Linear Equations and Inequalities in One Variable — Exercise 12.3

If the replacement set = {-7, -5, -3, -1, 0, 1, 3}, find the solution set of:

$x \gt -2$

Answer

$x \gt - 2$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are greater than - 2 are - 1, 0, 1 and 3.

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

If the replacement set = {-7, -5, -3, -1, 0, 1, 3}, find the solution set of:

$x \lt -2$

Answer

$x \lt - 2$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are less than - 2 are - 7, - 5 and - 3.

Hence, solution set = {-7, -5, -3}.

If the replacement set = {-7, -5, -3, -1, 0, 1, 3}, find the solution set of:

$x \gt 2$

Answer

$x \gt 2$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The only element of the replacement set which is greater than 2 is 3.

Hence, solution set = {3}.

If the replacement set = {-7, -5, -3, -1, 0, 1, 3}, find the solution set of:

$-5 \lt x \le 5$

Answer

$- 5 \lt x \le 5$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are greater than - 5 and less than or equal to 5 are - 3, - 1, 0, 1 and 3.

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

If the replacement set = {-7, -5, -3, -1, 0, 1, 3}, find the solution set of:

$-8 \lt x \lt 1$

Answer

$- 8 \lt x \lt 1$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are greater than - 8 and less than 1 are - 7, - 5, - 3, - 1 and 0.

Hence, solution set = {-7, -5, -3, -1, 0}.

If the replacement set = {-7, -5, -3, -1, 0, 1, 3}, find the solution set of:

$0 \le x \le 4$

Answer

$0 \le x \le 4$

∵ The replacement set = {-7, -5, -3, -1, 0, 1, 3}.

The elements of the replacement set which are greater than or equal to 0 and less than or equal to 4 are 0, 1 and 3.

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

Represent the solution of the following inequality graphically:

$x \le 4, x \in N$

Answer

$x \le 4$

∵ $x$ is a natural number.

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

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

Represent the solution of the following inequality graphically:

$x \lt 5, x \in W$

Answer

$x \lt 5$

∵ $x$ is a whole number.

Hence, solution set = {0, 1, 2, 3, 4}.

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

Represent the solution of the following inequality graphically:

$-3 \le x \lt 3, x \in I$

Answer

$- 3 \le x \lt 3$

∵ $x$ is an integer.

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

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

If the replacement set is {-6, -4, -2, 0, 2, 4, 6}, then represent the solution set of the inequality $-4 \le x \lt 4$ graphically.

Answer

$- 4 \le x \lt 4$

∵ The replacement set = {-6, -4, -2, 0, 2, 4, 6}.

The elements of the replacement set which are greater than or equal to - 4 and less than 4 are - 4, - 2, 0 and 2.

Hence, solution set = {-4, -2, 0, 2}.

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

Find the solution set of the inequality $x \lt 4$ if the replacement set is

{1, 2, 3, ..., 10}

Answer

$x \lt 4$

∵ The replacement set = {1, 2, 3, ..., 10}.

The elements of the replacement set which are less than 4 are 1, 2 and 3.

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

Find the solution set of the inequality $x \lt 4$ if the replacement set is

{-1, 0, 1, 2, 5, 8}

Answer

$x \lt 4$

∵ The replacement set = {-1, 0, 1, 2, 5, 8}.

The elements of the replacement set which are less than 4 are - 1, 0, 1 and 2.

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

Find the solution set of the inequality $x \lt 4$ if the replacement set is

{-5, 10}

Answer

$x \lt 4$

∵ The replacement set = {-5, 10}.

The only element of the replacement set which is less than 4 is - 5.

Hence, solution set = {-5}.

Find the solution set of the inequality $x \lt 4$ if the replacement set is

{5, 6, 7, 8, 9, 10}

Answer

$x \lt 4$

∵ The replacement set = {5, 6, 7, 8, 9, 10}.

No element of the replacement set is less than 4.

Hence, solution set = φ.

If the replacement set = {-6, -3, 0, 3, 6, 9, 12}, find the truth set of the following:

$2x - 3 \gt 7$

Answer

Solving,

$\Rightarrow 2x - 3 \gt 7 \\[1em] \Rightarrow 2x \gt 7 + 3 \\[1em] \Rightarrow 2x \gt 10 \\[1em] \Rightarrow x \gt \dfrac{10}{2} \\[1em] \Rightarrow x \gt 5$

∵ The replacement set = {-6, -3, 0, 3, 6, 9, 12}.

Hence, truth set = {6, 9, 12}.

If the replacement set = {-6, -3, 0, 3, 6, 9, 12}, find the truth set of the following:

$3x + 8 \le 2$

Answer

Solving,

$\Rightarrow 3x + 8 \le 2 \\[1em] \Rightarrow 3x \le 2 - 8 \\[1em] \Rightarrow 3x \le - 6 \\[1em] \Rightarrow x \le - \dfrac{6}{3} \\[1em] \Rightarrow x \le - 2$

∵ The replacement set = {-6, -3, 0, 3, 6, 9, 12}.

