Linear Inequations

Solving,

$$\begin{aligned}\Rightarrow x + 2 \lt 1 \\[1em] \Rightarrow x \lt 1 - 2 \\[1em] \Rightarrow x \lt -1\end{aligned}$$

The values of $x$ in the replacement set {-3, -2, -1, 0, 1, 2, 3} which are less than -1 are -3 and -2.

Hence, solution set = $\bm{{-3, -2}}$.

Solving,

$$\begin{aligned}\Rightarrow 2x - 1 \lt 4 \\[1em] \Rightarrow 2x \lt 4 + 1 \\[1em] \Rightarrow 2x \lt 5 \\[1em] \Rightarrow x \lt \frac{5}{2} \\[1em] \Rightarrow x \lt 2\frac{1}{2}\end{aligned}$$

The values of $x$ in the replacement set {-3, -2, -1, 0, 1, 2, 3} which are less than $2\frac{1}{2}$ are -3, -2, -1, 0, 1 and 2.

Hence, solution set = $\bm{{-3, -2, -1, 0, 1, 2}}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{2}{3}x \lt 1 \\[1em] \Rightarrow x \lt 1 \times \frac{3}{2} \\[1em] \Rightarrow x \lt \frac{3}{2} \\[1em] \Rightarrow x \lt 1\frac{1}{2}\end{aligned}$$

The values of $x$ in the replacement set {-3, -2, -1, 0, 1, 2, 3} which are less than $1\frac{1}{2}$ are -3, -2, -1, 0 and 1.

Hence, solution set = $\bm{{-3, -2, -1, 0, 1}}$.

Solving,

$$\begin{aligned}\Rightarrow 1 - x \gt 0 \\[1em] \Rightarrow -x \gt 0 - 1 \\[1em] \Rightarrow -x \gt -1 \\[1em] \Rightarrow x \lt 1\end{aligned}$$

The values of $x$ in the replacement set {-3, -2, -1, 0, 1, 2, 3} which are less than 1 are -3, -2, -1 and 0.

Hence, solution set = $\bm{{-3, -2, -1, 0}}$.

Solving,

$$\begin{aligned}\Rightarrow 3 - 5x \lt -1 \\[1em] \Rightarrow -5x \lt -1 - 3 \\[1em] \Rightarrow -5x \lt -4 \\[1em] \Rightarrow x \gt \frac{-4}{-5} \\[1em] \Rightarrow x \gt \frac{4}{5}\end{aligned}$$

The values of $x$ in the replacement set {-3, -2, -1, 0, 1, 2, 3} which are greater than $\frac{4}{5}$ are 1, 2 and 3.

Hence, solution set = $\bm{{1, 2, 3}}$.

Solving,

$$\begin{aligned}\Rightarrow 2 - 3x \gt 1 \\[1em] \Rightarrow -3x \gt 1 - 2 \\[1em] \Rightarrow -3x \gt -1 \\[1em] \Rightarrow x \lt \frac{-1}{-3} \\[1em] \Rightarrow x \lt \frac{1}{3}\end{aligned}$$

The values of $x$ in the replacement set {-3, -2, -1, 0, 1, 2, 3} which are less than $\frac{1}{3}$ are -3, -2, -1 and 0.

Hence, solution set = $\bm{{-3, -2, -1, 0}}$.

Solving,

$$\begin{aligned}\Rightarrow -6 \ge 2x - 4 \\[1em] \Rightarrow -6 + 4 \ge 2x \\[1em] \Rightarrow -2 \ge 2x \\[1em] \Rightarrow -1 \ge x \\[1em] \Rightarrow x \le -1\end{aligned}$$

The values of $x$ in the replacement set {-3, -2, -1, 0, 1, 2, 3} which are less than or equal to -1 are -3, -2 and -1.

Hence, solution set = $\bm{{-3, -2, -1}}$.

Solving,

$$\begin{aligned}\Rightarrow 3x - 5 \ge -12 \\[1em] \Rightarrow 3x \ge -12 + 5 \\[1em] \Rightarrow 3x \ge -7 \\[1em] \Rightarrow x \ge \frac{-7}{3} \\[1em] \Rightarrow x \ge -2\frac{1}{3}\end{aligned}$$

The values of $x$ in the replacement set {-3, -2, -1, 0, 1, 2, 3} which are greater than or equal to $-2\frac{1}{3}$ are -2, -1, 0, 1, 2 and 3.

