**1** **Solve $24x < 100,$ when$\\$ (i) x is a natural number $\\$ (ii) x is an integer.**

The given inequality is $24x < 100$$\\$ $24x < 100 \\ \implies \dfrac{24x}{24} < \dfrac{100}{24} \\ [\text{Dividing both sides by same positive number}] \\ \implies x < \dfrac{25}{6}$$\\$ (i) It is evident that $1,2,3,$ and $4$ are the only natural numbers less than$ \dfrac{25}{6}$$\\$ Thus, when x is a natural number, the solutions of the given inequality are $1, 2, 3$ and $4$ $\\$ Hence, in this case, the solution set is $\{1, 2,3, 4 \}$ .$\\$ (ii)The integers less than$ \dfrac{25}{6}$ are.....$-3,-2,-1,0,1,2,3,4.$$\\$ Thus, when x is an integer , the solutions of the given inequality are ... $- 3, - 2, - 1,0,1, 2,3, 4$$\\$ Hence, in this case, the solution set is $\{- 3, - 2, - 1,0,1,2,3,4 \}$

**2** **Solve $- 12 x > 30,$ when $\\$ (i) x is a natural number $\\$ (ii) x is an integer**

The given inequality is $-12 x > 30.$$\\$ $-12x >30.\\ \implies \dfrac{-12x}{-12} < \dfrac{30}{-12}\\ [\text{Dividing both sides by same negative number}]\\ \implies x< -\dfrac{5}{2}$$\\$ (i) There is no natural number less than $(-\dfrac{5}{2}).$ Thus, when x is a natural number, there is no solution of the given inequality.$\\$ (ii)The integers less than $(-\dfrac{5}{2})$ are......$-5,-4,-3.$$\\$ Thus, when x is an integer, the solutions of the given inequality are $...., - 5, - 4, - 3$$\\$ Hence, in this case, the solution set is $\{ ...., - 5, -4, -3 \}.$

**3** **Solve $5 x - 3 < 7 ,$ when $\\$(i) x is an integer$\\$ (ii) x is a real number**

The given inequality is $5 x - 3 < 7 .$$\\$ $5 x - 3 < 7 \\ \implies 5x-3+3 < 7+3\\ \implies 5x < 10\\ \implies \dfrac{5x}{5} < \dfrac{10}{5}\\ \implies x < 2$$\\$ (i) The integers less than $2$ are $..., - 4, - 3, - 2, - 1,0,1 .$$\\$ Thus, when $x$ is an integer, the solutions of the given inequality are $..., -4, - 3, - 2, - 1,0,1 .$$\\$ Hence, in this case, the solution set is $\{..., - 4, - 3, - 2, - 1,0,1 \}$ .$\\$ (ii) When $x$ is a real number, the solutions of the given inequality are given by $x < 2$, that is, all real numbers $x$ which are less than $2.$$\\$ Thus, the solution set of the given inequality is $x \in (-\infty , 2 ).$

**4** **Solve $3 x + 8 > 2 ,$ when$\\$ (i) x is an integer $\\$ (ii) x is a real number**

The given inequality is $ 3 x + 8 > 2$$\\$ $3 x + 8 > 2\\ \implies 3 x + 8 - 8 > 2 -8\\ \implies 3 x >- 6\\ \implies \dfrac{3x}{3}> \dfrac{-6}{3}\\ \implies x > -2$$\\$ (i) The integers greater than $- 2$ are $- 1,0,1, 2,...$$\\$ Thus, when x is an integer, the solutions of the given inequality are $- 1,0,1, 2,...$$\\$ Hence, in this case, the solution set is $\{ -1,0,1,2,... \}$ .$\\$ (ii) When $x$ is a real number, the solutions of the given inequality are all the real numbers, which are greater than $-2 $. Thus, in this case, the solution set is $(-2, \infty)$ .

**5** **Solve the given inequality for real $x : 4 x + 3 < 5 x + 7$**

$4 x + 3 < 5 x + 7\\ \implies 4 x + 3 - 7 < 5 x + 7 - 7 \\ \implies 4 x - 4 < 5 x\\ \implies 4 x - 4 - 4 x < 5 x - 4 x\\ \implies -4 < x $$\\$Thus, all real numbers x, which are greater than $- 4$ , are the solutions of the given inequality. Hence, the solution set of the given inequality is $(-4, \infty).$

**6** **Solve $24 x < 100$ , when (i) x is a natural number (ii) x is an integer.**

The given inequality is $24 x < 100$ $\\$ $24 x < 100 \\ \Rightarrow \dfrac{24 x}{24} < \dfrac{100}{24 } \quad [\text{Dividing both sides by same positive number}]\\ \Rightarrow x < \dfrac{25}{6}$$\\$ (i) It is evident that $1,2,3 $ and $4$ are the only natural numbers less than $\dfrac{25}{6}$$\\$ Thus, when x is a natural number, the solutions of the given inequality are $1, 2, 3$ and $4$$\\$ Hence, in this case, the solution set is $(1, 2,3, 4 )$ .$\\$ (ii) The integers less than $\dfrac{25}{6}$ are......$-3,-2,-1,0,1,2,3,4$$\\$ Thus, when $x$ is an integer , the solutions of the given inequality are ... $- 3, - 2, -1,0,1, 2,3, 4$$\\$ Hence, in this case, the solution set is $\{-3, - 2, -1,0,1,2,3,4 \}$

