Linear Inequations

Given,

$$\begin{aligned}3x-11 \lt 3 \\[0.5em] \Rightarrow 3x-11 + 11 \lt 3 + 11 \\[0.5em] \Rightarrow 3x \lt 14 \\[0.5em] \Rightarrow x \lt \frac{14}{3} \\[0.5em] \Rightarrow x \lt 4.66\end{aligned}$$

But x ∈ {1, 2, 3, ..........,, 10}

∴ Solution Set = {1, 2, 3, 4}.

The graph of the solution set is shown by thick dots on the number line.

Given,

$$\begin{aligned}2(x - 3) \lt 1 \\[0.5em] \Rightarrow 2x-6 \lt 1 \\[0.5em] \Rightarrow 2x-6 + 6 \lt 1 + 6 \\[0.5em] \Rightarrow 2x \lt 7 \\[0.5em] \Rightarrow x \lt \frac{7}{2} \\[0.5em] \Rightarrow x \lt 3.5\end{aligned}$$

But x ∈ {1, 2, 3, ……., 10}
Solution set is {1, 2, 3}.

Given,

$$\begin{aligned}5 - 4x \gt 2 - 3x \\[0.5em] \Rightarrow 5 - 2 \gt -3x + 4x \\[0.5em] \Rightarrow 3 \gt x \\[0.5em] \Rightarrow x \lt 3\end{aligned}$$

Hence, Solution set is {0, 1, 2}.

The graph of the solution set is shown by thick dots on the number line.

Given,

$$\begin{aligned}30 - 4(2x - 1) \lt 30 \\[0.5em] \Rightarrow 30-8x+4 \lt 30 \\[0.5em] \Rightarrow 34-8x \lt 30 \\[0.5em] \Rightarrow 34-30 \lt 8x \\[0.5em] \Rightarrow 8x \gt 4 \\[0.5em] \Rightarrow x \gt \frac{1}{2}\end{aligned}$$

Since , x is a positive integer
x = {1, 2, 3, 4, …..}.

Given,

$$\begin{aligned}2(x – 2) \lt 3x – 2 \\[0.5em] \Rightarrow 2x – 4 \lt 3x – 2 \\[0.5em] \Rightarrow 2x – 3x \lt – 2 + 4 \\[0.5em] \Rightarrow – x \lt 2 \\[0.5em] \Rightarrow x \gt – 2 \\[0.5em]\end{aligned}$$

Solution set = {–1, 0, 1, 2, 3} .

Given,

$$\begin{aligned}\frac{2}{3}+\frac{1}{3} (x + 1) \gt 0 \\[0.5em] \Rightarrow \frac{2}{3} +\frac{x}{3} + \frac{1}{3} \gt 0 \\[0.5em] \Rightarrow \frac{x}{3} + 1 \gt 0 \\[0.5em] \Rightarrow \frac{x}{3} \gt – 1 \\[0.5em] \Rightarrow x \gt -3\end{aligned}$$

x is a negative integer Solution set = {-2, –1}.

Given,

$$\begin{aligned}x - 3(2 + x) \gt 2(3x - 1) \\[0.5em] \Rightarrow x - 6 - 3x \gt 6x - 2 \\[0.5em] \Rightarrow x - 3x - 6x \gt - 2 + 6 \\[0.5em] \Rightarrow - 8x \gt 4 \\[0.5em] \Rightarrow 8x \lt -4 \\[0.5em] \Rightarrow x \lt -\frac{1}{2}\end{aligned}$$

Since, x ∈ { -3, -2, -1, 0, 1, 2}
Hence, Solution set = { -3, -2, -1}.

The graph of the solution set is shown by thick dots on the number line.

Given,

$$\begin{aligned}x - 3 \lt 2x - 1\\[0.5em] \Rightarrow x - 2x \lt - 1 + 3\\[0.5em] \Rightarrow - x \lt 2\\[0.5em] \Rightarrow x \gt - 2\end{aligned}$$

Since, x ∈ {1, 2, 3, 4, 5, 6, 7, 9}

Solution set = {1, 2, 3, 4, 5, 6, 7, 9}.

Given, $$\begin{aligned}\frac{1}{2} +8x \gt 5x - \frac{3}{2} \\[0.5em] \Rightarrow 8x-5x \gt - \frac{3}{2}-\frac{1}{2} \\[0.5em] \Rightarrow 3x \gt - \frac{4}{2} \\[0.5em] \Rightarrow 3x \gt -2 \\[0.5em] \Rightarrow x \gt -\frac{2}{3}\end{aligned}$$

Since , x ∈ Z
Solution set = {0, 1, 2, 3, 4…..}

Given,

$$\begin{aligned}\frac{11-2x}{5} \ge \frac{9-3x}{8} + \frac{3}{4} \\[0.5em] \Rightarrow \frac{11-2x}{5} \ge \frac{9 - 3x + 6}{8} \\[0.5em] \Rightarrow 8(11 - 2x) \ge 5(15 - 3x) \\[0.5em] \Rightarrow 88 - 16x \ge 75 - 15x \\[0.5em] \Rightarrow - 16x + 15x \ge 75 - 88 \\[0.5em] \Rightarrow - x \ge - 13 \\[0.5em] \Rightarrow x ≤ 13 \\[0.5em]\end{aligned}$$

Since, x ∈ N
Solution set = {1, 2, 3, 4, 5, ..... , 13}.