Hence, solution set = $\bm{{-2, -1, 0, 1, 2, 3}}$.

Solving,

$$\begin{aligned}\Rightarrow 14 - 2x \lt 6 \\[1em] \Rightarrow -2x \lt 6 - 14 \\[1em] \Rightarrow -2x \lt -8 \\[1em] \Rightarrow x \gt \frac{-8}{-2} \\[1em] \Rightarrow x \gt 4\end{aligned}$$

There is no value of $x$ in the replacement set {-3, -2, -1, 0, 1, 2, 3} which is greater than 4.

Hence, solution set = $\bm{φ}$.

Solving,

$$\begin{aligned}\Rightarrow 3x - 8 \lt 0 \\[1em] \Rightarrow 3x \lt 0 + 8 \\[1em] \Rightarrow 3x \lt 8 \\[1em] \Rightarrow x \lt \frac{8}{3} \\[1em] \Rightarrow x \lt 2\frac{2}{3}\end{aligned}$$

As $x$ ∈ N, the natural numbers less than $2\frac{2}{3}$ are 1 and 2.

Hence, solution set = $\bm{{1, 2}}$.

Solving,

$$\begin{aligned}\Rightarrow 7x + 3 \le 17 \\[1em] \Rightarrow 7x \le 17 - 3 \\[1em] \Rightarrow 7x \le 14 \\[1em] \Rightarrow x \le \frac{14}{7} \\[1em] \Rightarrow x \le 2\end{aligned}$$

As $x$ ∈ N, the natural numbers less than or equal to 2 are 1 and 2.

Hence, solution set = $\bm{{1, 2}}$.

Solving,

$$\begin{aligned}\Rightarrow 5 - x \gt 1 \\[1em] \Rightarrow -x \gt 1 - 5 \\[1em] \Rightarrow -x \gt -4 \\[1em] \Rightarrow x \lt 4\end{aligned}$$

As $x$ ∈ N, the natural numbers less than 4 are 1, 2 and 3.

Hence, solution set = $\bm{{1, 2, 3}}$.

Solving,

$$\begin{aligned}\Rightarrow 1 - 3x \gt -4 \\[1em] \Rightarrow -3x \gt -4 - 1 \\[1em] \Rightarrow -3x \gt -5 \\[1em] \Rightarrow x \lt \frac{-5}{-3} \\[1em] \Rightarrow x \lt \frac{5}{3} \\[1em] \Rightarrow x \lt 1\frac{2}{3}\end{aligned}$$

As $x$ ∈ N, the only natural number less than $1\frac{2}{3}$ is 1.

Hence, solution set = $\bm{{1}}$.

Given,

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

Multiplying both sides by 2, we get

$$\begin{aligned}\Rightarrow 2 \times \Big(\frac{3}{2} - \frac{x}{2}\Big) \gt 2 \times (-1) \\[1em] \Rightarrow \frac{2 \times 3}{2} - \frac{2x}{2} \gt -2 \\[1em] \Rightarrow 3 - x \gt -2 \\[1em] \Rightarrow -x \gt -2 - 3 \\[1em] \Rightarrow -x \gt -5 \\[1em] \Rightarrow x \lt 5\end{aligned}$$

As $x$ ∈ N, the natural numbers less than 5 are 1, 2, 3 and 4.

Hence, solution set = $\bm{{1, 2, 3, 4}}$.

Given,

$-\frac{1}{4} \le \frac{1}{2} - \frac{x}{3}$

L.C.M. of 4, 2 and 3 = 12

Multiplying both sides by 12, we get

$$\begin{aligned}\Rightarrow 12 \times \Big(-\frac{1}{4}\Big) \le 12 \times \Big(\frac{1}{2} - \frac{x}{3}\Big) \\[1em] \Rightarrow -\frac{12}{4} \le \frac{12}{2} - \frac{12x}{3} \\[1em] \Rightarrow -3 \le 6 - 4x \\[1em] \Rightarrow -3 - 6 \le -4x \\[1em] \Rightarrow -9 \le -4x \\[1em] \Rightarrow \frac{-9}{-4} \ge x \\[1em] \Rightarrow x \le \frac{9}{4} \\[1em] \Rightarrow x \le 2\frac{1}{4}\end{aligned}$$

As $x$ ∈ N, the natural numbers less than or equal to $2\frac{1}{4}$ are 1 and 2.

