Linear Inequations — Exercise 4(A)

If x ∈ W, then solution set of inequation -x > -7, is :

1. {8, 9, 10, .....}

2. {0, 1, 2, 3, 4, 5, 6}

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

4. {-8, -9, -10, ....}

Answer

Given,

-x > -7

x < 7

Since x ∈ W and x < 7:

The solution set is: {0, 1, 2, 3, 4, 5, 6}

Hence, Option 2 is the correct option.

The value of x, for 4(2x - 5) < 2x + 28, x ∈ R, is :

1. x > 8

2. x < 8

3. x > -8

4. x < -8

Answer

Given,

⇒ 4(2x - 5) < 2x + 28

⇒ 8x - 20 < 2x + 28

⇒ 8x - 2x < 28 + 20

⇒ 6x < 48

⇒ x < $\dfrac{48}{6}$

⇒ x < 8.

Hence, Option 2 is the correct option.

The solution set for the inequation -2x + 7 ≤ 3, x ∈ R is :

1. {x : x ∈ R, x < 2}

2. {x : x ∈ R, x > 2}

3. {x : x ∈ R, x ≤ 2}

4. {x : x ∈ R, x ≥ 2}

Answer

Given,

⇒ -2x + 7 ≤ 3

⇒ 2x ≥ 7 - 3

⇒ 2x ≥ 4

⇒ x ≥ $\dfrac{4}{2}$

⇒ x ≥ 2.

Since, x ∈ R.

Solution set = {x : x ∈ R, x ≥ 2}

Hence, Option 4 is the correct option.

For 7 - 3x < x - 5, the solution set is :

1. x > 3

2. x < 3

3. x ≥ 3

4. x ≤ 3

Answer

Given,

⇒ 7 - 3x < x - 5

⇒ x + 3x > 7 + 5

⇒ 4x > 12

⇒ x > $\dfrac{12}{4}$

⇒ x > 3.

Hence, Option 1 is the correct option.

x(8 - x) > 0 and x ∈ N gives :

1. 0 ≤ x < 8

2. 1 < x ≤ 8

3. 0 < x < 8

4. 0 ≤ x ≤ 8

Answer

For x(8 - x) > 0

Either,

⇒ x > 0 and (8 - x) > 0

⇒ x > 0 and x < 8 ...........(1)

or,

⇒ x < 0 and (8 - x) < 0

⇒ x < 0 and x > 8 .............(2)

But both conditions of equation (2) are not possible simultaneously.

From equation (1),

Solution set = {0 < x < 8}.

Hence, Option 3 is the correct option.

State, true or false :

(i) x < -y ⇒ -x > y

(ii) -5x ≥ 15 ⇒ x ≥ -3

(iii) 2x ≤ -7 ⇒ $\dfrac{2x}{-4} \ge \dfrac{-7}{-4}$

(iv) 7 > 5 ⇒ $\dfrac{1}{7} \lt \dfrac{1}{5}$

Answer

(i) Given,

x < -y

∴ -x > y [Using rule 5]

Hence, the statement is True.

(ii) Given,

-5x ≥ 15

Dividing both sides of the above inequation by -5,

⇒ x ≤ -3 [Using rule 4]

Hence, the statement is False.

(iii) Given,

2x ≤ -7

Dividing both sides of the above inequation by -4,

⇒ $\dfrac{2x}{-4} \ge \dfrac{-7}{-4}$ [Using rule 4]

Hence, the statement is True.

(iv) Given,

7 > 5

Taking reciprocals,

$\dfrac{1}{7} \lt \dfrac{1}{5}$ [Using rule 6]

Hence, the statement is True.

State, whether the following statements are true or false.

(i) If a < b, then a - c < b - c

(ii) If a > b, then a + c > b + c

(iii) If a < b, then ac > bc

(iv) If a > b, then $\dfrac{a}{c} \lt \dfrac{b}{c}$

(v) If a - c > b - d; then a + d > b + c

(vi) If a < b, and c > 0, then a - c > b - c

where a, b, c, and d are real numbers c ≠ 0.

Answer

(i) Given,

a < b

Subtracting both sides by c,

a - c < b - c.

Hence, the statement is True.

(ii) Given,

a > b

Adding both sides by c,

a + c > b + c.

