Linear Inequations — Test Yourself

The maximum value of x for the inequation 4x ≤ 12 + x is :

1. 5

2. 4

3. 3

4. 2.4

Answer

Given,

⇒ 4x ≤ 12 + x

⇒ 4x - x ≤ 12

⇒ 3x ≤ 12

⇒ x ≤ $\dfrac{12}{3}$

⇒ x ≤ 4

∴ Maximum value of x will be 4.

Hence, Option 2 is the correct option.

The minimum value of x for the inequation 5x - 4 ≥ 18 - 6x is :

1. 2

2. 22

3. -22

4. -2

Answer

Given,

⇒ 5x - 4 ≥ 18 - 6x

⇒ 5x + 6x ≥ 18 + 4

⇒ 11x ≥ 22

⇒ x ≥ $\dfrac{22}{11}$

⇒ x ≥ 2

∴ Minimum value of x will be 2.

Hence, Option 1 is the correct option.

If 1 ≤ -x < 5, x is an integer then the sum of smallest and greatest values of x is :

1. -5

2. -4

3. -3

4. 0

Answer

Given,

⇒ 1 ≤ -x < 5

⇒ -5 < x ≤ -1

Since, x is an integer

The possible integer values are: {-4, -3, -2, -1}

The sum of smallest and greatest values = -4 + (-1) = -5

Hence, Option 1 is the correct option.

The value of x for the inequation 3x + 15 < 5x + 13, x ∈ Z is :

1. > 1

2. < 1

3. = 1

4. ≥ 1

Answer

Given,

⇒ 3x + 15 < 5x + 13

⇒ 5x + 13 > 3x + 15

⇒ 5x - 3x > 15 - 13

⇒ 2x > 2

⇒ x > 1

Hence, Option 1 is the correct option.

The real number lines for two inequations A and B are as given below, A ∩ B is :

Answer

A = {x : x ∈ R and -3 < x ≤ 1}

B = {x : x ∈ R and -4 ≤ x < 0}

A ∩ B = {x : x ∈ R and -3 < x < 0}

Hence, Option 1 is the correct option.

For the inequations A and B [as given above in part (d)], A ∪ B is :

Answer

A = {x : x ∈ R and -3 < x ≤ 1}

B = {x : x ∈ R and -4 ≤ x < 0}

A ∪ B = {x : x ∈ R and -4 ≤ x ≤ 1}

Hence, Option 1 is the correct option.

$-\dfrac{3}{2} \le -\dfrac{2x}{3}$ where x ∈ R.

Assertion (A): The largest value of x is $\dfrac{9}{4}$.

Reason (R): When the signs of both the sides of an inequalities are changed, the sign of inequality reverses.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is the correct reason for A.

4. Both A and R are true and R is the incorrect reason for A.

Answer

Both A and R are true and R is the correct reason for A.

Reason

According to the assertion:

$\Rightarrow -\dfrac{3}{2} \le -\dfrac{2x}{3}\\[1em] \Rightarrow \dfrac{3}{2} \ge \dfrac{2x}{3}\\[1em] \Rightarrow x \le \dfrac{3 \times 3}{2 \times 2}\\[1em] \Rightarrow x \le \dfrac{9}{4}$

So, Assertion (A) is true.

According to the reason:

When you multiply or divide both sides of an inequality by a negative number, the direction of the inequality sign must be reversed to maintain the validity of the inequality

So, Reason (R) is true.

Hence, option 3 is correct.

Inequation 5 - 2x ≥ x - 10, where x ∈ N (Natural numbers)

Assertion (A): 5 - 2x ≥ x - 10 ⇒ -3x ≥ -15 ⇒ x ≥ 5

∴ Solution set = {5, 6, 7, 8, ..........}

Reason (R): 5 - 2x ≥ x - 10 ⇒ 5 + 10 ≥ 3x ⇒ x ≤ 5

∴ Solution set = {1, 2, 3, 4, 5}

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

A is false, R is true.