Hence, solution set = $\bm{{1, 2}}$.

Solving,

$$\begin{aligned}\Rightarrow 9x - 7 \le 25 + 3x \\[1em] \Rightarrow 9x - 3x \le 25 + 7 \\[1em] \Rightarrow 6x \le 32 \\[1em] \Rightarrow x \le \frac{32}{6} \\[1em] \Rightarrow x \le \frac{16}{3} \\[1em] \Rightarrow x \le 5\frac{1}{3}\end{aligned}$$

As $x$ ∈ Z, the integers less than or equal to $5\frac{1}{3}$ are ....., -1, 0, 1, 2, 3, 4 and 5.

Representing the solution set on the number line, we get

Hence, solution set = $\bm{{....., -1, 0, 1, 2, 3, 4, 5}}$.

Solving,

$$\begin{aligned}\Rightarrow -17 \lt 9x - 8 \\[1em] \Rightarrow -17 + 8 \lt 9x \\[1em] \Rightarrow -9 \lt 9x \\[1em] \Rightarrow \frac{-9}{9} \lt x \\[1em] \Rightarrow -1 \lt x \\[1em] \Rightarrow x \gt -1\end{aligned}$$

As $x$ ∈ Z, the integers greater than -1 are 0, 1, 2, 3, 4, 5, 6, 7, .....

Representing the solution set on the number line, we get

Hence, solution set = $\bm{{0, 1, 2, 3, 4, 5, 6, 7, .....}}$.

Solving,

$$\begin{aligned}\Rightarrow -4(x + 5) \gt 10 \\[1em] \Rightarrow -4x - 20 \gt 10 \\[1em] \Rightarrow -4x \gt 10 + 20 \\[1em] \Rightarrow -4x \gt 30 \\[1em] \Rightarrow x \lt \frac{30}{-4} \\[1em] \Rightarrow x \lt -\frac{15}{2} \\[1em] \Rightarrow x \lt -7\frac{1}{2}\end{aligned}$$

As $x$ ∈ Z, the integers less than $-7\frac{1}{2}$ are ....., -10, -9 and -8.

Representing the solution set on the number line, we get

Hence, solution set = $\bm{{....., -10, -9, -8}}$.

Solving,

$$\begin{aligned}\Rightarrow 4 - 3x \lt 13 + x \\[1em] \Rightarrow -3x - x \lt 13 - 4 \\[1em] \Rightarrow -4x \lt 9 \\[1em] \Rightarrow x \gt \frac{9}{-4} \\[1em] \Rightarrow x \gt -\frac{9}{4} \\[1em] \Rightarrow x \gt -2\frac{1}{4}\end{aligned}$$

As $x$ ∈ Z, the integers greater than $-2\frac{1}{4}$ are -2, -1, 0, 1, 2, 3, .....

Representing the solution set on the number line, we get

Hence, solution set = $\bm{{-2, -1, 0, 1, 2, 3, .....}}$.

Solving,

$$\begin{aligned}\Rightarrow 5 - 4x \lt 10 - x \\[1em] \Rightarrow -4x + x \lt 10 - 5 \\[1em] \Rightarrow -3x \lt 5 \\[1em] \Rightarrow x \gt \frac{5}{-3} \\[1em] \Rightarrow x \gt -\frac{5}{3} \\[1em] \Rightarrow x \gt -1\frac{2}{3}\end{aligned}$$

As $x$ ∈ Z, the integers greater than $-1\frac{2}{3}$ are -1, 0, 1, 2, 3, 4, .....

Representing the solution set on the number line, we get

Hence, solution set = $\bm{{-1, 0, 1, 2, 3, 4, .....}}$.

