# Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.2

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.2

Question 1.
Solve for x.

(i) |3 – x| < 7
Solution:
– 7 < 3 – x < 7
– 7 – 3 < – x < 7 – 3
– 10 < – x < 4 10 > x > – 4
– 4 < x < 10
∴ The solution set is x ∈ (-4, 10)

(ii) |4x – 5| ≥ -2
Solution: (iii) Solution: (iv) |x| – 10 < -3
Solution:
|x| < -3 + 10 (= 7)
|x| < 7 ⇒ -7 < x < 7 Question 2.
Solve $$\frac{1}{|2 x-1|}<6$$ and express the solution using the interval notation.
Solution: Question 3.
Solve -3|x| + 5 ≤ – 2 and graph the solution set in a number line.
Solution:
-3 |x| + 5 ≤ – 2
-3 |x| ≤ – 2 – 5
-3 |x| ≤ – 7
3 |x| ≥ 7
|x| ≥ $$\frac{7}{3}$$  Question 4.
Solve 2|x + 1| – 6 ≤ 7 and graph the solution set in a number line.
Solution: Question 5. Solution:  Question 6.
Solve |5x – 12| < -2
Solution: ### Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.2 Additional Questions

Question 1. Solution: Question 2. Solution: ⇒ x – 2 < – 1 (or) x – 2 > 1 and – 2 < x – 2 < 2
⇒ x < 1 (or) x > 3 and -2 + 2 < x < 2 + 2
⇒ x < 1 (or) x > 3 and 0 < x < 4 Hence, the required solution is (0, 1) ∪ (3, 4) Question 3. Solution: Question 4.
Solve: |x – 1| ≤ 5, |x| ≥ 2
Solution:
|x – 1| ≤ 5 and |x| ≥ 2
⇒ -5 ≤ x – 1 ≤ 5 and x ≤ -2 (or) x > 2
⇒ – 5 + 1 ≤ x ≤ 5 + 1
⇒ -4 ≤ x ≤ 6 and x ≤ -2 (or) x ≥ 2
Hence x < [-4, -2] ∪ [2, 6] Question 5. Solution: 