Given,

$$\begin{aligned}-2 \le \frac{1}{2} - \frac{2x}{3} \le 1\frac{5}{6} \\[0.5em] \Rightarrow -2 \le \frac{1}{2}-\frac{2x}{3} \le \frac{11}{6} \\[0.5em] \Rightarrow -2-\frac{1}{2} \le \frac{1}{2} - \frac{2x}{3} - \frac{1}{2} \le \frac{11}{6} - \frac{1}{2}\\[0.5em] \text {By subtracting } \frac{1}{2} \text{ on both sides of inequality in the above line.}\\[0.5em] \Rightarrow -\frac{5}{2} \le -\frac{2x}{3} \le \frac{8}{6} \\[0.5em] \Rightarrow -\frac{5}{2} \times 6 \le -\frac{2x}{3}\times 6 \le \frac{8}{6}\times 6 \text{ (Multiplying complete equation with 6) } \\[0.5em] \Rightarrow -15 \le -4x \le 8 \\[0.5em] \Rightarrow 15 \ge 4x \ge -8 \\[0.5em] \Rightarrow \frac{15}{4} \ge x \ge -2\end{aligned}$$

Since x ∈ N,

∴ Solution Set = {1, 2, 3}

The graph of the solution set is shown by thick dots on the number line.

Given,

$$\begin{aligned}\frac{3}{5}x-\frac{2x−1}{3} \gt 1\\[0.5em] \Rightarrow \frac{9x - 10x + 5}{15} \gt 1 \text{ [Taking LCM as 15]} \\[0.5em] -x + 5 \gt 15 \\[0.5em] \Rightarrow – x \gt 15 – 5\\[0.5em] \Rightarrow – x \gt 10\\[0.5em] \Rightarrow x \lt – 10\\[0.5em]\end{aligned}$$

But x ∈ W
Solution set = Φ
Hence it can't be represented on number line.

Given,

$$\begin{aligned}\frac{x}{2} +5 \le \frac{x}{3} + 6\\[0.5em] \Rightarrow \frac{x}{2} - \frac{x}{3} \le 6-5\\[0.5em] \Rightarrow \frac{3x-2x}{6} \le 1\\[0.5em] \Rightarrow \frac{x}{6} \le 1\\[0.5em] \Rightarrow x \le 6\end{aligned}$$

Since, x is a positive odd integer
Hence, x = {1, 3, 5}.

Given,

$$\begin{aligned}\frac{2x+3}{3} \ge \frac{3x−1}{4}\\[0.5em] \Rightarrow \frac{2x}{3} + \frac{3}{3} \ge \frac{3x}{4}-\frac{1}{4}\\[0.5em] \Rightarrow \frac{2x}{3} + 1 \ge \frac{3x}{4} - \frac{1}{4}\\[0.5em] \Rightarrow \frac{2x}{3} - \frac{3x}{4} \ge -\frac{1}{4}-1\\[0.5em] \Rightarrow \frac{8x-9x}{12} \ge - \frac{5}{4}\\[0.5em] \Rightarrow -\frac{x}{12} \ge -\frac{5}{4}\\[0.5em] \Rightarrow \frac{x}{12} \le \frac{5}{4}\\[0.5em] \Rightarrow x \le \frac{5}{4} \times 12\\[0.5em] \Rightarrow x \le 15\end{aligned}$$

Since, x is a positive even integer.
x = {2, 4, 6, 8, 10, 12, 14}.

Given,

$$\begin{aligned}3 \ge \frac{x−4}{2}+ \frac{x}{3} \ge 2\\[2em] \text{ Solving left side: }\\[1em] 3 \ge \frac{x−4}{2}+ \frac{x}{3} \\[0.5em] \Rightarrow 3 \ge + \frac{3x-12+2x}{6}\\[0.5em] \Rightarrow 3 \ge + \frac{5x-12}{6}\\[0.5em] \Rightarrow 18 \ge 5x-12\\[0.5em] \Rightarrow 5x-12 \le 18\\[0.5em] \Rightarrow 5x \le 30\\[0.5em] \Rightarrow x \le 6\\[1em] \text{Solving right side:} \\[1em] \frac{x−4}{2} + \frac{x}{3} ≥ 2\\[0.5em] \Rightarrow \frac{3x-12+2x}{6} \ge 2 \\[0.5em] \Rightarrow \frac{5x-12}{6} \ge 2\\[0.5em] \Rightarrow 5x-12 \ge 12 \\[0.5em] \Rightarrow 5x \ge 24\\[0.5em] \Rightarrow x \ge \frac{24}{5}\\[0.5em] \Rightarrow x \ge 4\frac{4}{5} \\[0.5em]\end{aligned}$$

∴ Solution Set = {5, 6}.

The graph of the solution set is shown by thick dots on the number line.

Given,

$$\begin{aligned}1 \ge 15 - 7x \gt 2x - 27\\[0.5em] \Rightarrow 1 \ge 15 - 7x \text{ and } 15 - 7x \gt 2x - 27\\[0.5em] \Rightarrow 7x \ge 15-1 \text{ and } -7x-2x \gt -27-15\\[0.5em] \Rightarrow 7x \ge 14 \text{ and } -9x \gt -42\\[0.5em] \Rightarrow x \ge 2 \text{ and } -x \gt -\frac{42}{9}\\[0.5em] \Rightarrow x \ge 2 \text{ and } x \lt \frac{14}{3}\\[0.5em] \Rightarrow 2 \le x \lt \frac{14}{3}\end{aligned}$$

Since x ∈ N,

Solution set = {2, 3, 4}.