Solving,

$$\begin{aligned}\Rightarrow 10 - 2(1 + 4x) \lt 20 \\[1em] \Rightarrow 10 - 2 - 8x \lt 20 \\[1em] \Rightarrow 8 - 8x \lt 20 \\[1em] \Rightarrow -8x \lt 20 - 8 \\[1em] \Rightarrow -8x \lt 12 \\[1em] \Rightarrow x \gt \frac{12}{-8} \\[1em] \Rightarrow x \gt -\frac{3}{2} \\[1em] \Rightarrow x \gt -1\frac{1}{2}\end{aligned}$$

As $x$ ∈ Z, the integers greater than $-1\frac{1}{2}$ are -1, 0, 1, 2, 3, .....

Representing the solution set on the number line, we get

Hence, solution set = $\bm{{-1, 0, 1, 2, 3, .....}}$.

Solving,

$$\begin{aligned}\Rightarrow 1 - 4x \ge -1 \\[1em] \Rightarrow -4x \ge -1 - 1 \\[1em] \Rightarrow -4x \ge -2 \\[1em] \Rightarrow x \le \frac{-2}{-4} \\[1em] \Rightarrow x \le \frac{1}{2}\end{aligned}$$

As $x$ ∈ N, there is no natural number less than or equal to $\frac{1}{2}$.

Hence, solution set = $\bm{φ}$.

We shall take up $-3 \le 4x + 1$ and $4x + 1 \lt 9$ separately.

Case I. Solving $-3 \le 4x + 1$,

$$\begin{aligned}\Rightarrow -3 \le 4x + 1 \\[1em] \Rightarrow -3 - 1 \le 4x \\[1em] \Rightarrow -4 \le 4x \\[1em] \Rightarrow \frac{-4}{4} \le x \\[1em] \Rightarrow x \ge -1\end{aligned}$$

⇒ Solution set of $-3 \le 4x + 1$ is A = {-1, 0, 1, 2, 3, .....}

Case II. Solving $4x + 1 \lt 9$,

$$\begin{aligned}\Rightarrow 4x + 1 \lt 9 \\[1em] \Rightarrow 4x \lt 9 - 1 \\[1em] \Rightarrow 4x \lt 8 \\[1em] \Rightarrow x \lt \frac{8}{4} \\[1em] \Rightarrow x \lt 2\end{aligned}$$

⇒ Solution set of $4x + 1 \lt 9$ is B = {....., -2, -1, 0, 1}

⇒ Solution set for both the given inequations = A ∩ B = {-1, 0, 1}

Hence, solution set = $\bm{{-1, 0, 1}}$.

We shall take up $0 \lt 2x - 5$ and $2x - 5 \lt 5$ separately.

Case I. Solving $0 \lt 2x - 5$,

$$\begin{aligned}\Rightarrow 0 \lt 2x - 5 \\[1em] \Rightarrow 0 + 5 \lt 2x \\[1em] \Rightarrow 5 \lt 2x \\[1em] \Rightarrow \frac{5}{2} \lt x \\[1em] \Rightarrow x \gt \frac{5}{2} \\[1em] \Rightarrow x \gt 2\frac{1}{2}\end{aligned}$$

⇒ Solution set of $0 \lt 2x - 5$ is A = {3, 4, 5, 6, .....}

Case II. Solving $2x - 5 \lt 5$,

$$\begin{aligned}\Rightarrow 2x - 5 \lt 5 \\[1em] \Rightarrow 2x \lt 5 + 5 \\[1em] \Rightarrow 2x \lt 10 \\[1em] \Rightarrow x \lt \frac{10}{2} \\[1em] \Rightarrow x \lt 5\end{aligned}$$

⇒ Solution set of $2x - 5 \lt 5$ is B = {0, 1, 2, 3, 4}

⇒ Solution set for both the given inequations = A ∩ B = {3, 4}

Hence, solution set = $\bm{{3, 4}}$.

We shall take up $-3 \lt \frac{x}{2} - 1$ and $\frac{x}{2} - 1 \lt 1$ separately.

