Linear Inequations

Solve the inequation 3x - 11 ≤ 3, where x ∈ {1, 2, 3, .........., 10}. Also represent its solution on number line.

Answer

Given,

$3x-11 \leq 3 \\[0.5em] \Rightarrow 3x-11 + 11 \leq 3 + 11 \\[0.5em] \Rightarrow 3x \leq 14 \\[0.5em] \Rightarrow x \leq \dfrac{14}{3} \\[0.5em] \Rightarrow x \leq 4.66$

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.

Solve 2(x - 3) < 1, x ∈ {1, 2, 3, .........., 10}.

Answer

Given,

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

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

Solve 5 - 4x > 2 - 3x, x ∈ W . Also represent its solution on number line.

Answer

Given,

$5 - 4x \gt 2 - 3x \\[0.5em] \Rightarrow 5 - 2 \gt -3x + 4x \\[0.5em] \Rightarrow 3 \gt x \\[0.5em] \Rightarrow x \lt 3$

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

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

List the solution set of 30 – 4(2x – 1) < 30, given that x is a positive integer.

Answer

Given,

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

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

Solve : 2(x – 2) < 3x – 2, x ∈ { –3, –2, –1, 0, 1, 2, 3} .

Answer

Given,

$2(x – 2) \lt 3x – 2 \\[0.5em] \Rightarrow 2x – 4 \lt 3x – 2 \\[0.5em] \Rightarrow 2x – 3x \lt – 2 + 4 \\[0.5em] \Rightarrow – x \lt 2 \\[0.5em] \Rightarrow x \gt – 2 \\[0.5em]$

Solution set = {–1, 0, 1, 2, 3} .

If x is a negative integer, find the solution set of $\dfrac{2}{3}+\dfrac{1}{3} (x + 1) \gt 0$.

Answer

Given,

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

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

Solve x – 3(2 + x) > 2(3x - 1), x ∈ {-3, -2, -1, 0, 1, 2}. Also represent its solution on the number line.

Answer

Given,

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

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 x ∈ {1, 2, 3, 4, 5, 6, 7, 9} solve x – 3 < 2x – 1.

Answer

Given,

$x - 3 \lt 2x - 1\\[0.5em] \Rightarrow x - 2x \lt - 1 + 3\\[0.5em] \Rightarrow - x \lt 2\\[0.5em] \Rightarrow x \gt - 2$

Since, x ∈ {1, 2, 3, 4, 5, 6, 7, 9}

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

List the solution set of the inequation

$\dfrac{1}{2} + 8x \gt 5x -\dfrac{3}{2}, x ∈ \bold{Z}$

Answer

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

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

List the solution set of:

$\dfrac{11-2x}{5} \ge \dfrac{9-3x}{8} + \dfrac{3}{4}, x ∈ \bold{N}.$

Answer

Given,

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

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

Find the values of x, which satisfy the inequation:

$−2 \le \dfrac{1}{2}-\dfrac{2x}{3} \le 1\dfrac{5}{6}$, x ∈ N.

Graph the solution set on the number line.

Answer

Given,

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

Since x ∈ N,

∴ Solution Set = {1, 2, 3}

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

If x ∈ W, find the solution set of $\dfrac{3}{5} x - \dfrac{2x−1}{3} \gt$ 1. Also graph the solution set on the number line, if possible.

Answer

Given,

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

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

Solve:

$\dfrac{x}{2} + 5 \le \dfrac{x}{3} + 6$ where x is a positive odd integer.

Answer

Given,

$\dfrac{x}{2} +5 \le \dfrac{x}{3} + 6\\[0.5em] \Rightarrow \dfrac{x}{2} - \dfrac{x}{3} \le 6-5\\[0.5em] \Rightarrow \dfrac{3x-2x}{6} \le 1\\[0.5em] \Rightarrow \dfrac{x}{6} \le 1\\[0.5em] \Rightarrow x \le 6$

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

Solve:

$\dfrac{2x+3}{3} \ge \dfrac{3x−1}{4}$ where x is positive even integer.

Answer

Given,

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

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

Given that x ∈ I, solve the inequation and graph the solution on the number line :

$3 \ge \dfrac{x−4}{2} + \dfrac{x}{3} \ge 2$

Answer

Given,

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

∴ Solution Set = {5, 6}.

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

Solve 1 ≥ 15 - 7x > 2x - 27, x ∈ N.

Answer

Given,

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

Since x ∈ N,

Solution set = {2, 3, 4}.

If x ∈ Z, solve 2 + 4x < 2x – 5 ≤ 3x. Also represent its solution on the number line.

Answer

Given,

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

Since x ∈ Z,

∴ Solution set = {-5, -4}.

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

Solve 3x - 5 ≤ 6x + 4 < 11 + x, when

(i) x ∈ W

(ii) x ∈ Z

Represent the solution set on a real number in each case.

Answer

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 ≥ -$\dfrac{9}{3}$

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

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

⇒ -3 ≤ x < $\dfrac{7}{5}$

⇒ -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}.

Solve : $\dfrac{4x-10}{3} \le \dfrac{5x-7}{2}, x ∈ \bold{R}$ and represent the solution set on the number line.

Answer

Given,

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

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

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

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

Solve $\dfrac{3x}{5} - \dfrac{2x-1}{3} \gt 1, x ∈ \bold{R}$ and represent the solution set on the number line.

Answer

Given,

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

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 that x ∈ R, solve the following inequation and graph the solution on the number line:

-1 ≤ 3 + 4x < 23

Answer

Given,

$-1 \le 3 + 4x \lt 23 \\[0.5em] \Rightarrow – 1 – 3 \le 4x \lt 23 – 3 \\[0.5em] \Rightarrow – 4 \le 4x \lt 20 \\[0.5em] \Rightarrow – 1 \le x \lt 5, x ∈ \bold{R} \\[0.5em]$

Solution Set = {x : x ∈ R, -1 ≤ x < 5}

The graph of the solution set is represented by thick black line starting from -1 ( including -1) till 5 ( not including 5 ) on the number line.

Solve the following inequation and represent the solution on the number line :

$\dfrac{3x}{5} + 2 \lt x + 4 \le \dfrac{x}{2} + 5$, x ∈ R

Answer

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

Solving L.H.S. of the inequation,

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

Solving R.H.S. of the inequation,

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

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}.

Solve the following inequation. Write down the solution set and represent it on the real number line.

-5(x - 9) ≥ 17 - 9x > x + 2, x ∈ R.

Answer

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 ≥ $-\dfrac{28}{4}$

⇒ x ≥ -7 ............(1)

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

⇒ 17 - 9x > x + 2

⇒ x + 9x < 17 - 2

⇒ 10x < 15

⇒ x < $\dfrac{15}{10}$

⇒ x < $\dfrac{3}{2}$

⇒ x < 1.5 .............(2)

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

Solution set : {x : -7 ≤ x < 1.5}

Solution set on the number line :

Solve the following inequation, write down the solution set and represent it on the real number line:

-2 + 10x ≤ 13x + 10 < 24 + 10x, x ∈ Z.

Answer

Given,

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

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

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

Solve the inequation 2x – 5 ≤ 5x + 4 < 11, where x ∈ I. Also represent the solution set on the number line.

Answer

Given,

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

∴ Solution set = {-3, -2, -1, 0, 1}

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

If x ∈ I, A is the solution set of 2(x - 1) < 3x - 1 and B is the solution set of 4x – 3 ≤ 8 + x, find A ∩ B.

Answer

Given,

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

Also,

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

∴ A ∩ B = {0, 1, 2, 3}

A = {x : 11x - 5 > 7x + 3, x ∈ R } and
B = {x : 18x - 9 ≥ 15 + 12x, x ∈ R }.
Find set A ∩ B and represent it on a number line.

Answer

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

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

Given: P = {x : 5 < 2x - 1 ≤ 11, x ∈ R}
Q = {x : - 1 ≤ 3 + 4x < 23, x ∈ I} where R = {real numbers}, I = {integers}.
Represent P and Q on number line. Write down the elements of P ∩ Q.