**7** **Solve $-12 x > 30$, when$\\$ (i) x is a natural number $\\$ (ii) x is an integer**

The given inequality is $-12 x > 30.$$\\$ $-12x > 30\\ \Rightarrow \dfrac{-12x}{-12} < \dfrac{30}{-12} [\text{Dividing both sides by same negative number}]\\ \Rightarrow x < -\dfrac{5}{2}$$\\$ (i) There is no natural number less than $(-\dfrac{5}{2})$$\\$ Thus, when x is a natural number, there is no solution of the given inequality.$\\$ (ii) The integers less than $(-\dfrac{5}{2})$ are .... $-5,-4,-3.$$\\$ Thus, when $x$ is an integer, the solutions of the given inequality are ....,$- 5, - 4, - 3$$\\$ Hence, in this case, the solution set is $\{...., - 5, - 4, - 3 \}$ .

**8** **Solve $5 x -3 < 7$ , when$\\$ (i) x is an integer$\\$ (ii) x is a real number**

The given inequality is $5 x - 3 < 7$$\\$ $ 5x-3 < 7\\ \Rightarrow 5x -3+3 < 7+3\\ \Rightarrow 5x < 10\\ \Rightarrow \dfrac{5x}{5} < \dfrac{10}{5}\\ \Rightarrow x< 2$$\\$ (i) The integers less than $2$ are ..., $-4, - 3, -2, - 1,0,1 .$$\\$ Thus, when x is an integer, the solutions of the given inequality are ..., $- 4, - 3, - 2, - 1,0,1 .$$\\$ Hence, in this case, the solution set is $\{ ..., - 4, - 3, - 2, - 1,0,1 \}$$\\$ . (ii) When $x$ is a real number, the solutions of the given inequality are given by $x < 2,$ that is, all real numbers x which are less than $2$.$\\$ Thus, the solution set of the given inequality is $x \in (- \infty, 2 )$ .

**9** **Solve $3 x + 8 > 2$ , when$\\$ (i) x is an integer$\\$ (ii) x is a real number**

The given inequality is $3 x + 8 > 2$$\\$ $3x+8 > 2\\ \Rightarrow 3x+8-8 > 2-8\\ \Rightarrow 3x > -6\\ \Rightarrow \dfrac{3x}{3} > \dfrac{-6}{3}\\ \Rightarrow x > -2$$\\$ (i) The integers greater than $-2$ are $-1,0,1, 2,...$$\\$ Thus, when x is an integer, the solutions of the given inequality are $- 1,0,1, 2,...$$\\$ Hence, in this case, the solution set is $\{-1,0,1,2,... \}$ .$\\$ (ii) When $x$ is a real number, the solutions of the given inequality are all the real numbers, which are greater than $- 2 $.$\\$ Thus, in this case, the solution set is $(-2, \infty)$ .

**10** **Solve the given inequality for real $x : 4 x + 3 < 5 x + 7$**

$4x+3 < 5x +7\\ \Rightarrow 4x+3-7 < 5x +7 -7\\ \Rightarrow 4x -4 < 5x\\ \Rightarrow 4x - 4 - 4x < 5x -4x \\ \Rightarrow -4 < x $$\\$ Thus, all real numbers $x$, which are greater than $-4$ , are the solutions of the given inequality.$\\$ Hence, the solution set of the given inequality is $(-4, \infty )$ .

**11** **Solve the given inequality for real x : 3 x - 7 > 5 x - 1**

3 x - 7 > 5 x - 1$\\$ 3 x - 7 + 7 > 5 x - 1 + 7$\\$ 3 x >5 x + 6$\\$ 3 x - 5 x > 5 x +6 - 5 x$\\$ - 2 x > 6$\\$ $\dfrac{-2x}{-2}<\dfrac{6}{-2}$$\\$ x<-3$\\$ Thus, all real numbers x, which are less than - 3 , are the solutions of the given inequality.$\\$ Hence, the solution set of the given inequality is (-$\infty$ , - 3 )

**12** **Solve the given inequality for real x : 3 (x -1 )$\leq$ 2 ( x - 3 )**

3 ( x - 1 )$\leq$ 2 ( x - 3 )$\\$ 3 x - 3 $\leq$ 2 x - 6$\\$ 3 x - 3 + 3 $\leq$ 2 x - 6 + 3$\\$ 3 x $\leq$ 2 x - 3$\\$ 3 x - 2 x $\leq$ 2 x - 3 - 2 x$\\$ x $\leq$ - 3$\\$ Thus, all real numbers x, which are less than or equal to - 3 , are the solutions of the given inequality.$\\$ Hence, the solution set of the given inequality is ($-\infty,- 3 ]$ .