Case I. Solving $-3 \lt \frac{x}{2} - 1$,

$$\begin{aligned}\Rightarrow -3 \lt \frac{x}{2} - 1 \\[1em] \Rightarrow -3 + 1 \lt \frac{x}{2} \\[1em] \Rightarrow -2 \lt \frac{x}{2} \\[1em] \Rightarrow 2 \times (-2) \lt 2 \times \frac{x}{2} \\[1em] \Rightarrow -4 \lt x \\[1em] \Rightarrow x \gt -4\end{aligned}$$

⇒ Solution set of $-3 \lt \frac{x}{2} - 1$ is A = {-3, -2, -1, 0, 1, 2, .....}

Case II. Solving $\frac{x}{2} - 1 \lt 1$,

$$\begin{aligned}\Rightarrow \frac{x}{2} - 1 \lt 1 \\[1em] \Rightarrow \frac{x}{2} \lt 1 + 1 \\[1em] \Rightarrow \frac{x}{2} \lt 2 \\[1em] \Rightarrow 2 \times \frac{x}{2} \lt 2 \times 2 \\[1em] \Rightarrow x \lt 4\end{aligned}$$

⇒ Solution set of $\frac{x}{2} - 1 \lt 1$ is B = {....., -1, 0, 1, 2, 3}

⇒ Solution set for both the given inequations = A ∩ B = {-3, -2, -1, 0, 1, 2, 3}

Hence, solution set = $\bm{{-3, -2, -1, 0, 1, 2, 3}}$.

We shall take up $-4 \lt \frac{2x}{5} + 1$ and $\frac{2x}{5} + 1 \lt -3$ separately.

Case I. Solving $-4 \lt \frac{2x}{5} + 1$,

$$\begin{aligned}\Rightarrow -4 \lt \frac{2x}{5} + 1 \\[1em] \Rightarrow -4 - 1 \lt \frac{2x}{5} \\[1em] \Rightarrow -5 \lt \frac{2x}{5} \\[1em] \Rightarrow 5 \times (-5) \lt 5 \times \frac{2x}{5} \\[1em] \Rightarrow -25 \lt 2x \\[1em] \Rightarrow \frac{-25}{2} \lt x \\[1em] \Rightarrow x \gt -\frac{25}{2} \\[1em] \Rightarrow x \gt -12\frac{1}{2}\end{aligned}$$

⇒ Solution set of $-4 \lt \frac{2x}{5} + 1$ is A = {-12, -11, -10, -9, .....}

Case II. Solving $\frac{2x}{5} + 1 \lt -3$,

$$\begin{aligned}\Rightarrow \frac{2x}{5} + 1 \lt -3 \\[1em] \Rightarrow \frac{2x}{5} \lt -3 - 1 \\[1em] \Rightarrow \frac{2x}{5} \lt -4 \\[1em] \Rightarrow 5 \times \frac{2x}{5} \lt 5 \times (-4) \\[1em] \Rightarrow 2x \lt -20 \\[1em] \Rightarrow x \lt \frac{-20}{2} \\[1em] \Rightarrow x \lt -10\end{aligned}$$

⇒ Solution set of $\frac{2x}{5} + 1 \lt -3$ is B = {....., -13, -12, -11}

⇒ Solution set for both the given inequations = A ∩ B = {-12, -11}

Hence, solution set = $\bm{{-12, -11}}$.

We shall take up $-1 \lt \frac{2x}{3} + 1$ and $\frac{2x}{3} + 1 \le 5$ separately.

Case I. Solving $-1 \lt \frac{2x}{3} + 1$,

$$\begin{aligned}\Rightarrow -1 \lt \frac{2x}{3} + 1 \\[1em] \Rightarrow -1 - 1 \lt \frac{2x}{3} \\[1em] \Rightarrow -2 \lt \frac{2x}{3} \\[1em] \Rightarrow 3 \times (-2) \lt 3 \times \frac{2x}{3} \\[1em] \Rightarrow -6 \lt 2x \\[1em] \Rightarrow -3 \lt x \\[1em] \Rightarrow x \gt -3\end{aligned}$$