Answer

Given,

P = {x : 5 < 2x - 1 ≤ 11, x ∈ R}

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

Solving left side,

$5 \lt 2x -1 \\[0.5em] \Rightarrow 6 \lt 2x \\[0.5em] \Rightarrow 3 \lt x \\[0.5em] \Rightarrow x \gt 3 \\[0.5em]$

Solving right side,

$2x -1 \le 11 \\[0.5em] \Rightarrow 2x \le 12 \\[0.5em] \Rightarrow x \le 6$

∴ 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}

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

∴ 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}

If x ∈ I, find the smallest value of x which satisfies the inequation

$2x +\dfrac{5}{2} \gt \dfrac{5x}{3} + 2$

Answer

Given,

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

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

Given 20 – 5x < 5(x + 8), find the smallest value of x, when

(i) x ∈ I

(ii) x ∈ W

(iii) x ∈ N

Answer

Given,

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

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

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

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

Solve the following inequation and represent the solution set on the number line:

$4x - 19 \lt \dfrac{3x}{5} - 2 \le - \dfrac{2}{5} + x, x ∈ \bold{R}$

Answer

Given,

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

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).

Solve the given inequation and graph the solution on the number line :

2y - 3 < y + 1 ≤ 4y + 7; y ∈ R

Answer

Given,

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

∴ 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).

Solve the following inequation, write down the solution set and represent it on the real number line.

-3 + x ≤ $\dfrac{7x}{2} + 2$ < 8 + 2x, x ∈ I.

Answer

Given,

-3 + x ≤ $\dfrac{7x}{2} + 2$ < 8 + 2x

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

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

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

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

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}.

Solve the following inequation, write the solution set and represent it on the real number line.

5x - 21 < $\dfrac{5x}{7} - 6 \le -3\dfrac{3}{7} + x$, x ∈ R.

Answer

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

Solving L.H.S. of the inequation :

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

Solving R.H.S. of the inequation :

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

From equation (1) and (2),

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

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

Find the greatest integer which is such that if 7 is added to its double, the resulting number becomes greater than three times the integer.

Answer

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

One-third of a bamboo pole is buried in mud, one-sixth of it is in water and the part above the water is greater than or equal to 3 metres. Find the length of the shortest pole.

Answer

Let the length of the shortest pole = x metre

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

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

Given,

$x-\Big[\dfrac{x}{3} + \dfrac{x}{6}\Big] \ge 3 \\[0.5em] \Rightarrow x - \Big(\dfrac{2x + x}{6}\Big) \ge 3 \\[0.5em] \Rightarrow x - \Big(\dfrac{3x}{6}\Big) \ge 3 \\[0.5em] \Rightarrow x - \dfrac{x}{2} \ge 3 \\[0.5em] \Rightarrow \dfrac{x}{2} \ge 3 \\[0.5em] \Rightarrow x \ge 6$

∴ Length of pole which is shortest = 6 meters.

If x ∈ {-3, -1, 0, 1, 3, 5}, then the solution set of the inequation 3x – 2 ≤ 8 is

1. {-3, -1, 1, 3}
2. {-3, -1, 0, 1, 3}
3. {-3, -2, -1, 0, 1, 2, 3}
4. {-3, -2, -1, 0, 1, 2}

Answer

Given,

x ∈ {-3, -1, 0, 1, 3, 5}

$3x - 2 \le 8 \\[0.5em] \Rightarrow 3x \le 8 + 2 \\[0.5em] \Rightarrow 3x \le 10 \\[0.5em] \Rightarrow x \le \dfrac{10}{3} \\[0.5em] \Rightarrow x \le 3\dfrac{1}{3}$

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

∴ Option 2 is the correct option.

If x ∈ W, then the solution set of the inequation 3x + 11 ≥ x + 8 is

1. {-2, -1, 0, 1, 2, …}
2. {-1, 0, 1, 2, …}
3. {0, 1, 2, 3, …}
4. {x : x ∈ R, x ≥ –$\dfrac{3}{2}$ }

Answer

x ∈ W

$3x + 11 \ge x + 8 \\[0.5em] \Rightarrow 3x - x \ge 8 – 11 \\[0.5em] \Rightarrow 2x \ge -3 \\[0.5em] \Rightarrow x \ge -\dfrac{3}{2} \\[0.5em]$

Solution set = {0, 1, 2, 3, …}

∴ Option 3 is the correct option.

