(i) Let A = {x : 2x – 9 < 7, x∈R}
∴ 2x – 9 < 7 ⇒ 2x < 16 ⇒ x < 8
⇒ A = {x : x < 8, x∈R}
Let B = {x : 3x + 9 < 25, x∈R}
∴ 3x + 9 < 25 ⇒ 3x < 16 ⇒ x < \(\frac{16}{3}\)
⇒ B = {x : x < \(\frac{16}{3}\) , x∈R}
∴ Required solution set = A ∩ B
= {x : x < 8, x∈R} ∩ {x : x < \(\frac{16}{3}\), x∈R}
= {x : x < \(\frac{16}{3}\), x∈R} = \(x\) ∈ \(\big(-∞,\frac{16}{3}\big)\)
This solution set can be shown on a graph as:
(ii) Let A = {x : 3x – 2 > 19, x∈R}
Then, 3x – 2 > 19 ⇒ 3x > 21 ⇒ x > 7 ⇒ A = {x : x > 7, x∈R}
Let B = {x : 3 – 2x > – 7, x∈R}
Then, 3 – 2x > – 7 ⇒ 10 > 2x ⇒ 2x < 10 ⇒ x < 5 ⇒ B = {x : x < 5, x∈R}
Required solution set = A or B = A ∪ B
= {x : x > 7, x∈R} ∪ {x : x < 5, x∈R}
= x > 7 or x < 5 = x∈ (7, ∞) or \(x\)∈ (–∞, 5]
This can be shown on a graph as: