Solve the following linear inequations in $\displaystyle R$

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

$\displaystyle \text{i) } x \in R \hspace{1.0cm} \text{ii) } x \in Z \hspace{1.0cm} \text{iii) } x \in N$

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

$\displaystyle \text{i) } \text{If } \Rightarrow x \in R$

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

$\displaystyle \text{ii) } \text{If } x \in Z$

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

$\displaystyle \text{iii) } \text{If } x \in N$

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

$\displaystyle \\$

Question 2: Solve: $\displaystyle -4x > 30$ when

$\displaystyle \text{i) } x \in R \hspace{1.0cm} \text{ii) } x \in Z \hspace{1.0cm} \text{iii) } x \in N$

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

$\displaystyle \text{i) } \text{If } \Rightarrow x \in R$

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

$\displaystyle \text{ii) } \text{If } x \in Z$

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

$\displaystyle \text{iii) } \text{If } x \in N$

$\displaystyle \text{Therefore the solution of the given inequation is} \phi$ This is null set.

$\displaystyle \\$

Question 3: Solve: $\displaystyle 4x-2 < 8$ when

$\displaystyle \text{i) } x \in R \hspace{1.0cm} \text{ii) } x \in Z \hspace{1.0cm} \text{iii) } x \in N$

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

$\displaystyle \text{i) } \text{If } \Rightarrow x \in R$

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

$\displaystyle \text{ii) } \text{If } x \in Z$

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

$\displaystyle \text{iii) } \text{If } x \in N$

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

$\displaystyle \\$

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

$\displaystyle \text{ Given } 3x-7 > x + 1 \Rightarrow 2x > 8 \Rightarrow x > 4$

$\displaystyle \text{Therefore the solution of the given inequation is} (4, \infty)$

$\displaystyle \\$

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

$\displaystyle \text{ Given } x+5 > 4x-10 \Rightarrow 3x < 15 \Rightarrow x < 5$

$\displaystyle \text{Therefore the solution of the given inequation is} (-\infty , 5)$

$\displaystyle \\$

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \Rightarrow 14x \geq 11$

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \Rightarrow x < -2$

$\displaystyle \text{Therefore the solution of the given inequation is} (- \infty , -2)$

$\displaystyle \\$

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

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

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

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

$\displaystyle \Rightarrow 9x < 2$

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \Rightarrow x \geq -44$

$\displaystyle \text{Therefore the solution of the given inequation is} [-44, \infty)$

$\displaystyle \\$

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

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

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

$\displaystyle \Rightarrow x \geq 3$

$\displaystyle \text{Therefore the solution of the given inequation is} [ 3, \infty)$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow 2x < - 100$

$\displaystyle \Rightarrow x < - 50$

$\displaystyle \text{Therefore the solution of the given inequation is} ( -\infty, -50 )$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow 2x < -13$

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \Rightarrow 5x > 40$

$\displaystyle \Rightarrow x > 8$

$\displaystyle \text{Therefore the solution of the given inequation is} ( 8, \infty )$

$\displaystyle \\$

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

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

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

$\displaystyle \Rightarrow x \geq -26$

$\displaystyle \text{Therefore the solution of the given inequation is} [ -26, \infty )$

$\displaystyle \\$

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

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

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

$\displaystyle \Rightarrow x > -1$

$\displaystyle \text{Therefore the solution of the given inequation is} ( -1, \infty )$

$\displaystyle \\$

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

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

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

$\displaystyle \Rightarrow -14 \leq 2x$

$\displaystyle \Rightarrow x \geq -7$

$\displaystyle \text{Therefore the solution of the given inequation is} [ -7, \infty )$

$\displaystyle \\$

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

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

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

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

This is not possible

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow x < 2$

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

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

$\displaystyle \Rightarrow x > 5$

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow x < 1$

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

This is not possible.

$\displaystyle \text{Therefore the solution set for the inequation is } ( -6, 3 )$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow x > 4$

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

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

$\displaystyle \Rightarrow x < 0$

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

$\displaystyle \\$

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

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

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

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

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

This is not possible.

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

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

$\displaystyle \text{Therefore the solution set for the inequation is } ( - 7, -3 )$

$\displaystyle \\$

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

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

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

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

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

This is not possible.

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

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

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow x > 5$

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

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

$\displaystyle \Rightarrow x < -5$

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