Hence, truth set = {-6, -3}.

If the replacement set = {-6, -3, 0, 3, 6, 9, 12}, find the truth set of the following:

$-3 \lt 1 - 2x$

Answer

Solving,

$\Rightarrow - 3 \lt 1 - 2x \\[1em] \Rightarrow 2x \lt 1 + 3 \\[1em] \Rightarrow 2x \lt 4 \\[1em] \Rightarrow x \lt \dfrac{4}{2} \\[1em] \Rightarrow x \lt 2$

∵ The replacement set = {-6, -3, 0, 3, 6, 9, 12}.

Hence, truth set = {-6, -3, 0}.

Solve the following inequation:

$4x + 1 \lt 17, x \in N$

Answer

Solving,

$\Rightarrow 4x + 1 \lt 17 \\[1em] \Rightarrow 4x \lt 17 - 1 \\[1em] \Rightarrow 4x \lt 16 \\[1em] \Rightarrow x \lt \dfrac{16}{4} \\[1em] \Rightarrow x \lt 4$

∵ $x$ is a natural number.

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

Solve the following inequation:

$4x + 1 \le 17, x \in W$

Answer

Solving,

$\Rightarrow 4x + 1 \le 17 \\[1em] \Rightarrow 4x \le 17 - 1 \\[1em] \Rightarrow 4x \le 16 \\[1em] \Rightarrow x \le \dfrac{16}{4} \\[1em] \Rightarrow x \le 4$

∵ $x$ is a whole number.

Hence, solution set = {0, 1, 2, 3, 4}.

Solve the following inequation:

$4 \gt 3x - 11, x \in N$

Answer

Solving,

$\Rightarrow 4 \gt 3x - 11 \\[1em] \Rightarrow 4 + 11 \gt 3x \\[1em] \Rightarrow 15 \gt 3x \\[1em] \Rightarrow x \lt \dfrac{15}{3} \\[1em] \Rightarrow x \lt 5$

∵ $x$ is a natural number.

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

Solve the following inequation:

$-17 \le 9x - 8, x \in Z$

Answer

Solving,

$\Rightarrow - 17 \le 9x - 8 \\[1em] \Rightarrow - 17 + 8 \le 9x \\[1em] \Rightarrow - 9 \le 9x \\[1em] \Rightarrow x \ge - \dfrac{9}{9} \\[1em] \Rightarrow x \ge - 1$

∵ $x$ is an integer.

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

Solve the following inequation:

$\dfrac{2y - 1}{5} \le 2, y \in N$

Answer

$\dfrac{2y - 1}{5} \le 2$

By cross multiplication, we get

$\Rightarrow 2y - 1 \le 2 \times 5 \\[1em] \Rightarrow 2y - 1 \le 10 \\[1em] \Rightarrow 2y \le 10 + 1 \\[1em] \Rightarrow 2y \le 11 \\[1em] \Rightarrow y \le \dfrac{11}{2} \\[1em] \Rightarrow y \le 5.5$

∵ $y$ is a natural number.

Hence, solution set = {1, 2, 3, 4, 5}.

Solve the following inequation:

$\dfrac{2y + 1}{3} + 1 \le 3, y \in W$

Answer

Solving,

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

By cross multiplication, we get

$\Rightarrow 2y + 1 \le 2 \times 3 \\[1em] \Rightarrow 2y + 1 \le 6 \\[1em] \Rightarrow 2y \le 6 - 1 \\[1em] \Rightarrow 2y \le 5 \\[1em] \Rightarrow y \le \dfrac{5}{2} \\[1em] \Rightarrow y \le 2.5$

∵ $y$ is a whole number.

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

Solve the following inequation:

$\dfrac{2}{3}p + 5 \lt 9, p \in W$

Answer

Solving,

$\Rightarrow \dfrac{2}{3}p + 5 \lt 9 \\[1em] \Rightarrow \dfrac{2}{3}p \lt 9 - 5 \\[1em] \Rightarrow \dfrac{2}{3}p \lt 4 \\[1em] \Rightarrow p \lt \dfrac{4 \times 3}{2} \\[1em] \Rightarrow p \lt \dfrac{12}{2} \\[1em] \Rightarrow p \lt 6$

∵ $p$ is a whole number.

Hence, solution set = {0, 1, 2, 3, 4, 5}.

Solve the following inequation:

$-2(p + 3) \gt 5, p \in I$

Answer

Solving,

$\Rightarrow - 2(p + 3) \gt 5 \\[1em] \Rightarrow - 2p - 6 \gt 5 \\[1em] \Rightarrow - 2p \gt 5 + 6 \\[1em] \Rightarrow - 2p \gt 11 \\[1em] \Rightarrow 2p \lt - 11 \\[1em] \Rightarrow p \lt - \dfrac{11}{2} \\[1em] \Rightarrow p \lt - 5.5$

∵ $p$ is an integer.

Hence, solution set = {........., -8, -7, -6}.

