Inequalities

Given,

Inequation : $4x - 3 \lt 2$

Adding 10 to each side of the inequation, we get :

$\Rightarrow 4x - 3 + 10 \lt 2 + 10$

$\Rightarrow 4x + 7 \lt 12$

Hence, the resulting inequation is $\bm{4x + 7 \lt 12}$.

Given,

Inequation : $2x + 5 \ge 8$

Subtracting 7 from each side of the inequation, we get :

$\Rightarrow 2x + 5 - 7 \ge 8 - 7$

$\Rightarrow 2x - 2 \ge 1$

Hence, the resulting inequation is $\bm{2x - 2 \ge 1}$.

Given,

Inequation : $3x - 5 \le 2$

Multiplying each side of the inequation by 4, we get :

$\Rightarrow (3x - 5) \times 4 \le 2 \times 4$

$\Rightarrow 12x - 20 \le 8$

Hence, the resulting inequation is $\bm{12x - 20 \le 8}$.

Given,

Inequation : $8x + 3 \gt 13$

Dividing each side of the inequation by 6, we get :

$$\begin{aligned}\Rightarrow \frac{8x + 3}{6} \gt \frac{13}{6}\\[1em] \Rightarrow \frac{8x}{6} + \frac{3}{6} \gt \frac{13}{6}\\[1em] \Rightarrow \frac{4x}{3} + \frac{1}{2} \gt \frac{13}{6}\end{aligned}$$

Hence, the resulting inequation is $\bm{\frac{4x}{3} + \frac{1}{2} \gt \frac{13}{6}}$.

The replacement set is {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5}.

(i) For $x \lt 4$, the values from the replacement set that are less than 4 are -4, -3, -2, -1, 0, 1, 2 and 3.

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

(ii) For $x \lt -1$, the values from the replacement set that are less than -1 are -4, -3 and -2.

Hence, the solution set is {-4, -3, -2}.

(iii) Solving,

$\Rightarrow x - 3 \ge 0$

$\Rightarrow x \ge 3$

The values from the replacement set that are greater than or equal to 3 are 3, 4 and 5.

Hence, the solution set is {3, 4, 5}.

(iv) Solving,

$$\begin{aligned}\Rightarrow 2x \gt 6\\[1em] \Rightarrow x \gt \frac{6}{2}\\[1em] \Rightarrow x \gt 3\end{aligned}$$

The values from the replacement set that are greater than 3 are 4 and 5.

Hence, the solution set is {4, 5}.

(v) Solving,

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

The values from the replacement set that are less than or equal to -1 are -4, -3, -2 and -1.

Hence, the solution set is {-4, -3, -2, -1}.

The whole numbers between -2 and 6 are 0, 1, 2, 3, 4 and 5.

∴ Replacement set = {0, 1, 2, 3, 4, 5}

(i) For $-5 \lt x \lt 4$, the values from the replacement set that lie between -5 and 4 are 0, 1, 2 and 3.

Hence, the solution set is {0, 1, 2, 3}.

(ii) Solving,

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

The values from the replacement set that are greater than or equal to 4 are 4 and 5.

Hence, the solution set is {4, 5}.

(iii) Solving,

$$\begin{aligned}\Rightarrow 3x - 5 \le 10\\[1em] \Rightarrow 3x \le 10 + 5\\[1em] \Rightarrow 3x \le 15\\[1em] \Rightarrow x \le \frac{15}{3}\\[1em] \Rightarrow x \le 5\end{aligned}$$

The values from the replacement set that are less than or equal to 5 are 0, 1, 2, 3, 4 and 5.

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

(iv) For $2 \lt x \le 7$, the values from the replacement set that are greater than 2 and less than or equal to 7 are 3, 4 and 5.

Hence, the solution set is {3, 4, 5}.

(v) For $0 \le x \le 6$, the values from the replacement set that are greater than or equal to 0 and less than or equal to 6 are 0, 1, 2, 3, 4 and 5.