⇒ Solution set of $-1 \lt \frac{2x}{3} + 1$ is A = {$x$ ∈ Q : $x \gt -3$}

Case II. Solving $\frac{2x}{3} + 1 \le 5$,

$$\begin{aligned}\Rightarrow \frac{2x}{3} + 1 \le 5 \\[1em] \Rightarrow \frac{2x}{3} \le 5 - 1 \\[1em] \Rightarrow \frac{2x}{3} \le 4 \\[1em] \Rightarrow 3 \times \frac{2x}{3} \le 3 \times 4 \\[1em] \Rightarrow 2x \le 12 \\[1em] \Rightarrow x \le 6\end{aligned}$$

⇒ Solution set of $\frac{2x}{3} + 1 \le 5$ is B = {$x$ ∈ Q : $x \le 6$}

⇒ Solution set for both the given inequations = A ∩ B = {$x$ ∈ Q : $-3 \lt x \le 6$}

Hence, solution set = $\bm{{x ∈ Q : -3 \lt x \le 6}}$.

We shall take up $-\frac{1}{4} \le \frac{1}{2} - \frac{x}{3}$ and $\frac{1}{2} - \frac{x}{3} \lt 2$ separately.

Case I. Solving $-\frac{1}{4} \le \frac{1}{2} - \frac{x}{3}$,

L.C.M. of 4, 2 and 3 = 12

Multiplying both sides by 12, we get

$$\begin{aligned}\Rightarrow 12 \times \Big(-\frac{1}{4}\Big) \le 12 \times \Big(\frac{1}{2} - \frac{x}{3}\Big) \\[1em] \Rightarrow -\frac{12}{4} \le \frac{12}{2} - \frac{12x}{3} \\[1em] \Rightarrow -3 \le 6 - 4x \\[1em] \Rightarrow -3 - 6 \le -4x \\[1em] \Rightarrow -9 \le -4x \\[1em] \Rightarrow \frac{-9}{-4} \ge x \\[1em] \Rightarrow x \le \frac{9}{4} \\[1em] \Rightarrow x \le 2\frac{1}{4}\end{aligned}$$

⇒ Solution set of $-\frac{1}{4} \le \frac{1}{2} - \frac{x}{3}$ is A = {....., 0, 1, 2}

Case II. Solving $\frac{1}{2} - \frac{x}{3} \lt 2$,

L.C.M. of 2 and 3 = 6

Multiplying both sides by 6, we get

$$\begin{aligned}\Rightarrow 6 \times \Big(\frac{1}{2} - \frac{x}{3}\Big) \lt 6 \times 2 \\[1em] \Rightarrow \frac{6}{2} - \frac{6x}{3} \lt 12 \\[1em] \Rightarrow 3 - 2x \lt 12 \\[1em] \Rightarrow -2x \lt 12 - 3 \\[1em] \Rightarrow -2x \lt 9 \\[1em] \Rightarrow x \gt \frac{9}{-2} \\[1em] \Rightarrow x \gt -\frac{9}{2} \\[1em] \Rightarrow x \gt -4\frac{1}{2}\end{aligned}$$

⇒ Solution set of $\frac{1}{2} - \frac{x}{3} \lt 2$ is B = {-4, -3, -2, -1, 0, 1, 2, .....}

⇒ Solution set for both the given inequations = A ∩ B = {-4, -3, -2, -1, 0, 1, 2}

Hence, solution set = $\bm{{-4, -3, -2, -1, 0, 1, 2}}$.

Given,

$3 + \frac{x}{4} \lt \frac{2x}{3} + 5$

L.C.M. of 4 and 3 = 12

Multiplying both sides by 12, we get

$$\begin{aligned}\Rightarrow 12 \times \Big(3 + \frac{x}{4}\Big) \lt 12 \times \Big(\frac{2x}{3} + 5\Big) \\[1em] \Rightarrow 12 \times 3 + \frac{12x}{4} \lt \frac{12 \times 2x}{3} + 12 \times 5 \\[1em] \Rightarrow 36 + 3x \lt 8x + 60 \\[1em] \Rightarrow 36 - 60 \lt 8x - 3x \\[1em] \Rightarrow -24 \lt 5x \\[1em] \Rightarrow \frac{-24}{5} \lt x \\[1em] \Rightarrow x \gt -\frac{24}{5} \\[1em] \Rightarrow x \gt -4\frac{4}{5}\end{aligned}$$

Hence, solution set = $\bm{{x ∈ Q : x \gt -4\frac{4}{5}}}$.

