Question 1: Solve the following inequation and graph the solution on a number line $\displaystyle 2x- 5 \leq 5x+4 < 11$ , where $\displaystyle x \in I$ . $\displaystyle 2x-5 \leq 5x+4 < 11$ $\displaystyle 2x-5 \leq 5x+4 \text{ or } -9 \leq 3x \text{ or } -3 \leq x$ $\displaystyle 5x+4 < 11 \text{ or } 5x < 7 \text{ or } x < \frac{7}{5}$ $\displaystyle -3 \leq x <\frac{7}{5}$

Therefore $\displaystyle x \in \{-3, -2, -1, 0, 1 \}$  $\displaystyle \\$

Question 2: Given that $\displaystyle x \in I$ , solve the inequation and graph it on a number line: $\displaystyle 3 \geq \frac{x-4}{2}+\frac{x}{3} \geq 2$ . $\displaystyle 3 \geq \frac{x-4}{2}+\frac{x}{3} \geq 2$ $\displaystyle 18 \geq 3(x-4)+2x \geq 12$ $\displaystyle 30 \geq 5x \geq 24$ $\displaystyle 6 \geq x \geq 4.8$

Therefore $\displaystyle x \in \{5, 6 \}$ $\displaystyle \\$

Question 3: Given $\displaystyle A = \{x: 11x-5 > 7x + 3, x \in R \}$ , $\displaystyle B = \{x: 18x-9 \geq 15+12x , x \in R \}$ . Find the range of the set $\displaystyle A \cap B$ and represent it on a number line. $\displaystyle A: 11x-5 > 7x+3$ $\displaystyle 4x >8 \text{ or } x >2$ $\displaystyle B: 18x-9 \geq 15+12x$ $\displaystyle 6x \geq 24 \text{ or } x \geq 4$ $\displaystyle A \cap B = \{ x: x \geq 4, x \in R \}$ $\displaystyle \\$

Question 4: Solve the given inequation and graph it on a number line: $\displaystyle 2y-3 < y+1 \leq 4y+7, y \in R$ . $\displaystyle 2y-3 < y+1 \leq 4y+7$ $\displaystyle 2y-3 < y+1 \text{ or } y < 4$ $\displaystyle y+1 \leq 4y+7 \text{ or } -6 \leq 3y \text{ or } -2 \leq y$

Hence $\displaystyle \{ x: -2 \leq y < 4, x \in R \}$ $\displaystyle \\$

Question 5: Solve the given inequation and graph it on a number line: $\displaystyle -3 < -\frac{1}{2}-\frac{2x}{3} \leq \frac{5}{6}, x \in R$ . $\displaystyle -3 < -\frac{1}{2}-\frac{2x}{3} \leq \frac{5}{6}$ $\displaystyle -3 < -\frac{1}{2}-\frac{2x}{3}$ $\displaystyle -18 < -3 -4x$ $\displaystyle 4x < 15 \text{ or } x < \frac{15}{4}$ $\displaystyle -\frac{1}{2}-\frac{2x}{3} \leq \frac{5}{6} \text{ or } -3-4x \leq 5$ $\displaystyle -8 \leq 4x \text{ or } -2 \leq x$

Therefore $\displaystyle \{ x: -2 \leq x < \frac{15}{4}, x \in R \}$  $\displaystyle \\$

Question 6: Solve the given inequation and graph it on a number line: $\displaystyle 4x-19 < \frac{3x}{5}-2 \leq -\frac{2}{5}+x, x \in R$ . $\displaystyle 4x-19 < \frac{3x}{5}-2 \leq -\frac{2}{5}+x$ $\displaystyle 4x-19 < \frac{3x}{5}-2 \text{ or } 20x-95 < 3x-10 \text{ or } 17x < 85 \text{ or } x < 5$ $\displaystyle \frac{3x}{5}-2 \leq -\frac{2}{5}+x \text{ or } 3x-10 \leq -2 +5x \text{ or } -8 \leq 2x \text{ or } -4 \leq x$

Therefore $\displaystyle \{x : -4 \leq x < 5, x \in R \}$  $\displaystyle \\$

Question 7: Solve the given inequation and graph it on a number line: $\displaystyle -\frac{x}{3} \leq \frac{x}{2}-1\frac{1}{3} <\frac{1}{6}.x \in R$ . $\displaystyle -\frac{x}{3} \leq \frac{x}{2}-1\frac{1}{3} < \frac{1}{6}$ $\displaystyle -\frac{x}{3} \leq \frac{x}{2}-1\frac{1}{3} < \frac{1}{6}$ $\displaystyle -2x \leq 3x-8 < 1 \text{ or } -2x \leq 3x-8 \text{ or } 8 \leq 5x$ $\displaystyle \frac{8}{5} \leq x \text{ or } 3x-8 < 1 \text{ or } 3x < 9 \text{ or } x < 3$

Therefore $\displaystyle \{ x : \frac{8}{5} \leq x < 3, x \in R \}$  $\displaystyle \\$

Question 8: Find the value of $\displaystyle x$ which satisfies the inequation: $\displaystyle -2\frac{5}{6} < \frac{1}{2} - \frac{2x}{3} \leq 2, x \in W$ . $\displaystyle -2\frac{5}{6} < \frac{1}{2} - \frac{2x}{3} \leq 2$ $\displaystyle -\frac{17}{6} < \frac{1}{2} -\frac{2x}{3} \leq 2$ $\displaystyle -17 < 3-4x \leq 12 \text{ or } -17 < 3-4x \text{ or } 4x < 20 \text{ or } x < 5$ $\displaystyle 3-4x \leq 12 \text{ or } -9 \leq 4x \text{ or } -2.25 \leq x$

Therefore $\displaystyle \{x : -2.25 \leq x < 5, x \in W \}$ $\displaystyle x \in \{0, 1, 2, 3, 4\}$ Question 9: Solve the inequation: $\displaystyle 3-2x \geq x-12$ given that $\displaystyle x \in N$ $\displaystyle 3-2x \geq x-12$ $\displaystyle \Rightarrow 3x \leq 15$ $\displaystyle \Rightarrow x \leq 5 \ or \ x \in \{1, 2, 3, 4, 5 \}$ $\displaystyle \\$

Question 10: Solve the inequation: $\displaystyle 12+1\frac{5}{6}x \leq 5+3x$ and $\displaystyle x \in R$ . $\displaystyle 12+1\frac{5}{6}x \leq 5+3x$ $\displaystyle \Rightarrow 12+\frac{11}{6}x \leq 5+3x$ $\displaystyle \Rightarrow 7 \leq \frac{7}{6}x$ $\displaystyle \Rightarrow x \geq 6 \ or\ \{x: x\in R \ and \ x \geq 6 \}$