(i) 5x < 20 ⇒ x ……… 4

As, 5x < 20

Then,

Dividing both the sides by 5

x < 4

Therefore,

5x < 20 ⇒ x < 4

(ii) -3x > 9 ⇒ x ……… -3

As, -3x > 9

Then, Dividing both the sides by 3

x > -3

Therefore,

-3x > 9 ⇒ x > -3

(iii) 4x > -16 ⇒ x ……… -4

As, 4x > -16

Then, Dividing both the sides by 4

x > -4

Therefore,

4x > -16 ⇒ x > -4

(iv) -6x ≤ -18 ⇒ x ……… 3

As -6x ≤ -18

Then, Dividing both the sides by 6

-x ≤ -3

Now multiplying by -1 on both sides

-x(-1) ≤ -3(-1)

x ≥ 3 (inequality sign reversed)

Therefore,

-6x ≤ -18 ⇒ x ≥ 3

(v) x > -3 ⇒ -2x ……… 6

As, x > -3

Multiplying both sides by 2

Then,

2x > -6

Now multiplying both the sides by -1

2x(-1) < 6(-1)

-2x > 6

Therefore,

x > -3 ⇒ -2x > 6

(vi) a < b and c > 0 ⇒ ………

As,

a < b …(1)

c > 0

Dividing both sides by c in equation (1)

Then,

Therefore,

a < b and c > 0 ⇒

(vii) p – q = -3 ⇒ p ……… q

As,

p – q = -3

p = q - 3

From the above equation it is clear that p would always be less than q

Therefore,

p – q = -3 ⇒ p < q

(viii) u – v = 2 ⇒ u ……… v

As,

u – v = 2

u = v + 2

From the above equation it is clear that u would always be greater than v

Therefore,

u – v = 2 ⇒ u > v