If x ∈ W, then the solution set of the inequation 5 - 4x ≤ 2 - 3x is

1. {…, -2, -1, 0, 1, 2, 3}
2. {1, 2, 3}
3. {0, 1, 2, 3}
4. {x : x ∈ R, x ≤ 3}

Answer

x ∈ W

$5 - 4x \le 2 - 3x \\[0.5em] \Rightarrow -4x + 3x \le 2 - 5 \\[0.5em] \Rightarrow -x \le -3 \\[0.5em] \Rightarrow x \ge 3$

Solution set = {3, 4, 5, ..... }.

∴ No option is correct .

If x ∈ I, then the solution set of the inequation 1 < 3x + 5 ≤ 11 is

1. { -1, 0, 1, 2}
2. { -2, -1, 0, 1}
3. { -1, 0, 1}
4. {x : x ∈ R, -$\dfrac{4}{3}$ $\lt$ x $\le$ 2}

Answer

x ∈ I

Given,
1 < 3x + 5 ≤ 11

Solving left side,

$\Rightarrow 1 \lt 3x + 5 \\[0.5em] \Rightarrow 1 - 5 \lt 3x \\[0.5em] \Rightarrow -4 \lt 3x \\[0.5em] \Rightarrow 3x \gt -4 \\[0.5em] \Rightarrow x \gt -\dfrac{4}{3}$

Solving right side,

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

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

∴ Option 1 is the correct option.

If x ∈ R, the solution set of 6 ≤ -3(2x - 4) < 12 is

1. {x : x ∈ R, 0 < x ≤ 1}
2. {x : x ∈ R, 0 ≤ x < 1}
3. {0, 1}
4. none of these

Answer

x ∈ R

Given,
6 ≤ -3(2x - 4) < 12

Solving left side,

$6 \le -3(2x - 4) \\[0.5em] \Rightarrow 6 \le - 6x + 12 \\[0.5em] \Rightarrow 6-12 \le -6x \\[0.5em] \Rightarrow -6 \le -6x \\[0.5em] \Rightarrow 6x \le 6 \\[0.5em] \Rightarrow x \le 1$

Solving right side,

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

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

∴ Option 1 is the correct option.

The solution set for the inequation 2x + 4 ≤ 14, x ∈ W is :

1. {1, 2, 3, 4, 5}

2. {0, 1, 2, 3, 4, 5}

3. {1, 2, 3, 4}

4. {0, 1, 2, 3, 4}

Answer

Solving,

⇒ 2x + 4 ≤ 14

⇒ 2x ≤ 14 - 4

⇒ 2x ≤ 10

⇒ x ≤ $\dfrac{10}{2}$

⇒ x ≤ 5.

Since, x ∈ W,

∴ x = {0, 1, 2, 3, 4, 5}.

Hence, Option 2 is the correct option.

Given, x + 2 ≤ $\dfrac{x}{3} + 3$ and x is a prime number. The solution set for x is :

1. ∅

2. {0}

3. {1}

4. {0, 1}

Answer

Solving the given equation :

$\Rightarrow x + 2 \le \dfrac{x}{3} + 3 \\[1em] \Rightarrow x - \dfrac{x}{3} \le 3 - 2 \\[1em] \Rightarrow \dfrac{3x - x}{3} \le 1 \\[1em] \Rightarrow \dfrac{2x}{3} \le 1 \\[1em] \Rightarrow x \le \dfrac{3}{2} \\[1em] \Rightarrow x \le 1.5$

Since, x is a prime number less than 1.5

Solution set is empty.

Hence, Option 1 is the correct option.

If x ≤ -5, x ∈ W

Assertion (A): The above inequation has no solution.

Reason (R): The whole numbers are always positive.

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

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

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

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.

If -5 < x and x ≤ 6, x ∈ R

Assertion (A): The above inequation has no solution.

Reason (R): Infinitely many real numbers lie between -5 and 6.

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

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

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

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.

If -3 ≤ x < $5\dfrac{2}{3}$, x ∈ Z

Assertion (A): x has nine values.

Reason (R): x = 5 is included in the solution set.