Given,

$$\begin{aligned}2 + 4x \lt 2x - 5 \le 3x\\[0.5em] \Rightarrow 2 + 4x \lt 2x - 5 \text{ and } 2x - 5 \le 3x \\[0.5em] \Rightarrow 4x - 2x \lt - 5 - 2 \text{ and } 2x - 3x \le 5\\[0.5em] \Rightarrow 2x \lt -7 \text{ and } -x \le 5\\[0.5em] \Rightarrow x \lt -\frac{7}{2} \text{ and } x \ge -5\\[0.5em] \Rightarrow -5\le x \lt -\frac{7}{2}\end{aligned}$$

Since x ∈ Z,

∴ Solution set = {-5, -4}.

The graph of the solution set is shown by thick dots on the number line.

Given : 3x - 5 ≤ 6x + 4 < 11 + x

Solving L.H.S. of the inequation, we get :

⇒ 3x - 5 ≤ 6x + 4

⇒ 3x - 6x - 5 ≤ 4

⇒ -3x - 5 ≤ 4

⇒ -3x ≤ 4 + 5

⇒ -3x ≤ 9

⇒ 3x ≥ -9

⇒ x ≥ -$\frac{9}{3}$

⇒ x ≥ -3 .......................(1)

Solving R.H.S. of the inequation, we get :

⇒ 6x + 4 < 11 + x

⇒ 6x + 4 - x < 11

⇒ 5x + 4 < 11

⇒ 5x < 11 - 4

⇒ 5x < 7

⇒ x < $\frac{7}{5}$ ......................(2)

From (1) and (2), we get :

⇒ -3 ≤ x < $\frac{7}{5}$

⇒ -3 ≤ x < 1.4

(i) Since,

x ∈ W and -3 ≤ x < 1.4

Solution set = {0, 1}

Hence, solution set = {0, 1}.

(ii) Since,

x ∈ Z and -3 ≤ x < 1.4

Solution set = {-3, -2, -1, 0, 1}

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

Given,

$$\begin{aligned}\frac{4x−10}{3} \le \frac{5x−7}{2}\\[0.5em] \Rightarrow \frac{4x-10}{3} \times 6 \le \frac{5x-7}{2} \times 6 \text{ ( Multiplying both sides by 6) }\\[0.5em] \Rightarrow 8x - 20 \le 15x - 21\end{aligned}$$

$$\begin{aligned}\Rightarrow 8x-15x \le -21 +20\\[0.5em] \Rightarrow -7x \le -1\\[0.5em] \Rightarrow 7x \ge 1\\[0.5em] \Rightarrow x \ge \frac{1}{7}\end{aligned}$$

∴ Solution Set = {x : x ∈ R, x ≥ $\frac{1}{7}$}

The graph of the solution set is represented by thick black line starting from and including $\frac{1}{7}$ on the number line.

Given,

$$\begin{aligned}\frac{3x}{5} - \frac{2x-1}{3} \gt 1 \\[0.5em] \Rightarrow 9x - (10x - 5) \gt 15 \\[0.5em] \Rightarrow 9x - 10x + 5 \gt 15 \\[0.5em] \Rightarrow - x \gt 15 - 5 \\[0.5em] \Rightarrow - x \gt 10 \\[0.5em] \Rightarrow x \lt -10\end{aligned}$$

x ∈ R.
Hence , Solution set = {x : x ∈ R, x < –10}.

The graph of the solution set is represented by thick black line starting from -10 (not including -10) on the number line.

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

Solving L.H.S. of the inequation,

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

Solving R.H.S. of the inequation,

$$\begin{aligned}\Rightarrow x + 4 \le \frac{x}{2} + 5\\[1em] \Rightarrow x + 4 \le \frac{x + 10}{2}\\[1em] \Rightarrow 2(x + 4) \le x + 10\\[1em] \Rightarrow 2x + 8 \le x + 10\\[1em] \Rightarrow 2x + 8 - x \le 10\\[1em] \Rightarrow x + 8 \le 10\\[1em] \Rightarrow x \le 10 - 8\\[1em] \Rightarrow x \le 2 ...............................(2)\end{aligned}$$

From (1) and (2), we get

-5 < x ≤ 2

Since, x ∈ R

The solution set of x = {x : x ∈ R, -5 < x ≤ 2}

Hence, solution set = {x : x ∈ R, -5 < x ≤ 2}.