Solve the following inequation:

$2x - 3 \lt x + 2, x \in N$

Answer

Solving,

$\Rightarrow 2x - 3 \lt x + 2 \\[1em] \Rightarrow 2x - x \lt 2 + 3 \\[1em] \Rightarrow x \lt 5$

∵ $x$ is a natural number.

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

Solve the following inequation:

$3 - x \le 5 - 3x, x \in W$

Answer

Solving,

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

∵ $x$ is a whole number.

Hence, solution set = {0, 1}.

Solve the following inequation:

$3(x - 2) \lt 2(x - 1), x \in W$

Answer

Solving,

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

∵ $x$ is a whole number.

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

Solve the following inequation:

$\dfrac{3}{2} - \dfrac{x}{2} \gt -1, x \in N$

Answer

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

Multiplying both sides by 2, we get:

$\Rightarrow \Big[\dfrac{3}{2} - \dfrac{x}{2}\Big] \times 2 \gt -1 \times 2 \\[1em] \Rightarrow \dfrac{3}{2} \times 2 - \dfrac{x}{2} \times 2 \gt - 2 \\[1em] \Rightarrow 3 - x \gt - 2 \\[1em] \Rightarrow - x \gt - 2 - 3 \\[1em] \Rightarrow - x \gt - 5 \\[1em] \Rightarrow x \lt 5$

∵ $x$ is a natural number.

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

If the replacement set is {-3, -2, -1, 0, 1, 2, 3}, solve the inequation $\dfrac{3x - 1}{2} \lt 2$. Also represent its solution on the number line.

Answer

$\dfrac{3x - 1}{2} \lt 2$

By cross multiplication, we get:

$\Rightarrow 3x - 1 \lt 2 \times 2 \\[1em] \Rightarrow 3x - 1 \lt 4 \\[1em] \Rightarrow 3x \lt 4 + 1 \\[1em] \Rightarrow 3x \lt 5 \\[1em] \Rightarrow x \lt \dfrac{5}{3} \\[1em] \Rightarrow x \lt 1.66.....$

∵ The replacement set = {-3, -2, -1, 0, 1, 2, 3}.

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

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

Solve $\dfrac{x}{3} + \dfrac{1}{4} \lt \dfrac{x}{6} + \dfrac{1}{2}, x \in W$. Also represent its solution on the number line.

Answer

$\dfrac{x}{3} + \dfrac{1}{4} \lt \dfrac{x}{6} + \dfrac{1}{2}$

L.C.M. of 3, 4, 6 and 2 = 12

Multiplying both sides by 12, we get:

$\Rightarrow \Big[\dfrac{x}{3} + \dfrac{1}{4}\Big] \times 12 \lt \Big[\dfrac{x}{6} + \dfrac{1}{2}\Big] \times 12 \\[1em] \Rightarrow \dfrac{x}{3} \times 12 + \dfrac{1}{4} \times 12 \lt \dfrac{x}{6} \times 12 + \dfrac{1}{2} \times 12 \\[1em] \Rightarrow 4x + 3 \lt 2x + 6 \\[1em] \Rightarrow 4x - 2x \lt 6 - 3 \\[1em] \Rightarrow 2x \lt 3 \\[1em] \Rightarrow x \lt \dfrac{3}{2} \\[1em] \Rightarrow x \lt 1.5$

∵ $x$ is a whole number.

Hence, solution set = {0, 1}.

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

Solve the following inequation and graph its solution on a number line:

$-4 \le 4x \lt 14, x \in N$

Answer

$- 4 \le 4x \lt 14$

Solving L.H.S., we get:

$\Rightarrow - 4 \le 4x \\[1em] \Rightarrow x \ge - \dfrac{4}{4} \\[1em] \Rightarrow x \ge - 1 \quad \text{...(1)}$

Solving R.H.S., we get:

$\Rightarrow 4x \lt 14 \\[1em] \Rightarrow x \lt \dfrac{14}{4} \\[1em] \Rightarrow x \lt 3.5 \quad \text{...(2)}$

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

⇒ $- 1 \le x \lt 3.5$

∵ $x$ is a natural number.

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

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

Solve the following inequation and graph its solution on a number line:

$-1 \lt \dfrac{x}{2} + 1 \le 3, x \in I$

Answer

$-1 \lt \dfrac{x}{2} + 1 \le 3$

Solving L.H.S., we get:

$\Rightarrow - 1 \lt \dfrac{x}{2} + 1 \\[1em] \Rightarrow - 1 - 1 \lt \dfrac{x}{2} \\[1em] \Rightarrow - 2 \lt \dfrac{x}{2} \\[1em] \Rightarrow - 4 \lt x \quad \text{...(1)}$

Solving R.H.S., we get:

$\Rightarrow \dfrac{x}{2} + 1 \le 3 \\[1em] \Rightarrow \dfrac{x}{2} \le 3 - 1 \\[1em] \Rightarrow \dfrac{x}{2} \le 2 \\[1em] \Rightarrow x \le 4 \quad \text{...(2)}$

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

⇒ $- 4 \lt x \le 4$

∵ $x$ is an integer.

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

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