Reason

According to Assertion: 5 - 2x ≥ x - 10

⇒ 5 - 2x + 10 ≥ x

⇒ -2x + 15 ≥ x

⇒ 15 ≥ x + 2x

⇒ 15 ≥ 3x

⇒ x ≤ $\dfrac{15}{3}$

⇒ x ≤ 5

∴ Solution set = {1, 2, 3, 4, 5}

So, Assertion (A) is false.

According to Reason:

Solution set = {1, 2, 3, 4, 5}

So, Reason (R) is true.

Hence, A is false, R is true.

x ∈ W, x ≥ -3 and x < 5.

Statement (1) : There will be no solution for the given inequalities.

Statement (2) : The real number line for the given inequations is :

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Statement 1 is false, and statement 2 is true.

Reason

x ≥ -3

Solution set of x = {-3, -2, -1, 0, 1, 2, ..........} .......... (1)

And, x < 5

Solution set of x = {.........., 1, 2, 3, 4} .......... (2)

From (1) and (2), we get

Solution set = {-3, -2, -1, 0, 1, 2, 3, 4}

So, statement 1 is false.

The real number line for the given inequations is :

So, statement 2 is true.

Hence, option 4 is correct.

5 + x ≤ 2x < x - 2, x ∈ R.

Statement (1) : There is no value of x ∈ R that satisfies the given inequation.

Statement (2) : 5 + x - x ≤ 2x - x < x - 2 - x ⇒ 5 ≤ x < -2

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Both the statements are true.

Reason

Given,

Equation : 5 + x ≤ 2x < x - 2

⇒ 5 + x ≤ 2x

⇒ 5 ≤ 2x - x

⇒ 5 ≤ x .......... (1)

And, 2x < x - 2

⇒ 2x - x < -2

⇒ x < -2 .......... (2)

There can be no real number which is greater than or equal to 5 and less than -2 at the same time.

∴ Both the statements are true.

Hence, option 1 is correct.

Solve the inequation :

$12 + 1\dfrac{5}{6}x \le 5 + 3x$ and x ∈ R.

Answer

Given,

$\Rightarrow 12 + 1\dfrac{5}{6}x \le 5 + 3x \\[1em] \Rightarrow 12 + \dfrac{11}{6}x \le 5 + 3x \\[1em] \Rightarrow 3x - \dfrac{11}{6}x \ge 12 - 5 \\[1em] \Rightarrow \dfrac{18x - 11x}{6} \ge 7 \\[1em] \Rightarrow \dfrac{7x}{6} \ge 7 \\[1em] \Rightarrow x \ge \dfrac{7 \times 6}{7} \\[1em] \Rightarrow x \ge 6.$

∴ Solution set = {x : x ∈ R and x ≥ 6}.

Given x ∈ {whole numbers}, find the solution set of :

-1 ≤ 3 + 4x < 23

Answer

Given,

-1 ≤ 3 + 4x < 23

Solving L.H.S. of the equation,

⇒ -1 ≤ 3 + 4x

⇒ 4x ≥ -1 - 3

⇒ 4x ≥ -4

⇒ x ≥ -1 ........(i)

Solving R.H.S. of the equation,

⇒ 3 + 4x < 23

⇒ 4x < 23 - 3

⇒ 4x < 20

⇒ x < 5 .........(ii)

From (i) and (ii) we get,

-1 ≤ x < 5

Since, x ∈ {whole numbers},

∴ Solution set = {0, 1, 2, 3, 4}.

Find the set of values of x, satisfying :

7x + 3 ≥ 3x - 5 and $\dfrac{x}{4} - 5 \le \dfrac{5}{4} - x$, where x ∈ N.

Answer

Solving,

⇒ 7x + 3 ≥ 3x - 5

⇒ 7x - 3x ≥ -5 - 3

⇒ 4x ≥ -8

⇒ x ≥ -2 .......(i)

Solving,

$\Rightarrow \dfrac{x}{4} - 5 \le \dfrac{5}{4} - x \\[1em] \Rightarrow \dfrac{x}{4} + x \le \dfrac{5}{4} + 5 \\[1em] \dfrac{x + 4x}{4} \le \dfrac{25}{4} \\[1em] \dfrac{5x}{4} \le \dfrac{25}{4} \\[1em] x \le \dfrac{25}{4} \times \dfrac{4}{5} \\[1em] x \le 5 ........(ii)$

From (i) and (ii) we get,

-2 ≤ x ≤ 5

Since, x ∈ N

∴ Solution set = {1, 2, 3, 4, 5}.

Solve :

(i) $\dfrac{x}{2} + 5 \le \dfrac{x}{3} + 6$, where x is a positive odd integer

(ii) $\dfrac{2x + 3}{3} \ge \dfrac{3x - 1}{4}$, where x is a positive even integer

Answer

(i) Solving,

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

Since, x is a positive odd integer

∴ Solution set = {1, 3, 5}.

(ii) Solving,

$\Rightarrow \dfrac{2x + 3}{3} \ge \dfrac{3x - 1}{4} \\[1em] \Rightarrow \dfrac{2x + 3}{3} - \dfrac{3x - 1}{4} \ge 0 \\[1em] \Rightarrow \dfrac{4(2x + 3) - 3(3x - 1)}{12} \ge 0 \\[1em] \Rightarrow 8x + 12 - 9x + 3 \ge 0 \\[1em] \Rightarrow -x + 15 \ge 0 \\[1em] \Rightarrow x \le 15.$

Since, x is a positive even integer

∴ Solution set = {2, 4, 6, 8, 10, 12, 14}.

Solve the inequation :

$-2\dfrac{1}{2} + 2x \le \dfrac{4x}{5} \le \dfrac{4}{3} + 2x$, x ∈ W.

Graph the solution set on the number line.

Answer

Given,

$-2\dfrac{1}{2} + 2x \le \dfrac{4x}{5} \le \dfrac{4}{3} + 2x$

Solving L.H.S. of the equation,

$\Rightarrow -2\dfrac{1}{2} + 2x \le \dfrac{4x}{5} \\[1em] \Rightarrow -\dfrac{5}{2} + 2x \le \dfrac{4x}{5} \\[1em] \Rightarrow 2x - \dfrac{4x}{5} \le \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{10x - 4x}{5} \le \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{6x}{5} \le \dfrac{5}{2} \\[1em] \Rightarrow x \le \dfrac{5}{2} \times \dfrac{5}{6} \\[1em] \Rightarrow x \le \dfrac{25}{12} \\[1em] \Rightarrow x \le 2\dfrac{1}{12} ........(i)$

Solving R.H.S. of the equation,

$\Rightarrow \dfrac{4x}{5} \le \dfrac{4}{3} + 2x \\[1em] \Rightarrow \dfrac{4x}{5} - 2x \le \dfrac{4}{3} \\[1em] \Rightarrow \dfrac{4x - 10x}{5} \le \dfrac{4}{3} \\[1em] \Rightarrow -\dfrac{6x}{5} \le \dfrac{4}{3} \\[1em] \Rightarrow \dfrac{6x}{5} \ge -\dfrac{4}{3} \\[1em] \Rightarrow x \ge -\dfrac{4}{3} \times \dfrac{5}{6} \\[1em] \Rightarrow x \ge -\dfrac{10}{9} \\[1em] \Rightarrow x \ge -1\dfrac{1}{9} .......(ii)$

From (i) and (ii) we get,

$-1\dfrac{1}{9} \le x \le 2\dfrac{1}{12}$

Since, x ∈ W

∴ Solution set = {0, 1, 2}.

Solution on the number line is :

Find three consecutive largest positive integers such that the sum of one-third of first, one-fourth of second and one-fifth of third is at most 20.

Answer

Let three consecutive positive integers be x, x + 1 and x + 2.

Given, sum of one-third of first, one-fourth of second and one-fifth of third is at most 20

$\therefore \dfrac{1}{3}x + \dfrac{1}{4}(x + 1) + \dfrac{1}{5}(x + 2) \le 20 \\[1em] \Rightarrow \dfrac{x}{3} + \dfrac{x + 1}{4} + \dfrac{x + 2}{5} \le 20 \\[1em] \Rightarrow \dfrac{20x + 15(x + 1) + 12(x + 2)}{60} \le 20 \\[1em] \Rightarrow \dfrac{20x + 15x + 15 + 12x + 24}{60} \le 20 \\[1em] \Rightarrow 47x + 39 \le 1200 \\[1em] \Rightarrow 47x \le 1161 \\[1em] \Rightarrow x \le \dfrac{1161}{47} \\[1em] \Rightarrow x \le 24.702$

∴ x = 24, x + 1 = 25, x + 2 = 26.

Hence, three consecutive numbers are 24, 25 and 26.

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

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

Answer

Given,

4x - 19 < $\dfrac{3x}{5} - 2 \le -\dfrac{2}{5} + x$

Solving L.H.S. of the equation,

$\Rightarrow 4x - 19 \lt \dfrac{3x}{5} - 2 \\[1em] \Rightarrow 4x - 19 \lt \dfrac{3x - 10}{5} \\[1em] \Rightarrow 5(4x - 19) \lt 3x - 10 \\[1em] \Rightarrow 20x - 95 \lt 3x - 10 \\[1em] \Rightarrow 20x - 3x \lt 95 - 10 \\[1em] \Rightarrow 17x \lt 85 \\[1em] \Rightarrow x \lt 5 ........(i)$

Solving R.H.S. of the equation,

$\Rightarrow \dfrac{3x}{5} - 2 \le -\dfrac{2}{5} + x \\[1em] \Rightarrow x - \dfrac{3x}{5} \ge -2 + \dfrac{2}{5} \\[1em] \Rightarrow \dfrac{5x - 3x}{5} \ge \dfrac{-10 + 2}{5} \\[1em] \Rightarrow \dfrac{2x}{5} \ge \dfrac{-8}{5} \\[1em] \Rightarrow x \ge -\dfrac{8}{5} \times \dfrac{5}{2} \\[1em] \Rightarrow x \ge -4 ........(ii)$

From (i) and (ii) we get,

-4 ≤ x < 5

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

Solution on the number line is :

Find the greatest value of x ∈ Z, so that :

-1 ≤ 3 + 4x < 23

Answer

Given,

-1 ≤ 3 + 4x < 23

Solving L.H.S. of the inequation,

⇒ -1 ≤ 3 + 4x

⇒ -1 - 3 ≤ 4x

⇒ -4 ≤ 4x

⇒ $\dfrac{-4}{4}$ ≤ x

⇒ -1 ≤ x

⇒ x ≥ -1 ....(1)

Solving R.H.S. of the inequation,

⇒ 3 + 4x < 23

⇒ 4x < 23 - 3

⇒ 4x < 20

⇒ x < $\dfrac{20}{4}$

⇒ x < 5 ....(2)

From (1) and (2) we get,

-1 ≤ x < 5

Since x ∈ Z,

x = {-1, 0, 1, 2, 3, 4}

Hence, greatest value of x is 4.

If 7 ≥ -2x + 1 > -7 ; find the sum of greatest and smallest values of x ∈ I.

Answer

Given,

7 ≥ -2x + 1 > -7

Solving L.H.S. of the inequation,

⇒ 7 ≥ -2x + 1

⇒ 7 - 1 ≥ -2x

⇒ 6 ≥ -2x

⇒ 2x ≥ -6

⇒ x ≥ $\dfrac{-6}{2}$

⇒ x ≥ -3 ....(1)

Solving R.H.S. of the inequation,

⇒ -2x + 1 > -7

⇒ -2x > -7 - 1

⇒ -2x > -8

⇒ 2x < 8

⇒ x < $\dfrac{8}{2}$

⇒ x < 4 ....(2)

From (1) and (2) we get,

-3 ≤ x < 4

Since x ∈ I,

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

Greatest value of x is 3 and smallest value of x is -3

The sum of greatest and smallest values of x = -3 + 3 = 0

Hence, sum of greatest and smallest values of x = 0.