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

Solving,

$$\begin{aligned}\Rightarrow x - 5 \le 2\\[1em] \Rightarrow x \le 2 + 5\\[1em] \Rightarrow x \le 7\end{aligned}$$

As $x$ ∈ W, the solution set is {0, 1, 2, 3, 4, 5, 6, 7}.

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

Solving,

$$\begin{aligned}\Rightarrow 2x - 3 \lt 7\\[1em] \Rightarrow 2x \lt 7 + 3\\[1em] \Rightarrow 2x \lt 10\\[1em] \Rightarrow x \lt \frac{10}{2}\\[1em] \Rightarrow x \lt 5\end{aligned}$$

As $x$ ∈ N, the solution set is {1, 2, 3, 4}.

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

Solving,

$$\begin{aligned}\Rightarrow 5x + 12 \gt -13\\[1em] \Rightarrow 5x \gt -13 - 12\\[1em] \Rightarrow 5x \gt -25\\[1em] \Rightarrow x \gt \frac{-25}{5}\\[1em] \Rightarrow x \gt -5\end{aligned}$$

As $x$ ∈ I, the solution set is {-4, -3, -2, -1, 0, 1, 2, 3, 4, .....}.

The solution set is shown by thick dots on the number line and the dark arrow head on the right side shows that the solution set continues towards the right.

Solving,

$$\begin{aligned}\Rightarrow 3x - 15 \ge 15\\[1em] \Rightarrow 3x \ge 15 + 15\\[1em] \Rightarrow 3x \ge 30\\[1em] \Rightarrow x \ge \frac{30}{3}\\[1em] \Rightarrow x \ge 10\end{aligned}$$

As $x$ ∈ N, the solution set is {10, 11, 12, 13, 14, .....}.

The solution set is shown by thick dots on the number line and the dark arrow head on the right side shows that the solution set continues towards the right.

Solving,

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

As $x$ ∈ I, the solution set is {....., -7, -6, -5, -4}.

The solution set is shown by thick dots on the number line and the dark arrow head on the left side shows that the solution set continues towards the left.

Solving,

$$\begin{aligned}\Rightarrow \frac{3x - 2}{4} \le 4\\[1em] \Rightarrow 3x - 2 \le 4 \times 4\\[1em] \Rightarrow 3x - 2 \le 16\\[1em] \Rightarrow 3x \le 16 + 2\\[1em] \Rightarrow 3x \le 18\\[1em] \Rightarrow x \le \frac{18}{3}\\[1em] \Rightarrow x \le 6\end{aligned}$$

As $x$ ∈ W, the whole numbers less than or equal to 6 are 0, 1, 2, 3, 4, 5 and 6.

Hence, the solution set is {0, 1, 2, 3, 4, 5, 6}.

Solving,

$$\begin{aligned}\Rightarrow \frac{3x + 2}{5} + 5 \lt 0\\[1em] \Rightarrow \frac{3x + 2}{5} \lt -5\\[1em] \Rightarrow 3x + 2 \lt -5 \times 5\\[1em] \Rightarrow 3x + 2 \lt -25\\[1em] \Rightarrow 3x \lt -25 - 2\\[1em] \Rightarrow 3x \lt -27\\[1em] \Rightarrow x \lt \frac{-27}{3}\\[1em] \Rightarrow x \lt -9\end{aligned}$$

As $x$ ∈ N, there is no natural number which is less than -9.

Hence, there is no solution i.e. the solution set is φ.

Solving,

$$\begin{aligned}\Rightarrow 5 - 2x \ge x - 12\\[1em] \Rightarrow 5 + 12 \ge x + 2x\\[1em] \Rightarrow 17 \ge 3x \\[1em] \Rightarrow \frac{17}{3} \ge x\\[1em] \Rightarrow x \le \frac{17}{3}\\[1em] \Rightarrow x \le 5\frac{2}{3}\end{aligned}$$

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

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

Solving,

$$\begin{aligned}\Rightarrow 2(x + 2) + 9 \gt 3(x - 1)\\[1em] \Rightarrow 2x + 4 + 9 \gt 3x - 3\\[1em] \Rightarrow 2x + 13 \gt 3x - 3\\[1em] \Rightarrow 13 + 3 \gt 3x - 2x\\[1em] \Rightarrow 16 \gt x\\[1em] \Rightarrow x \lt 16\end{aligned}$$

As $x$ ∈ N, the natural numbers less than 16 are 1, 2, 3, ....., 15.

