Chapter 6: Linear Inequalities Miscellaneous Exercise

According to the question,

The inequality given is

2 ≤ 3x – 4 ≤ 5

⇒ 2 ≤ 3x – 4 ≤ 5

⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4

⇒ 6 ≤ 3x ≤ 9

⇒ 6/3 ≤ 3x/3 ≤ 9/3

⇒ 2 ≤ x ≤ 3

Hence, all real numbers x greater than or equal to 2 but less than or equal to 3 are solutions of given equality.

x ∈ [2, 3]

According to the question,

The inequality given is

6 ≤ –3 (2x – 4) < 12

⇒ 6 ≤ -3 (2x – 4) < 12

Dividing the inequality by 3, we get

⇒ 2 ≤ – (2x – 4) < 4

Multiplying the inequality by -1,

⇒ -2 ≥ 2x – 4 > -4 [multiplying the inequality with -1 changes the inequality sign.]

⇒ -2 + 4 ≥ 2x – 4 + 4 > -4 + 4

⇒ 2 ≥ 2x > 0

Dividing the inequality by 2,

⇒ 0 < x ≤ 1

Hence, all real numbers x greater than 0 but less than or equal to 1 are solutions of given equality.

x ∈ (0, 1]

According to the question,

The inequality given is

– 3 ≤ 4 – 7x/2 ≤ 18

⇒ – 3 – 4 ≤ 4 – 7x/2 – 4 ≤ 18 – 4

⇒ – 7 ≤ – 7x/2 ≤ 18 – 14

Multiplying the inequality by -2,

⇒ 14 ≥ 7x ≥ -28

⇒ -28 ≤ 7x ≤ 14

Dividing the inequality by 7,

⇒ -4 ≤ x ≤ 2

Hence, all real numbers x greater than or equal to -4 but less than or equal to 2 are solutions of given equality.

x ∈ [-4, 2]

According to the question,

The inequality given is

– 15 ≤ 3(x – 2)/5 ≤ 0

⇒ – 15 < 3(x – 2)/5 ≤ 0

Multiplying the inequality by 5,

⇒ -75 < 3(x – 2) ≤ 0

Dividing the inequality by 3, we get

⇒ -25 < x – 2 ≤ 0

⇒ – 25 + 2 < x – 2 + 2 ≤ 0 + 2

⇒ – 23 < x ≤ 2

Hence, all real numbers x greater than -23 but less than or equal to 2 are solutions of given equality.

x ∈ (-23, 2]

According to the question,

The inequality given is

Hence, all real numbers x greater than -80/3 but less than or equal to -10/3 are solutions of given equality.

x ∈ (-80/3, -10/3]

According to the question,

The inequality given is

⇒ 14 ≤ 3x + 11 ≤ 22

⇒ 14 – 11 ≤ 3x + 11 – 11 ≤ 22 – 11

⇒ 3 ≤ 3x ≤ 11

⇒ 1 ≤ x ≤ 11/3

Hence, all real numbers x greater than or equal to -4 but less than or equal to 2 are solutions of given equality.

x ∈ [1, 11/3]

Solve the inequalities in Exercises 7 to 11 and represent the solution graphically on the number line.

According to the question,

The inequalities given are

5x + 1 > -24 and 5x – 1 < 24

5x + 1 > -24

⇒ 5x > -24 – 1

⇒ 5x > -25

⇒ x > -5 ……… (i)

5x – 1 < 24

⇒ 5x < 24 + 1

⇒ 5x < 25

⇒ x < 5 ……….(ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (-5, 5).

According to the question,

The inequalities given are

2 (x – 1) < x + 5 and 3 (x + 2) > 2 – x

2 (x – 1) < x + 5

⇒ 2x – 2 < x + 5

⇒ 2x – x < 5 + 2

⇒ x < 7 ……… (i)

3 (x + 2) > 2 – x

⇒ 3x + 6 > 2 – x

⇒ 3x + x > 2 – 6

⇒ 4x > -4

⇒ x > -1 ………. (ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (-1, 7).

According to the question,

The inequalities given are

3x – 7 > 2(x – 6) and 6 – x > 11 – 2x

3x – 7 > 2(x – 6)

⇒ 3x – 7 > 2x – 12

⇒ 3x – 2x > 7 – 12

⇒ x > -5 ………… (i)

6 – x > 11 – 2x

⇒ 2x – x > 11 – 6

⇒ x > 5 ……….(ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (5, ∞).

According to the question,

The inequalities given are

5(2x – 7) – 3(2x + 3) ≤ 0 and 2x + 19 ≤ 6x + 47

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ 4x ≤ 44

⇒ x ≤ 11 ……(i)

2x + 19 ≤ 6x +47

⇒ 6x – 2x ≥ 19 – 47

⇒ 4x ≥ -28

⇒ x ≥ -7 ……….(ii)

From equations (i) and (ii),

We can infer that the solution of given inequalities is (-7, 11).