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

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

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given,

⇒ -3 ≤ x < $5\dfrac{2}{3}$

⇒ -3 ≤ x < $\dfrac{17}{3}$

⇒ -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.

Given below is the graphical representation of an inequality on number line.

Assertion (A): It represents real numbers lying between -5 and $\dfrac{3}{2}$ but not -5 and $\dfrac{3}{2}$.

Reason (R): The hole represents absence of the number -5.

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

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

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

If the number is represented by x, then :

From the number line, we get :

$-5 \lt x \le \dfrac{3}{2}$ and x ∈ R.

∴ It represents real numbers lying between -5 and $\dfrac{3}{2}$ 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.

Solve the inequation : 5x - 2 ≤ 3(3 - x) where x ∈ { -2, -1, 0, 1, 2, 3, 4}. Also represent its solution on the number line.

Answer

Given,

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

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.

Solve the inequations : 6x - 5 < 3x + 4, x ∈ I

Answer

Given,

$6x - 5 \lt 3x + 4 \\[0.5em] \Rightarrow 6x - 3x \lt 4 + 5 \\[0.5em] \Rightarrow 3x \lt 9 \\[0.5em] \Rightarrow x \lt 3$

x ∈ I

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

Find the solution set of the inequation x + 5 ≤ 2x + 3 ; x ∈ R. Graph the solution set on the number line.

Answer

Given,

$x + 5 \le 2x + 3 \\[0.5em] \Rightarrow x - 2x \le 3 - 5\\[0.5em] \Rightarrow -x \le -2\\[0.5em] \Rightarrow x \ge 2$

∴ Solution set = {x : x ∈ R, x ≥ 2 }.

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

If x ∈ R (real numbers) and -1 < 3 - 2x ≤ 7, find solution set and represent it on a number line.

Answer

Given,

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

∴ 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.

Solve the inequation :

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

Answer

Given,

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

Multiplying both sides by 35

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

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

Find the range of values of x, which satisfy 7 ≤ –4x + 2 < 12, x ∈ R. Graph these values of x on the real number line.

Answer

Given,
7 ≤ –4x + 2 < 12

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

Solving left side,

$7 \le - 4x + 2 \\[0.5em] \Rightarrow 4x \le 2-7 \\[0.5em] \Rightarrow 4x \le -5 \\[0.5em] \Rightarrow x \le \dfrac{-5}{4}$

Solving right side,

$-4x + 2 \lt 12 \\[0.5em] \Rightarrow -4x \lt 12 - 2 \\[0.5em] \Rightarrow -4x \lt 10 \\[0.5em] \Rightarrow 4x \gt -10 \\[0.5em] \Rightarrow x \gt \dfrac{-5}{2}$

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

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

If x ∈ R, solve 3 - 2x ≥ x + $\dfrac{1 - x}{3}$ > $\dfrac{2}{5}x$. Also represent the solution on the number line.

Answer

To prove:

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

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

⇒ 3 - 2x ≥ x + $\dfrac{1 - x}{3}$

⇒ 3 - 2x ≥ $\dfrac{3x + 1 - x}{3}$

⇒ 3(3 - 2x) ≥ 2x + 1

⇒ 9 - 6x ≥ 2x + 1

⇒ 2x + 6x ≤ 9 - 1

⇒ 8x ≤ 8

⇒ x ≤ $\dfrac{8}{8}$

⇒ x ≤ 1 ............(1)

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

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

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

⇒ 5(2x + 1) > 3 × 2x

⇒ 10x + 5 > 6x

⇒ 10x - 6x > -5

⇒ 4x > -5

⇒ x > $-\dfrac{5}{4}$ ...........(2)

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

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

Representation of solution set on real number line is :

Find positive integers which are such that if 6 is subtracted from five times the integer then the resulting number cannot be greater than four times the integer.

Answer

Let the positive integer = x
According to the problem,

$5x - 6 \lt 4x \\[0.5em] \Rightarrow 5x - 4x \lt 6 \\[0.5em] \Rightarrow x \lt 6$

∴ Solution set = { 1, 2, 3, 4, 5, 6}.

Find three smallest consecutive natural numbers such that the difference between one-third of the largest and one-fifth of the smallest is at least 3.

Answer

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

Given,

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

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