Given, equation :

-5(x - 9) ≥ 17 - 9x > x + 2

Solving L.H.S. of the given equation :

⇒ -5(x - 9) ≥ 17 - 9x

⇒ -5x + 45 ≥ 17 - 9x

⇒ -5x + 9x ≥ 17 - 45

⇒ 4x ≥ -28

⇒ x ≥ $-\frac{28}{4}$

⇒ x ≥ -7 ............(1)

Solving L.H.S. of the given equation :

⇒ 17 - 9x > x + 2

⇒ x + 9x < 17 - 2

⇒ 10x < 15

⇒ x < $\frac{15}{10}$

⇒ x < $\frac{3}{2}$

⇒ x < 1.5 .............(2)

From equation (1) and (2), we get :

Solution set : {x : -7 ≤ x < 1.5}

Solution set on the number line :

Given,

$$\begin{aligned}−2 + 10x \le 13x + 10 \lt 24 + 10x \\[1em] \text{ Solving left side } \\[0.5em] -2 + 10x \le 13x + 10 \\[0.5em] \Rightarrow 10x -13x \le 10 +2\\[0.5em] \Rightarrow -3x \le 12 \\[0.5em] \Rightarrow 3x \ge -12 \\[0.5em] \Rightarrow x \ge -4 \\[1em] \text { Solving right side } \\[0.5em] 13x + 10 \lt 24 + 10x \\[0.5em] \Rightarrow 13x -10x \lt 24 -10 \\[0.5em] \Rightarrow 3x \lt 14 \\[0.5em] \Rightarrow x \lt \frac{14}{3}\end{aligned}$$

∴ Solution Set = {x : x ∈ Z , -4 ≤ x < $\frac{14}{3}$} = {-4, -3, -2, -1, 0, 1, 2, 3, 4}

The graph of the solution set is represented by thick black dots.

Given,

$$\begin{aligned}2x - 5 \le 5x + 4 \lt 11 \\[0.5em] \Rightarrow 2x - 5 \le 5x + 4 \text { and } 5x + 4 \lt 11 \\[0.5em] \Rightarrow 2x - 5 - 4 \le 5x \text { and } 5x + 4 \lt 11 \\[0.5em] \Rightarrow 2x - 9 \le 5x \text{ and } 5x \lt 11 - 4 \\[0.5em] \Rightarrow 2x - 5x \le 9 \text{ and } 5x \lt 7 \\[0.5em] \Rightarrow -3x \le 9 \text{ and } x \lt \frac{7}{5} \\[0.5em] \Rightarrow 3x \ge - 9 \text{ and } x \lt \frac{7}{5} \\[0.5em] \Rightarrow x \ge - 3 \text{ and } x \lt 1.4 \\[0.5em] \therefore -3 \le x \lt 1.4\end{aligned}$$

∴ Solution set = {-3, -2, -1, 0, 1}

The graph of the solution set is represented by thick black dots on the number line.

Given,

$$\begin{aligned}2 (x - 1) \lt 3 x - 1 \\[0.5em] \Rightarrow 2x - 2 \lt 3x - 1 \\[0.5em] \Rightarrow 2x - 3x \lt - 1 + 2 \\[0.5em] \Rightarrow - x \lt 1 \\[0.5em] \Rightarrow x \gt - 1 \\[0.5em] \therefore \text{A} = \lbrace 0, 1, 2, 3, ....\rbrace\end{aligned}$$

Also,

$$\begin{aligned}4x - 3 \le 8 + x \\[0.5em] \Rightarrow 4x - x \le 8 + 3 \\[0.5em] \Rightarrow 3x \le 11 \\[0.5em] \Rightarrow x \le \frac{11}{3} \\[0.5em] \therefore \text{B} = \lbrace....., -1, 0, 1, 2, 3\rbrace\end{aligned}$$

∴ A ∩ B = {0, 1, 2, 3}

$$\begin{aligned}\text{A} = {x : 11x - 5 \gt 7x + 3, x ∈ \bold{R}} \\[0.5em] \text{B} = {x : 18x - 9 \ge 15 + 12x, x ∈ \bold{R}} \\[1em] \text{Now, A} = 11x - 5 \gt 7x + 3 \\[0.5em] \Rightarrow 11x - 7x \gt 3 + 5 \\[0.5em] \Rightarrow 4x \gt 8 \\[0.5em] \Rightarrow x \gt 2, x ∈ \bold{R} \\[1em] B= 18x - 9 \ge 15 + 12x \\[0.5em] \Rightarrow 18x -12x \ge 15 + 9 \\[0.5em] \Rightarrow 6x \ge 24 \\[0.5em] \Rightarrow x \ge 4 \\[0.5em] \therefore \text{A} ∩ \text{B} = {x : x ∈ \bold{R}, x \ge 4}\end{aligned}$$

The graph is represented by a thick back line starting from 4 (including 4)

Given,

P = {x : 5 < 2x - 1 ≤ 11, x ∈ R}

$\Rightarrow 5 \lt 2x - 1 \le 11$

Solving left side,

$$\begin{aligned}5 \lt 2x -1 \\[0.5em] \Rightarrow 6 \lt 2x \\[0.5em] \Rightarrow 3 \lt x \\[0.5em] \Rightarrow x \gt 3 \\[0.5em]\end{aligned}$$

Solving right side,

$$\begin{aligned}2x -1 \le 11 \\[0.5em] \Rightarrow 2x \le 12 \\[0.5em] \Rightarrow x \le 6\end{aligned}$$

∴ P = {x : x ∈ R, 3 < x ≤ 6}.

The graph of the solution set of P is represented by thick black line starting from 3 (not including 3) till 6 (including 6).

