$ $\Big)$

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Question 8: $\frac{3x-2}{5}$ $\leq$ $\frac{4x-3}{2}$

$\frac{3x-2}{5}$ $\leq$ $\frac{4x-3}{2}$

$\Rightarrow 6x-4 \leq 20x-15$

$\Rightarrow 14x \geq 11$

$\Rightarrow x \geq$ $\frac{11}{14}$

Therefore the solution of the given inequation is $\Big[$ $\frac{11}{14}$ $, \infty \Big)$

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Question 9: $-(x-3) +4 < 5 - 2x$

$-(x-3) +4 < 5 - 2x$

$\Rightarrow -x+3+4 <5 -2x$

$\Rightarrow x < -2$

Therefore the solution of the given inequation is $(- \infty , -2)$

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Question 10: $\frac{x}{5}$ $<$ $\frac{3x-2}{4}$ $-$ $\frac{5x-3}{5}$

$\frac{x}{5}$ $<$ $\frac{3x-2}{4}$ $-$ $\frac{5x-3}{5}$

$\frac{x}{5}$ $<$ $\frac{15x-10-20x+12}{20}$

$\Rightarrow 4x < -5x+2$

$\Rightarrow 9x < 2$

$\Rightarrow x <$ $\frac{2}{9}$

Therefore the solution of the given inequation is $\Big[ - \infty,$ $\frac{2}{9}$ $, \Big)$

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Question 11: $\frac{2(x-1)}{5}$ $\leq$ $\frac{3(2+x)}{7}$

$\frac{2(x-1)}{5}$ $\leq$ $\frac{3(2+x)}{7}$

$\Rightarrow$ $\frac{2x-2}{5}$ $\leq$ $\frac{6+3x}{7}$

$\Rightarrow 14x-14 \leq 30+15x$

$\Rightarrow x \geq -44$

Therefore the solution of the given inequation is $[-44, \infty)$

$\\$

Question 12: $\frac{5x}{2}$ $+$ $\frac{3x}{4}$ $\geq$ $\frac{39}{4}$

$\frac{5x}{2}$ $+$ $\frac{3x}{4}$ $\geq$ $\frac{39}{4}$

$\Rightarrow 10x + 3x \geq 39$

$\Rightarrow x \geq 3$

Therefore the solution of the given inequation is $[ 3, \infty)$

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Question 13: $\frac{x-1}{3}$ $+ 4 <$ $\frac{x-5}{5}$ $-2$

$\frac{x-1}{3}$ $+ 4 <$ $\frac{x-5}{5}$ $-2$

$\Rightarrow$ $\frac{x-1+12}{3}$ $<$ $\frac{x-5-10}{5}$

$\Rightarrow 5(x+11) < 3 ( x- 15)$

$\Rightarrow 5x + 55 < 3x - 45$

$\Rightarrow 2x < - 100$

$\Rightarrow x < - 50$

Therefore the solution of the given inequation is $( -\infty, -50 )$

$\\$

Question 14: $\frac{2x+3}{4}$ $- 3 <$ $\frac{x-4}{3}$ $-2$

$\frac{2x+3}{4}$ $- 3 <$ $\frac{x-4}{3}$ $-2$

$\Rightarrow$ $\frac{2x+3-12}{4}$ $<$ $\frac{x-4-6}{3}$

$\Rightarrow$ $\frac{2x-9}{4}$ $<$ $\frac{x-10}{3}$

$\Rightarrow 6x-27 < 4x-40$

$\Rightarrow 2x < -13$

$\Rightarrow x <$ $\frac{-13}{2}$

Therefore the solution of the given inequation is $\Big( -\infty,$ $\frac{-13}{2}$ $\Big)$

$\\$

Question 15: $\frac{5-2x}{3}$ $<$ $\frac{x}{6}$ $-5$

$\frac{5-2x}{3}$ $<$ $\frac{x}{6}$ $-5$

$\Rightarrow 10-4x < x - 30$

$\Rightarrow 5x > 40$

$\Rightarrow x > 8$

Therefore the solution of the given inequation is $( 8, \infty )$

$\\$

Question 16: $\frac{4+2x}{3}$ $\geq$ $\frac{x}{2}$ $-3$

$\frac{4+2x}{3}$ $\geq$ $\frac{x}{2}$ $-3$

$\Rightarrow 8+4x \geq 3x-18$

$\Rightarrow x \geq -26$

Therefore the solution of the given inequation is $[ -26, \infty )$

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Question 17: $\frac{2x+3}{5}$ $- 2 <$ $\frac{3(x-2)}{5}$

$\frac{2x+3}{5}$ $- 2 <$ $\frac{3(x-2)}{5}$

$\Rightarrow 2x+3-10 < 3x-6$

$\Rightarrow x > -1$

Therefore the solution of the given inequation is $( -1, \infty )$

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Question 18: $x-2 \leq$ $\frac{5x+8}{3}$

$x-2 \leq$ $\frac{5x+8}{3}$

$\Rightarrow 3x-6 \leq 5x + 8$

$\Rightarrow -14 \leq 2x$

$\Rightarrow x \geq -7$

Therefore the solution of the given inequation is $[ -7, \infty )$

$\\$

Question 19: $\frac{6x-5}{4x+1}$ $< 0$

$\frac{6x-5}{4x+1}$ $< 0$

Case I:  $6x-5 > 0 \hspace{1.0cm} \& \hspace{1.0cm} 4x+1 < 0$

$\Rightarrow x >$ $\frac{5}{6}$                  $\& \hspace{1.0cm} x <$ $\frac{-1}{4}$

This is not possible

Case II: $6x-5 < 0 \hspace{1.0cm} \& \hspace{1.0cm} 4x+1 > 0$

$\Rightarrow x <$ $\frac{5}{6}$                   $\& \hspace{1.0cm} x >$ $\frac{-1}{4}$

Therefore the solution set for the inequation is $\Big($ $\frac{-1}{4}$ $,$ $\frac{5}{6}$ $\Big)$

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Question 20: $\frac{2x-3}{3x-7}$ $> 0$

$\frac{2x-3}{3x-7}$ $> 0$

Case I:  $2x-3 > 0 \hspace{1.0cm} \& \hspace{1.0cm} 3x-7 > 0$

$\Rightarrow x >$ $\frac{3}{2}$                  $\& \hspace{1.0cm} x >$ $\frac{7}{3}$

$\Rightarrow x >$ $\frac{7}{3}$

Case II:   $2x-3 < 0 \hspace{1.0cm} \& \hspace{1.0cm} 3x-7 < 0$

$\Rightarrow x <$ $\frac{3}{2}$                   $\& \hspace{1.0cm} x <$ $\frac{7}{3}$

$\Rightarrow x <$ $\frac{3}{2}$

Therefore the solution set for the inequation is $\Big( -\infty,$ $\frac{3}{2}$ $\Big) \cup \Big($ $\frac{7}{3}$ $, \infty \Big)$

$\\$

Question 21: $\frac{3}{x-2}$ $< 1$

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

$\Rightarrow$ $\frac{3}{x-2}$ $- 1 < 0 \ \ \Rightarrow$ $\frac{3-x+2}{x-2}$ $< 0 \ \ \Rightarrow$ $\frac{-x+5}{x-2}$ $< 0$

Case I:  $-x+5 > 0 \hspace{1.0cm} \& \hspace{1.0cm} x-2 < 0$

$\Rightarrow x < 5$                   $\& \hspace{1.0cm} x < 2$

$\Rightarrow x < 2$

Case II:   $-x+5 < 0 \hspace{1.0cm} \& \hspace{1.0cm} x-2 > 0$

$\Rightarrow x > 5$                   $\& \hspace{1.0cm} x > 2$

$\Rightarrow x > 5$

Therefore the solution set for the inequation is $( - \infty, 2 ) \cup ( 5, \infty)$