Hence, the statement is True.

(iii) Given,

a < b

If c is a positive number,

Multiplying both sides by c we get,

ac < bc

If c is a negative number,

Multiplying both sides by c we get,

ac > bc [Using rule 4]

Hence, the statement is False.

(iv) Given,

a > b

If c is a positive number,

Dividing both sides by c we get,

$\dfrac{a}{c} \gt \dfrac{b}{c}$

If c is a negative number,

Dividing both sides by c we get,

$\dfrac{a}{c} \lt \dfrac{b}{c}$ [Using rule 4]

Hence, the statement is False.

(v) Given,

a - c > b - d

Adding both sides by (c + d) we get,

⇒ a - c + (c + d) > b - d + (c + d)

⇒ a - c + c + d > b + c - d + d

⇒ a + d > b + c

Hence, the statement is True.

(vi) Given,

a < b and c > 0

Subtracting both sides by c we get,

a - c < b - c [As c is a positive number.]

Hence, the statement is False.

Solve the inequation :

3 - 2x ≥ x - 12 given that x ∈ N.

Answer

Given,

⇒ 3 - 2x ≥ x - 12

⇒ x + 2x ≤ 3 + 12

⇒ 3x ≤ 15

Dividing both sides by 3 we get,

⇒ x ≤ 5

Since, x ∈ N

∴ Solution set = {1, 2, 3, 4, 5}.

If 25 - 4x ≤ 16, find :

(i) the smallest value of x when x is a real number,

(ii) the smallest value of x when x is an integer.

Answer

Given,

⇒ 25 - 4x ≤ 16

⇒ -4x ≤ 16 - 25

⇒ -4x ≤ -9

Multiplying both sides by -1 we get,

⇒ 4x ≥ 9 (As on multiplying by negative no. the sign reverses.)

Dividing both sides by 4 we get,

⇒ x ≥ $\dfrac{9}{4}$

⇒ x ≥ 2.25

(i) Given,

x ≥ 2.25

Hence, smallest value of x when x is a real number is 2.25

(ii) Given,

x ≥ 2.25

Hence, smallest value of x when x is an integer is 3.

If the replacement set is the set of real numbers, solve :

(i) -4x ≥ -16

(ii) 8 - 3x ≤ 20

Answer

(i) -4x ≥ -16

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

∴ Solution set = {x : x ∈ R and x ≤ 4}

(ii) 8 - 3x ≤ 20

⇒ -3x ≤ 20 - 8

⇒ -3x ≤ 12

Dividing both sides by -3 we get.

⇒ x ≥ -4 (As on dividing by negative no. the sign reverses.)

∴ Solution set = {x : x ∈ R and x ≥ -4}.

Find the smallest value of x for which 5 - 2x < $5\dfrac{1}{2} - \dfrac{5}{3}x$, where x is an integer.

Answer

Given,

$\Rightarrow 5 - 2x \lt 5\dfrac{1}{2} - \dfrac{5}{3}x \\[1em] \Rightarrow 5 - 2x \lt \dfrac{11}{2} - \dfrac{5}{3}x \\[1em] \Rightarrow 5 - \dfrac{11}{2} \lt -\dfrac{5}{3}x + 2x \\[1em] \Rightarrow \dfrac{10 - 11}{2} \lt \dfrac{-5x + 6x}{3} \\[1em] \Rightarrow -\dfrac{1}{2} \lt \dfrac{x}{3} \\[1em] \Rightarrow \dfrac{x}{3} \times 3 \gt -\dfrac{1}{2} \times 3 \\[1em] \Rightarrow x \gt -\dfrac{3}{2} \\[1em] \Rightarrow x \gt -1.5$

Since, x is an integer.

Hence, smallest value of x = -1.

Find the largest value of x for which

2(x - 1) ≤ 9 - x and x ∈ W.

Answer

Given,

⇒ 2(x - 1) ≤ 9 - x

⇒ 2x - 2 ≤ 9 - x

⇒ 2x + x ≤ 9 + 2

⇒ 3x ≤ 11

⇒ x ≤ $\dfrac{11}{3}$

⇒ x ≤ 3.67

Since, x ∈ W

Hence, largest value of x for which 2(x - 1) ≤ 9 - x is 3.