Hence, the solution set is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}.

Solving,

$$\begin{aligned}\Rightarrow 5 \le 3(x - 2) + 4\\[1em] \Rightarrow 5 \le 3x - 6 + 4\\[1em] \Rightarrow 5 \le 3x - 2\\[1em] \Rightarrow 5 + 2 \le 3x\\[1em] \Rightarrow 7 \le 3x \\[1em] \Rightarrow \frac{7}{3} \le x\\[1em] \Rightarrow x \ge \frac{7}{3}\\[1em] \Rightarrow x \ge 2\frac{1}{3}\end{aligned}$$

As $x$ ∈ W, the whole numbers greater than or equal to $2\frac{1}{3}$ are 3, 4, 5, 6, .....

∴ Solution set is {3, 4, 5 .....}

The solution set is shown by thick dots on the number line and the dark arrow head on the right side shows that the solution set continues towards the right.

Solving,

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

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

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

The solution set is shown by thick dots on the number line and the dark arrow head on the right side shows that the solution set continues towards the right.

Solving,

$$\begin{aligned}\Rightarrow 8 - x \ge 3\\[1em] \Rightarrow -x \ge 3 - 8\\[1em] \Rightarrow -x \ge -5\\[1em] [\text{Multiplying by -1 and reversing the symbol}] \\[1em] \Rightarrow x \le 5\end{aligned}$$

As $x$ ∈ N, the solution set is {1, 2, 3, 4, 5}.

Hence, Option 2 is the correct answer.

For $x \lt 3$ and $x$ ∈ W, the solution set is {0, 1, 2}.

The number line showing thick dots at 0, 1 and 2 only, represents the solution set.

Hence, Option 3 is the correct answer.

Solving,

$$\begin{aligned}\Rightarrow \frac{2x - 1}{3} \le 4\\[1em] \Rightarrow 2x - 1 \le 12\\[1em] \Rightarrow 2x \le 12 + 1\\[1em] \Rightarrow 2x \le 13\\[1em] \Rightarrow x \le \frac{13}{2}\\[1em] \Rightarrow x \le 6\frac{1}{2}\end{aligned}$$

As $x$ ∈ W, the solution set is {0, 1, 2, 3, 4, 5, 6}.

Hence, Option 2 is the correct answer.

Solving,

$$\begin{aligned}\Rightarrow 19 - 3x \ge x - 2\\[1em] \Rightarrow 19 + 2 \ge x + 3x\\[1em] \Rightarrow 21 \ge 4x\\[1em] \Rightarrow \frac{21}{4} \ge x\\[1em] \Rightarrow x \le 5\frac{1}{4}\end{aligned}$$

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

Hence, Option 1 is the correct answer.

Solving,

$$\begin{aligned}\Rightarrow 2 - 3(5 - x) \ge -1\\[1em] \Rightarrow 2 - 15 + 3x \ge -1\\[1em] \Rightarrow -13 + 3x \ge -1\\[1em] \Rightarrow 3x \ge -1 + 13\\[1em] \Rightarrow 3x \ge 12\\[1em] \Rightarrow x \ge \frac{12}{3}\\[1em] \Rightarrow x \ge 4\end{aligned}$$

As $x$ ∈ N, the solution set is {4, 5, 6, 7, .......}.

Hence, Option 4 is the correct answer.

Solving,

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

As $x$ ∈ W, the solution set is {0, 1}.

Hence, Option 3 is the correct answer.