Given,

Q = {x : - 1 ≤ 3 + 4x < 23, x ∈ I}

$$\begin{aligned}\Rightarrow -1 \le 3 + 4x \lt 23 \\[0.5em] \Rightarrow -1 \le 3 + 4x \text{ and } 3 + 4x \lt 23 \\[0.5em] \Rightarrow -4x \le 3 + 1 \text{ and } 4x \lt 20 \\[0.5em] \Rightarrow -4x \le 4 \text{ and } x \lt 5 \\[0.5em] \Rightarrow -x \le 1 \text{ and } x \lt 5 \\[0.5em] \Rightarrow x \ge -1 \text{ and } x \lt 5\end{aligned}$$

∴ Q = {x : x ∈ I, -1 ≤ x < 5} = {-1, 0, 1, 2, 3, 4}.

The graph of the solution set of Q is represented by thick black dots.

∴ P ∩ Q = {4}

Given,

$$\begin{aligned}2x + \frac{5}{2} \gt \frac{5x}{3} + 2 \\[0.5em] \Rightarrow 2x - \frac{5x}{3} \gt 2 - \frac{5}{2} \\[0.5em] \Rightarrow 12x - 10x \gt 12 - 15 \text{ (Multiplying both sides by 6)} \\[0.5em] \Rightarrow 2x \gt - 3 \\[0.5em] \Rightarrow x \gt - \frac{3}{2}\end{aligned}$$

∴ Smallest value of x which satisfies this inequation is -1

Given,

$$\begin{aligned}20 - 5 x \lt 5 (x + 8) \\[0.5em] \Rightarrow 20 - 5x \lt 5x + 40 \\[0.5em] \Rightarrow - 5x - 5x \lt 40 - 20 \\[0.5em] \Rightarrow - 10x \lt 20 \\[0.5em] \Rightarrow - x \lt 2 \\[0.5em] \Rightarrow x \gt - 2 \\[0.5em]\end{aligned}$$

(i) When x ∈ I, then smallest value = -1.

(ii) When x ∈ W, then smallest value = 0.

(iii) When x ∈ N, then smallest value = 1.

Given,

$$\begin{aligned}4x-19 \lt \frac{3x}{5} - 2 \le - \frac{2}{5} + x, x ∈ \bold{R} \\[1em] 4x - 19 \lt \frac{3x}{5} -2 \text{ and } \frac{3x}{5} - 2 \le - \frac{2}{5} + x \\[0.5em] \Rightarrow 4x -\frac{3x}{5} \lt -2 + 19 \text{ and } \frac{3x}{5} - 2 \le - \frac{2}{5} + x \\[0.5em] \Rightarrow \frac{20x-3x}{5} \lt 17 \text{ and } -2 + \frac{2}{5} \le x - \frac{3x}{5} \\[0.5em] \Rightarrow \frac{17x}{5} \lt 17 \text{ and } \frac{-8}{5} \le \frac{2x}{5} \\[0.5em] \Rightarrow \frac{x}{5} \lt 1 \text{ and } x \ge -4 \\[0.5em] \Rightarrow x \lt 5 \text{ and } x \ge -4 \\[0.5em] \Rightarrow -4 \le x \lt 5 , x ∈ \bold{R}\end{aligned}$$

Hence, Solution set = {x : x ∈ R, -4 ≤ x < 5}.

The graph of the solution set is represented by thick line starting from -4 (including -4) till 5 (not including 5).

Given,

$$\begin{aligned}2y - 3 \lt y + 1 \le 4y + 7; y ∈ \bold{R}. \\[0.5em] \text{ (a) Solving left side } \\[0.5em] 2y - 3 \lt y + 1 \\[0.5em] \Rightarrow 2y - y \lt 1 + 3 \\[0.5em] \Rightarrow y \lt 4 \\[0.5em] \Rightarrow 4 \gt y \\[0.5em] \text{(b) Solving right side } \\[0.5em] y + 1 \le 4y + 7 \\[0.5em] \Rightarrow y - 4y \le 7 - 1 \\[0.5em] \Rightarrow -3y \le 6 \\[0.5em] \Rightarrow 3y \ge -6 \\[0.5em] \Rightarrow y \ge -2\end{aligned}$$

∴ Solution Set = {y : y ∈ R, -2 ≤ y < 4}

The graph of the solution set is represented by a thick black line starting from -2 (including -2) till 4 (not including 4).

Given,

-3 + x ≤ $\frac{7x}{2} + 2$ < 8 + 2x

Solving L.H.S. of the above inequation :

$$\begin{aligned}\Rightarrow -3 + x \le \frac{7x}{2} + 2 \\[1em] \Rightarrow \frac{7x}{2} -x \ge -3 - 2 \\[1em] \Rightarrow \frac{7x - 2x}{2} \ge -5 \\[1em] \Rightarrow \frac{5x}{2} \ge -5 \\[1em] \Rightarrow x \ge \frac{-5 \times 2}{5} \\[1em] \Rightarrow x \ge -2 \text{ ..........(1)}\end{aligned}$$

Solving R.H.S. of the above inequation :

$$\begin{aligned}\Rightarrow \frac{7x}{2} + 2 \lt 8 + 2x \\[1em] \Rightarrow \frac{7x}{2} - 2x \lt 8 - 2 \\[1em] \Rightarrow \frac{7x - 4x}{2} \lt 6 \\[1em] \Rightarrow \frac{3x}{2} \lt 6 \\[1em] \Rightarrow x \lt \frac{6 \times 2}{3} \\[1em] \Rightarrow x \lt 4 \text{ ..........(2)}\end{aligned}$$

From inequation (1) and (2), we get :

-2 ≤ x < 4.