Given,

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

L.C.M. of 4 and 3 = 12

Multiplying both sides by 12, we get

$$\begin{aligned}\Rightarrow 12 \times \frac{3x + 1}{4} \le 12 \times \frac{5x - 2}{3} \\[1em] \Rightarrow \frac{12(3x + 1)}{4} \le \frac{12(5x - 2)}{3} \\[1em] \Rightarrow 3(3x + 1) \le 4(5x - 2) \\[1em] \Rightarrow 9x + 3 \le 20x - 8 \\[1em] \Rightarrow 3 + 8 \le 20x - 9x \\[1em] \Rightarrow 11 \le 11x \\[1em] \Rightarrow \frac{11}{11} \le x \\[1em] \Rightarrow x \ge 1\end{aligned}$$

Hence, solution set = $\bm{{x ∈ Q : x \ge 1}}$.

Given,

$\frac{1}{3}(4x - 1) + 3 \le 4 + \frac{2}{5}(6x + 2)$

L.C.M. of 3 and 5 = 15

Multiplying both sides by 15, we get

$$\begin{aligned}\Rightarrow 15 \times \Big[\frac{1}{3}(4x - 1) + 3\Big] \le 15 \times \Big[4 + \frac{2}{5}(6x + 2)\Big] \\[1em] \Rightarrow \frac{15}{3}(4x - 1) + 15 \times 3 \le 15 \times 4 + \frac{15 \times 2}{5}(6x + 2) \\[1em] \Rightarrow 5(4x - 1) + 45 \le 60 + 6(6x + 2) \\[1em] \Rightarrow 20x - 5 + 45 \le 60 + 36x + 12 \\[1em] \Rightarrow 20x + 40 \le 36x + 72 \\[1em] \Rightarrow 40 - 72 \le 36x - 20x \\[1em] \Rightarrow -32 \le 16x \\[1em] \Rightarrow \frac{-32}{16} \le x \\[1em] \Rightarrow x \ge -2\end{aligned}$$

Hence, solution set = $\bm{{x ∈ Q : x \ge -2}}$.

Solving,

$$\begin{aligned}\Rightarrow -2x + 1 \gt 9 \\[1em] \Rightarrow -2x \gt 9 - 1 \\[1em] \Rightarrow -2x \gt 8 \\[1em] \Rightarrow x \lt \frac{8}{-2} \\[1em] \Rightarrow x \lt -4\end{aligned}$$

So, Assertion (A) is true.

Dividing each side of an inequality by a negative number reverses the inequality.

So, Reason (R) is true.

Here, both the sides were divided by -2, which is a negative number, and this is exactly why the sign > changed to <.

So, Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 6x + 15 \gt 0 \\[1em] \Rightarrow 6x \gt 0 - 15 \\[1em] \Rightarrow 6x \gt -15 \\[1em] \Rightarrow x \gt \frac{-15}{6} \\[1em] \Rightarrow x \gt -\frac{5}{2} \\[1em] \Rightarrow x \gt -2\frac{1}{2}\end{aligned}$$

As $x$ ∈ N, the natural numbers greater than $-2\frac{1}{2}$ are 1, 2, 3, .....

⇒ Solution set = {1, 2, 3, .....}

As 0 is not a natural number, it cannot belong to the solution set.

So, Assertion (A) is false.

Solution set is the subset of the replacement set consisting of all those values of the variable which satisfy the given inequation.

So, Reason (R) is true.

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 3 - x \lt 4 \\[1em] \Rightarrow -x \lt 4 - 3 \\[1em] \Rightarrow -x \lt 1 \\[1em] \Rightarrow x \gt -1\end{aligned}$$

The positive integers greater than -1 are 1, 2, 3, .....

So, the least positive integer value of $x$ is 1.

Hence, option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 2x + 6 \ge 0 \\[1em] \Rightarrow 2x \ge 0 - 6 \\[1em] \Rightarrow 2x \ge -6 \\[1em] \Rightarrow x \ge \frac{-6}{2} \\[1em] \Rightarrow x \ge -3\end{aligned}$$

The integers satisfying both $x \ge -3$ and $x \lt 0$ are -3, -2 and -1.