Solving,

$$\begin{aligned}\Rightarrow x + 14 \le 8x - 7\\[1em] \Rightarrow 14 + 7 \le 8x - x\\[1em] \Rightarrow 21 \le 7x\\[1em] \Rightarrow \frac{21}{7} \le x\\[1em] \Rightarrow 3 \le x\\[1em] \Rightarrow x \ge 3\end{aligned}$$

As $x$ ∈ Z, the solution set is {3, 4, 5, 6, .....}, which is shown by thick dots at 3, 4, 5, 6, ..... with the dark arrow head on the right side of the number line.

Hence, Option 2 is the correct answer.

Set A is a set of integers greater than equal to -1 and is shown by thick dots on the number line, whereas set B is a set of real numbers less than equal to -3 and is shown by dark arrow on the left side of the number line.

A ∪ B is the set of all the elements which belong either to set A or to set B or to both.

∴ A ∪ B is represented by the dark arrow of set B together with the thick dots of set A.

Hence, Option 2 is the correct answer.

The given inequation is $-1 \le x \lt 4$, where $x$ ∈ R.

As $x$ ∈ R (real numbers), the solution set contains all the real numbers from -1 to 4, so it is represented by a thick (continuous) line from -1 to 4 on the number line.

Since -1 is included in the solution set (as $-1 \le x$), it is marked with a solid (filled) circle, and since 4 is not included in the solution set (as $x \lt 4$), it is marked with a hollow (open) circle.

Hence, Option 4 is the correct answer.

The given set is ${x : x \in R, x \ge -5 \text{ and } x \lt 2}$, i.e. $-5 \le x \lt 2$, where $x$ ∈ R.

As $x$ ∈ R (real numbers), the set contains all the real numbers from -5 to 2, so it is represented by a thick (continuous) line from -5 to 2 on the number line.

Since -5 is included in the set (as $x \ge -5$), it is marked with a solid (filled) circle, and since 2 is not included in the set (as $x \lt 2$), it is marked with a hollow (open) circle.

Hence, Option 1 is the correct answer.

Solving,

$\Rightarrow 3x \gt 18$

$\Rightarrow x \gt 6$ [Dividing both sides by 3]

Since there is no element greater than 6 in the replacement set {0, 1, 2, 3, 4, 5, 6}, the solution set is a null set i.e. φ.

∴ Statement 1 is true.

Each element of the solution set belongs to the replacement set.

∴ Statement 2 is true.

Hence, Option 1 is the correct answer.

Solving,

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

As the replacement set is the set of negative integers, the solution set is {-1, -2} only and not {-2, -1, .....}.

∴ Statement 1 is false.

On multiplying both the sides of a linear inequation by a negative number, the direction of the inequality symbol changes (reverses).

∴ Statement 2 is true.

Hence, Option 4 is the correct answer.

Given,,

$\Rightarrow 15x - 10 \lt 4x + 2$

Transposing $4x$ to L.H.S. and -10 to R.H.S.,

$\Rightarrow 15x - 4x \lt 2 + 10$

∴ Assertion (A) is true.

Adding the same number or expression to both the sides of an inequation does not change the inequality.

∴ Reason (R) is true.

Hence, Option 3 is the correct answer.

$3x + 5 \lt 65$ is a mathematical statement showing that the two expressions are not equal and it involves the variable $x$.

∴ $3x + 5 \lt 65$ is an in-equation.

∴ Assertion (A) is false.

The solution set of an inequation is chosen from the replacement set. So, if the replacement set is not defined, the solution of the inequality cannot be represented on a number line.

∴ Reason (R) is true.

Hence, Option 2 is the correct answer.

Given, $3 \gt -x \ge -5$

Multiplying each side by -1, the inequality gets reversed.

$\Rightarrow -3 \lt x \le 5$

∴ Assertion (A) is true.

Subtracting the same quantity from both the sides of an inequation does not change the sign of inequality.

∴ Reason (R) is true.

Hence, Option 3 is the correct answer.