Since, x ∈ I.

x = {-2, -1, 0, 1, 2, 3}.

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

Given, inequation : 5x - 21 < $\frac{5x}{7} - 6 \le -3\frac{3}{7} + x$

Solving L.H.S. of the inequation :

$$\begin{aligned}\Rightarrow 5x - 21 \lt \frac{5x}{7} - 6\\[1em] \Rightarrow 5x - \frac{5x}{7} \lt 21 - 6 \\[1em] \Rightarrow \frac{35x - 5x}{7} \lt 15 \\[1em] \Rightarrow \frac{30x}{7} \lt 15 \\[1em] \Rightarrow x \lt \frac{7 \times 15}{30} \\[1em] \Rightarrow x \lt \frac{7}{2} \text{ ..........(1)}\end{aligned}$$

Solving R.H.S. of the inequation :

$$\begin{aligned}\Rightarrow \frac{5x}{7} - 6 \le -3\frac{3}{7} + x \\[1em] \Rightarrow \frac{5x}{7} - 6 \le -\frac{24}{7} + x \\[1em] \Rightarrow x - \frac{5x}{7} \ge -6 + \frac{24}{7} \\[1em] \Rightarrow \frac{7x - 5x}{7} \ge \frac{-42 + 24}{7} \\[1em] \Rightarrow \frac{2x}{7} \ge \frac{-18}{7} \\[1em] \Rightarrow 2x \ge -18 \\[1em] \Rightarrow x \ge \frac{-18}{2} \\[1em] \Rightarrow x \ge -9 \text{ ..........(2)}\end{aligned}$$

From equation (1) and (2),

Solution set = {x : -9 ≤ x < $\frac{7}{2}$, x ∈ R}

Hence, solution set = {x : -9 ≤ x < $\frac{7}{2}$, x ∈ R}.

$$\begin{aligned}\text{Let the greatest integer} = x \\[0.5em] \text{According to the condition}, \\[0.5em] 2x + 7 \gt 3x \\[0.5em] \Rightarrow 2x - 3x \gt - 7 \\[0.5em] \Rightarrow -x \gt - 7 \\[0.5em] \Rightarrow x \lt 7 \\[0.5em] \therefore \text{Value of x which is greatest = 6}\end{aligned}$$

Let the length of the shortest pole = x metre

Length of pole which is burried in mud = $\frac{x}{3}$

Length of pole which is in the water = $\frac{x}{6}$

Given,

$$\begin{aligned}x-\Big[\frac{x}{3} + \frac{x}{6}\Big] \ge 3 \\[0.5em] \Rightarrow x - \Big(\frac{2x + x}{6}\Big) \ge 3 \\[0.5em] \Rightarrow x - \Big(\frac{3x}{6}\Big) \ge 3 \\[0.5em] \Rightarrow x - \frac{x}{2} \ge 3 \\[0.5em] \Rightarrow \frac{x}{2} \ge 3 \\[0.5em] \Rightarrow x \ge 6\end{aligned}$$

∴ Length of pole which is shortest = 6 meters.

Given,

x ∈ {-3, -1, 0, 1, 3, 5}

$$\begin{aligned}3x - 2 \le 8 \\[0.5em] \Rightarrow 3x \le 8 + 2 \\[0.5em] \Rightarrow 3x \le 10 \\[0.5em] \Rightarrow x \le \frac{10}{3} \\[0.5em] \Rightarrow x \le 3\frac{1}{3}\end{aligned}$$

Solution set = {-3, -1, 0, 1, 3}

∴ Option 2 is the correct option.

x ∈ W

$$\begin{aligned}3x + 11 \ge x + 8 \\[0.5em] \Rightarrow 3x - x \ge 8 – 11 \\[0.5em] \Rightarrow 2x \ge -3 \\[0.5em] \Rightarrow x \ge -\frac{3}{2} \\[0.5em]\end{aligned}$$

Solution set = {0, 1, 2, 3, …}

∴ Option 3 is the correct option.

x ∈ W

$$\begin{aligned}\Rightarrow 5 - 4x \ge 2 - 3x \\[0.5em] \Rightarrow -3x + 4x \le 5 - 2 \\[0.5em] \Rightarrow x \le 3.\end{aligned}$$

Since, x ∈ W

Solution set = {0, 1, 2, 3}.

∴ Option 3 is the correct option.

x ∈ I

Given,
1 < 3x + 5 ≤ 11

Solving left side,

$$\begin{aligned}\Rightarrow 1 \lt 3x + 5 \\[0.5em] \Rightarrow 1 - 5 \lt 3x \\[0.5em] \Rightarrow -4 \lt 3x \\[0.5em] \Rightarrow 3x \gt -4 \\[0.5em] \Rightarrow x \gt -\frac{4}{3}\end{aligned}$$

Solving right side,

$$\begin{aligned}3x + 5 \le 11 \\[0.5em] \Rightarrow 3x \le 11-5 \\[0.5em] \Rightarrow 3x \le 6 \\[0.5em] \Rightarrow x \le 2 \\[1.5em] \therefore -\frac{4}{3} \lt x \le 2\end{aligned}$$

Solution set = {-1, 0, 1, 2}.