$\\$

Question 22: $\frac{1}{x-1}$ $\leq 2$

$\frac{1}{x-1}$ $\leq 2$

$\Rightarrow$ $\frac{1}{x-1}$ $- 2 \leq 0 \ \ \Rightarrow$ $\frac{1-2x+2}{x-1}$ $\leq 0 \ \ \Rightarrow$ $\frac{-2x+3}{x-1}$ $\leq 0$

Case I:  $-2x+3 \geq 0 \hspace{1.0cm} \& \hspace{1.0cm} x-1 < 0$

$\Rightarrow x \leq$ $\frac{3}{2}$                   $\& \hspace{1.0cm} x < 1$

$\Rightarrow x < 1$

Case II:  $-2x+3 \leq 0 \hspace{1.0cm} \& \hspace{1.0cm} x-1 > 0$

$\Rightarrow x \geq$ $\frac{3}{2}$                   $\& \hspace{1.0cm} x > 1$

$\Rightarrow x \geq$ $\frac{3}{2}$

Therefore the solution set for the inequation is $( - \infty, 1 ) \cup \Big[$ $\frac{3}{2}$ $, \infty \Big)$

$\\$

Question 23: $\frac{4x+3}{2x-5}$ $< 6$

$\frac{4x+3}{2x-5}$ $< 6$

$\Rightarrow$ $\frac{4x+3}{2x-5}$ $- 6 < 0 \ \ \Rightarrow$ $\frac{4x+3-12x+30}{2x-5}$ $< 0 \ \ \Rightarrow$ $\frac{-8x+33}{2x-5}$ $< 0 \ \ \Rightarrow$ $\frac{8x-33}{2x-5}$ $> 0$

Case I:  $8x-33 < 0 \hspace{1.0cm} \& \hspace{1.0cm} 2x-5 < 0$

$\Rightarrow x <$ $\frac{33}{8}$                   $\& \hspace{1.0cm} x <$ $\frac{5}{2}$

$\Rightarrow x <$ $\frac{5}{2}$

Case II: $8x-33 > 0 \hspace{1.0cm} \& \hspace{1.0cm} 2x-5 > 0$

$\Rightarrow x >$ $\frac{33}{8}$                   $\& \hspace{1.0cm} x >$ $\frac{5}{2}$

$\Rightarrow x >$ $\frac{33}{8}$

Therefore the solution set for the inequation is $\Big( - \infty,$ $\frac{5}{2}$ $\Big) \cup \Big($ $\frac{33}{8}$ $, \infty \Big)$

$\\$

Question 24: $\frac{5x-6}{x+6}$ $< 1$

$\frac{5x-6}{x+6}$ $< 1$

$\Rightarrow$ $\frac{5x-6}{x+6}$ $- 1 < 0 \ \ \Rightarrow$ $\frac{5x-6-x-6}{x+6}$ $< 0 \ \ \Rightarrow$ $\frac{4x-12}{x+6}$ $< 0$

Case I:  $4x-12 < 0 \hspace{1.0cm} \& \hspace{1.0cm} x+6 > 0$

$\Rightarrow x < 3$                   $\& \hspace{1.0cm} x > -6$

Case II:  $4x-12 > 0 \hspace{1.0cm} \& \hspace{1.0cm} x+6 < 0$

$\Rightarrow x > 3$                   $\& \hspace{1.0cm} x < -6$

This is not possible.