As $x \ne -2$, the value -2 has to be left out.

So, the two integer values of $x$ are -3 and -1.

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 5m \lt 23 \\[1em] \Rightarrow m \lt \frac{23}{5} \\[1em] \Rightarrow m \lt 4\frac{3}{5}\end{aligned}$$

The integers less than $4\frac{3}{5}$ are ....., 1, 2, 3 and 4.

So, the largest integer $m$ is 4.

Hence, option 3 is the correct option.

Given,

$x$ is an integer such that $-8 \le x \lt 4$.

⇒ $x$ ∈ {-8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3}

So, the greatest value of $x$ is 3.

Given, $y$ is an integer such that $-3 \le y \le 2$.

⇒ $y$ ∈ {-3, -2, -1, 0, 1, 2}

So, the greatest value of $y$ is 2.

The sum $(x + y)$ is greatest when both $x$ and $y$ are greatest.

⇒ Greatest value of $(x + y)$ = 3 + 2 = 5

Hence, option 2 is the correct option.

Given,

$2 \ge y \ge -3$ and $y$ ∈ I.

⇒ $y$ ∈ {-3, -2, -1, 0, 1, 2}

Given, $4 \gt z \ge -8$ and $z$ ∈ I.

⇒ $z$ ∈ {-8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3}

The product $yz$ is least when it is negative and numerically greatest.

This happens either when $y$ is the greatest positive value and $z$ is the smallest (most negative) value, or when $y$ is the smallest (most negative) value and $z$ is the greatest positive value.

⇒ Taking $y = 2$ and $z = -8$, $yz$ = 2 × (-8) = -16

⇒ Taking $y = -3$ and $z = 3$, $yz$ = -3 × 3 = -9

As -16 < -9, the least value of $yz$ is -16.

Hence, option 1 is the correct option.

We shall take up $-8 \lt 2x + 3$ and $2x + 3 \lt 11$ separately.

Case I. Solving $-8 \lt 2x + 3$,

$$\begin{aligned}\Rightarrow -8 \lt 2x + 3 \\[1em] \Rightarrow -8 - 3 \lt 2x \\[1em] \Rightarrow -11 \lt 2x \\[1em] \Rightarrow \frac{-11}{2} \lt x \\[1em] \Rightarrow x \gt -\frac{11}{2} \\[1em] \Rightarrow x \gt -5\frac{1}{2}\end{aligned}$$

⇒ Solution set of $-8 \lt 2x + 3$ is A = {-5, -4, -3, -2, .....}

Case II. Solving $2x + 3 \lt 11$,

$$\begin{aligned}\Rightarrow 2x + 3 \lt 11 \\[1em] \Rightarrow 2x \lt 11 - 3 \\[1em] \Rightarrow 2x \lt 8 \\[1em] \Rightarrow x \lt \frac{8}{2} \\[1em] \Rightarrow x \lt 4\end{aligned}$$

⇒ Solution set of $2x + 3 \lt 11$ is B = {....., 1, 2, 3}

⇒ Solution set for both the given inequations = A ∩ B = {-5, -4, -3, -2, -1, 0, 1, 2, 3}

So, the least integer satisfying both the inequations is -5.

Hence, option 3 is the correct option.

Solving $2x + 7 \lt 3$,

$$\begin{aligned}\Rightarrow 2x + 7 \lt 3 \\[1em] \Rightarrow 2x \lt 3 - 7 \\[1em] \Rightarrow 2x \lt -4 \\[1em] \Rightarrow x \lt \frac{-4}{2} \\[1em] \Rightarrow x \lt -2\end{aligned}$$

⇒ Solution set of $2x + 7 \lt 3$ is A = {....., -5, -4, -3}

⇒ Solution set of $x \ge -4$ is B = {-4, -3, -2, -1, 0, .....}

⇒ Solution set for both the given inequations = A ∩ B = {-4, -3}

⇒ Sum of all the integers satisfying both the inequations = -4 + (-3) = -7

Hence, option 1 is the correct option.