∴ Option 1 is the correct option.

x ∈ R

Given,
6 ≤ -3(2x - 4) < 12

Solving left side,

$$\begin{aligned}6 \le -3(2x - 4) \\[0.5em] \Rightarrow 6 \le - 6x + 12 \\[0.5em] \Rightarrow 6-12 \le -6x \\[0.5em] \Rightarrow -6 \le -6x \\[0.5em] \Rightarrow 6x \le 6 \\[0.5em] \Rightarrow x \le 1\end{aligned}$$

Solving right side,

$$\begin{aligned}-3(2x - 4) \lt 12 \\[0.5em] \Rightarrow -6x + 12 \lt 12 \\[0.5em] \Rightarrow -6x \lt 12-12 \\[0.5em] \Rightarrow -6x \lt 0 \\[0.5em] \Rightarrow x \gt 0 \\[1.5em] \therefore 0 \lt x \le 1\end{aligned}$$

Solution set = {x : x ∈ R, 0 $\lt x \le$ 1}

∴ Option 1 is the correct option.

Solving,

⇒ 2x + 4 ≤ 14

⇒ 2x ≤ 14 - 4

⇒ 2x ≤ 10

⇒ x ≤ $\frac{10}{2}$

⇒ x ≤ 5.

Since, x ∈ W,

∴ x = {0, 1, 2, 3, 4, 5}.

Hence, Option 2 is the correct option.

Solving the given equation :

$$\begin{aligned}\Rightarrow x + 2 \le \frac{x}{3} + 3 \\[1em] \Rightarrow x - \frac{x}{3} \le 3 - 2 \\[1em] \Rightarrow \frac{3x - x}{3} \le 1 \\[1em] \Rightarrow \frac{2x}{3} \le 1 \\[1em] \Rightarrow x \le \frac{3}{2} \\[1em] \Rightarrow x \le 1.5\end{aligned}$$

Since, x is a prime number less than 1.5

Solution set is empty.

Hence, Option 1 is the correct option.

Given, inequation :

⇒ x ≤ -5

⇒ x = {-5, -6, -7, -8, ........}

We are looking for whole numbers that are less than or equal to −5.

But since whole numbers start from 0 and go upward, there are no whole numbers ≤ -5.

So, assertion (A) is true.

Whole numbers include 0, which is neither positive nor negative.

So whole numbers are non-negative, not always positive.

So, reason (R) is false.

Thus, Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

Given; -5 < x and x ≤ 6

⇒ -5 < x ≤ 6

Since, x ∈ R

The inequality does have a solution. In fact, it has infinitely many solutions as infinitely many real numbers lie between -5 and 6.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given,

⇒ -3 ≤ x < $5\frac{2}{3}$

⇒ -3 ≤ x < $\frac{17}{3}$

⇒ -3 ≤ x < 5.66.....

Since, x ∈ Z and -3 ≤ x < 5.66.....

⇒ x = {-3, -2, -1, 0, 1, 2, 3, 4, 5}

Thus, x has nine values.

So, assertion (A) is true.

x = 5 is included in the solution set.

So, reason (R) is true but, reason (R) does not clearly explains assertion (A).

Thus, both Assertion (A) and Reason (R) are correct, and Reason (R) is the incorrect reason for Assertion (A).

Hence, option 4 is the correct option.

If the number is represented by x, then :

From the number line, we get :

$-5 \lt x \le \frac{3}{2}$ and x ∈ R.

∴ It represents real numbers lying between -5 and $\frac{3}{2}$ but not -5.

So, assertion (A) is false.

The hole on the number line represents absence of the number -5.

So, reason (R) is true.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given,

$$\begin{aligned}5x - 2 \le 3(3 - x) \\[0.5em] \Rightarrow 5x - 2 \le 9 - 3x \\[0.5em] \Rightarrow 5x + 3x \le 9 + 2 \\[0.5em] \Rightarrow 8x \le 11 \\[0.5em] \Rightarrow x \le \frac{11}{8}\end{aligned}$$

Since, x ∈ { -2, -1, 0, 1, 2, 3, 4}.

∴ Solution set ={-2, -1, 0, 1}.

The graph of the solution set is represented by thick black dots.

Given,

$$\begin{aligned}6x - 5 \lt 3x + 4 \\[0.5em] \Rightarrow 6x - 3x \lt 4 + 5 \\[0.5em] \Rightarrow 3x \lt 9 \\[0.5em] \Rightarrow x \lt 3\end{aligned}$$

x ∈ I

∴ Solution Set = {..., -2, -1, 0, 1, 2}.

Given,

$$\begin{aligned}x + 5 \le 2x + 3 \\[0.5em] \Rightarrow x - 2x \le 3 - 5\\[0.5em] \Rightarrow -x \le -2\\[0.5em] \Rightarrow x \ge 2\end{aligned}$$

∴ Solution set = {x : x ∈ R, x ≥ 2 }.

The graph of the solution set is represented by thick line starting from and including 2.

