Inequalities — Exercise 15

Find the resulting inequation, when:

10 is added to each side of $4x - 3 \lt 2$.

Answer

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 $\bm{4x + 7 \lt 12}$.

Find the resulting inequation, when:

7 is subtracted from each side of $2x + 5 \ge 8$.

Answer

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 $\bm{2x - 2 \ge 1}$.

Find the resulting inequation, when:

each side of $3x - 5 \le 2$ is multiplied by 4.

Answer

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 $\bm{12x - 20 \le 8}$.

Find the resulting inequation, when:

each side of $8x + 3 \gt 13$ is divided by 6.

Answer

Given,

Inequation : $8x + 3 \gt 13$

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

$\Rightarrow \dfrac{8x + 3}{6} \gt \dfrac{13}{6}\\[1em] \Rightarrow \dfrac{8x}{6} + \dfrac{3}{6} \gt \dfrac{13}{6}\\[1em] \Rightarrow \dfrac{4x}{3} + \dfrac{1}{2} \gt \dfrac{13}{6}$

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

If the replacement set = {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5}, find the solution set, if:

(i) $x \lt 4$

(ii) $x \lt -1$

(iii) $x - 3 \ge 0$

(iv) $2x \gt 6$

(v) $\dfrac{3}{2} + x \le \dfrac{1}{2}$

Answer

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,

$\Rightarrow 2x \gt 6\\[1em] \Rightarrow x \gt \dfrac{6}{2}\\[1em] \Rightarrow x \gt 3$

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

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

(v) Solving,

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

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}.

If the replacement set = Set of whole numbers between -2 and 6, find the solution set for each of the following inequations:

(i) $-5 \lt x \lt 4$

(ii) $2x - 3 \ge 5$

(iii) $3x - 5 \le 10$

(iv) $2 \lt x \le 7$

(v) $0 \le x \le 6$

Answer

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,

$\Rightarrow 2x - 3 \ge 5\\[1em] \Rightarrow 2x \ge 5 + 3\\[1em] \Rightarrow 2x \ge 8\\[1em] \Rightarrow x \ge \dfrac{8}{2}\\[1em] \Rightarrow x \ge 4$

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,

$\Rightarrow 3x - 5 \le 10\\[1em] \Rightarrow 3x \le 10 + 5\\[1em] \Rightarrow 3x \le 15\\[1em] \Rightarrow x \le \dfrac{15}{3}\\[1em] \Rightarrow x \le 5$

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}.

Represent the solution set for each of the following inequations on the number line:

$x - 5 \le 2, x$ ∈ W

Answer

Solving,

$\Rightarrow x - 5 \le 2\\[1em] \Rightarrow x \le 2 + 5\\[1em] \Rightarrow x \le 7$

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.

Represent the solution set for each of the following inequations on the number line:

$2x - 3 \lt 7, x$ ∈ N

Answer

Solving,

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

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

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

Represent the solution set for each of the following inequations on the number line:

$5x + 12 \gt -13, x$ ∈ I

Answer

Solving,

$\Rightarrow 5x + 12 \gt -13\\[1em] \Rightarrow 5x \gt -13 - 12\\[1em] \Rightarrow 5x \gt -25\\[1em] \Rightarrow x \gt \dfrac{-25}{5}\\[1em] \Rightarrow x \gt -5$

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.

Represent the solution set for each of the following inequations on the number line:

$3x - 15 \ge 15, x$ ∈ N

Answer

Solving,

$\Rightarrow 3x - 15 \ge 15\\[1em] \Rightarrow 3x \ge 15 + 15\\[1em] \Rightarrow 3x \ge 30\\[1em] \Rightarrow x \ge \dfrac{30}{3}\\[1em] \Rightarrow x \ge 10$

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.

Represent the solution set for each of the following inequations on the number line:

$2x + 5 \le -3, x$ ∈ I

Answer

Solving,

$\Rightarrow 2x + 5 \le -3\\[1em] \Rightarrow 2x \le -3 - 5\\[1em] \Rightarrow 2x \le -8\\[1em] \Rightarrow x \le \dfrac{-8}{2}\\[1em] \Rightarrow x \le -4$

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.

Solve $\dfrac{3x - 2}{4} \le 4$, where $x$ ∈ W.

Answer

Solving,

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

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}.

Solve :

$\dfrac{3x + 2}{5} + 5 \lt 0$, where $x$ ∈ N.

Answer

Solving,

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

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.

Solve $5 - 2x \ge x - 12, x$ ∈ W.

Answer

Solving,

$\Rightarrow 5 - 2x \ge x - 12\\[1em] \Rightarrow 5 + 12 \ge x + 2x\\[1em] \Rightarrow 17 \ge 3x \\[1em] \Rightarrow \dfrac{17}{3} \ge x\\[1em] \Rightarrow x \le \dfrac{17}{3}\\[1em] \Rightarrow x \le 5\dfrac{2}{3}$

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

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

Solve $2(x + 2) + 9 \gt 3(x - 1), x$ ∈ N.

Answer

Solving,

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

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}.

Solve $5 \le 3(x - 2) + 4, x$ ∈ W. Draw a suitable number line to represent the solution obtained.

Answer

Solving,

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

As $x$ ∈ W, the whole numbers greater than or equal to $2\dfrac{1}{3}$ 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.

Solve the inequation $-10 \lt -5 + 2x, x$ ∈ I and represent the solution set on a number line.

Answer

Solving,

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

As $x$ ∈ I, the integers greater than $-2\dfrac{1}{2}$ 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.