Therefore the solution set for the inequation is $( -6, 3 )$

$\\$

Question 25: $\frac{5x+8}{4-x}$ $< 2$

$\frac{5x+8}{4-x}$ $< 2$

$\Rightarrow$ $\frac{5x+8}{4-x}$ $- 2 < 0 \ \ \Rightarrow$ $\frac{5x+8-8+2x}{4-x}$ $< 0 \ \ \Rightarrow$ $\frac{7x}{4-x}$ $< 0$

Case I:  $7x > 0 \hspace{1.0cm} \& \hspace{1.0cm} 4-x < 0$

$\Rightarrow x > 0$                   $\& \hspace{1.0cm} x > 4$

$\Rightarrow x > 4$

Case II: $7x < 0 \hspace{1.0cm} \& \hspace{1.0cm} 4-x > 0$

$\Rightarrow x < 0$                   $\& \hspace{1.0cm} x < 4$

$\Rightarrow x < 0$

Therefore the solution set for the inequation is $( - \infty, 0 ) \cup ( 4, \infty)$

$\\$

Question 26: $\frac{x-1}{x+3}$ $> 2$

$\frac{x-1}{x+3}$ $> 2$

$\Rightarrow$ $\frac{x-1}{x+3}$ $- 2 > 0 \ \ \Rightarrow$ $\frac{x-1-2x-6}{x+3}$ $> 0 \ \ \Rightarrow$ $\frac{-x-7}{x+3}$ $> 0$

Case I:  $-(x+7) > 0 \hspace{1.0cm} \& \hspace{1.0cm} x+3 > 0$

$\Rightarrow x < -7$                   $\& \hspace{1.0cm} x > -3$

This is not possible.

Case II:  $-(x+7) < 0 \hspace{1.0cm} \& \hspace{1.0cm} x+3 < 0$

$\Rightarrow x > -7$                   $\& \hspace{1.0cm} x < -3$

Therefore the solution set for the inequation is $( - 7, -3 )$

$\\$

Question 27: $\frac{7x-5}{8x+3}$ $> 4$

$\frac{7x-5}{8x+3}$ $> 4$

$\Rightarrow$ $\frac{7x-5}{8x+3}$ $- 4 > 0 \ \ \Rightarrow$ $\frac{7x-5-32x-12}{8x+3}$ $> 0 \ \ \Rightarrow$ $\frac{-25x-17}{8x+3}$ $> 0$

Case I:  $-(25x+17) > 0 \hspace{1.0cm} \& \hspace{1.0cm} 8x+3 > 0$

$\Rightarrow x <$ $\frac{-17}{25}$                   $\& \hspace{1.0cm} x >$ $\frac{-3}{8}$

This is not possible.

Case II: $-(25x+17) < 0 \hspace{1.0cm} \& \hspace{1.0cm} 8x+3 < 0$

$\Rightarrow x >$ $\frac{-17}{25}$                   $\& \hspace{1.0cm} x <$ $\frac{-3}{8}$

Therefore the solution set for the inequation is $\Big($ $\frac{-17}{25}$ $,$ $\frac{-3}{8}$ $\Big)$

$\\$

Question 28: $\frac{x}{x-5}$ $>$ $\frac{1}{2}$

$\frac{x}{x-5}$ $>$ $\frac{1}{2}$

$\Rightarrow$ $\frac{x}{x-5}$ $-$ $\frac{1}{2}$ $> 0 \ \ \Rightarrow$ $\frac{2x-x+5}{2x-10}$ $> 0 \ \ \Rightarrow$ $\frac{x+5}{2x-10}$ $> 0$

Case I:  $x+5 > 0 \hspace{1.0cm} \& \hspace{1.0cm} 2x-7 > 0$

$\Rightarrow x > -5$                   $\& \hspace{1.0cm} x > 5$

$\Rightarrow x > 5$

Case II:  $x+5 < 0 \hspace{1.0cm} \& \hspace{1.0cm} 2x-10 < 0$

$\Rightarrow x < -5$                   $\& \hspace{1.0cm} x < 5$

$\Rightarrow x < -5$

Therefore the solution set for the inequation is $( - \infty, -5 ) \cup ( 5, \infty)$