Given,

$$\begin{aligned}-1 \lt 3 - 2x \le 7 \\[0.5em] \Rightarrow -1 \lt 3 - 2x \text{ and } 3 - 2x \le 7 \\[0.5em] \Rightarrow 2x \lt 3 + 1 \text{ and } -2x \le 7 - 3 \\[0.5em] \Rightarrow 2x \lt 4 \text{ and } -2x \le 4 \\[0.5em] \Rightarrow x \lt 2 \text{ and } -x \le 2 \\[0.5em] \Rightarrow x \lt 2 \text{ and } x \ge -2\end{aligned}$$

∴ Solution set = {x : x ∈ R, -2 ≤ x < 2}.

The graph of this inequation is represented by thick black line starting from -2 (including -2) till (not including) 2.

Given,

$\frac{5x+1}{7} - 4\Big(\frac{x}{7}+ \frac{2}{5}\Big) \le 1\frac{3}{5}+\frac{3x-1}{7}, x ∈ \bold{R}.$

Multiplying both sides by 35

$$\begin{aligned}\Rightarrow 25x + 5-4(5x + 14) \le 56 + 15x -5 \\[0.5em] \Rightarrow 25x + 5 - 20x - 56 \le 56 + 15x -5 \\[0.5em] \Rightarrow 5x - 51 \le 51 + 15x \\[0.5em] \Rightarrow 5x - 15x \le 51 + 51 \\[0.5em] \Rightarrow -10x \le 102 \\[0.5em] \Rightarrow x \ge -\frac{102}{10} \\[0.5em] \Rightarrow x \ge -\frac{51}{5}\end{aligned}$$

∴ Solution set = {x : x ∈ R, x $\ge -\frac{51}{5}$ }.

Given,
7 ≤ –4x + 2 < 12

$$\begin{aligned}\Rightarrow 7 \le - 4x + 2 \text{ and } - 4x + 2 \lt 12 \\[0.5em]\end{aligned}$$

Solving left side,

$$\begin{aligned}7 \le - 4x + 2 \\[0.5em] \Rightarrow 4x \le 2-7 \\[0.5em] \Rightarrow 4x \le -5 \\[0.5em] \Rightarrow x \le \frac{-5}{4}\end{aligned}$$

Solving right side,

$$\begin{aligned}-4x + 2 \lt 12 \\[0.5em] \Rightarrow -4x \lt 12 - 2 \\[0.5em] \Rightarrow -4x \lt 10 \\[0.5em] \Rightarrow 4x \gt -10 \\[0.5em] \Rightarrow x \gt \frac{-5}{2}\end{aligned}$$

∴ Solution set = {x : x ∈ R, $-\frac{5}{2}$ < x ≤ $-\frac{5}{4}$}.

The graph of the inequation is represented by thick black line starting from $-\frac{5}{2}$ (excluding $-\frac{5}{2}$) till $-\frac{5}{4}$ (including $-\frac{5}{4}$).

To prove:

3 - 2x ≥ x + $\frac{1 - x}{3}$ > $\frac{2x}{5}$

Solving L.H.S. of the above inequation, we get :

⇒ 3 - 2x ≥ x + $\frac{1 - x}{3}$

⇒ 3 - 2x ≥ $\frac{3x + 1 - x}{3}$

⇒ 3(3 - 2x) ≥ 2x + 1

⇒ 9 - 6x ≥ 2x + 1

⇒ 2x + 6x ≤ 9 - 1

⇒ 8x ≤ 8

⇒ x ≤ $\frac{8}{8}$

⇒ x ≤ 1 ............(1)

Solving R.H.S. of the above equation, we get :

⇒ x + $\frac{1 - x}{3} \gt \frac{2x}{5}$

⇒ $\frac{3x + 1 - x}{3} \gt \frac{2x}{5}$

⇒ 5(2x + 1) > 3 × 2x

⇒ 10x + 5 > 6x

⇒ 10x - 6x > -5

⇒ 4x > -5

⇒ x > $-\frac{5}{4}$ ...........(2)

From equation (1) and (2), we get :

Solution set = {x : $-\frac{5}{4}$ < x ≤ 1, x ∈ R}

Representation of solution set on real number line is :

Let the positive integer = x
According to the problem,

$$\begin{aligned}5x - 6 \lt 4x \\[0.5em] \Rightarrow 5x - 4x \lt 6 \\[0.5em] \Rightarrow x \lt 6\end{aligned}$$

∴ Solution set = { 1, 2, 3, 4, 5}.

Let first least natural number = x
then, second number = x + 1
and third number = x + 2

Given,

$$\begin{aligned}\frac{1}{3}(x+2) - \frac{1}{5}(x) \ge 3 \\[0.5em] \Rightarrow \frac{x}{3} - \frac{x}{5} \ge 3 - \frac{2}{3} \\[0.5em] \Rightarrow \frac{5x- 3x}{15} \ge \frac{9-2}{3} \\[0.5em] \Rightarrow \frac{2x}{15} \ge \frac{7}{3} \\[0.5em] \Rightarrow 2x \ge \frac{7 \times 15}{3} \\[0.5em] \Rightarrow 2x \ge 35 \\[0.5em] \Rightarrow x \ge \frac{35}{2} \\[0.5em] \Rightarrow x \ge 17\frac{1}{2}\end{aligned}$$

Since the three consecutive numbers should be natural numbers
∴ x = 18
    x + 1 = 19
    x + 2 = 20

Hence, the three smallest consecutive natural numbers are 18, 19, 20