RD Chapter 15 Linear Inequations Ex 15.2 Solutions
Question - 1 : - Solve each of the following system of equations in R.x + 3 > 0, 2x < 14
Answer - 1 : -
Given: x + 3 > 0 and 2x < 14
Let us consider the first inequality.
x + 3 > 0
x + 3 – 3 > 0 – 3
x > –3
Now, let us consider the second inequality.
2x < 14
Divide both the sides by 2 we get,
2x/2 < 14/2
x < 7
∴ The solution of the given system of inequation is (–3, 7).
Question - 2 : - 2x – 7 > 5 – x, 11 – 5x ≤ 1
Answer - 2 : -
Given:
2x – 7 > 5 – x and 11 – 5x ≤ 1
Let us consider the first inequality.
2x – 7 > 5 – x
2x – 7 + 7 > 5 – x + 7
2x > 12 – x
2x + x > 12 – x + x
3x > 12
Divide both the sides by 3 we get,
3x/3 > 12/3
x > 4
∴ x ∈ ( 4, ∞) … (1)
Now, let us consider the second inequality.
11 – 5x ≤ 1
11 – 5x – 11 ≤ 1 – 11
–5x ≤ –10
Divide both the sides by 5 we get,
-5x/5 ≤ -10/5
–x ≤ –2
x ≥ 2
∴ x ∈ [2, ∞) … (2)
From (1) and (2) we get
x ∈ (4, ∞) ∩ [2, ∞)
x ∈ (4, ∞)
∴ The solution of the given system of inequations is (4, ∞).
Question - 3 : - x – 2 > 0, 3x < 18
Answer - 3 : -
Given:
x – 2 > 0 and 3x < 18
Let us consider the first inequality.
x – 2 > 0
x – 2 + 2 > 0 + 2
x > 2
∴ x ∈ ( 2, ∞) …(1)
Now, let us consider the second inequality.
3x < 18
Divide both the sides by 3 we get,
3x/3 < 18/3
x < 6
∴ x ∈ (–∞, 6) …(2)
From (1) and (2), we get
x ∈ (2, ∞) ∩ (–∞, 6)
x ∈ ( 2, 6)
∴ The solution of the given system of inequations is (2, 6).
Question - 4 : - 2x + 6 ≥ 0, 4x – 7 < 0
Answer - 4 : -
Given:
2x + 6 ≥ 0 and 4x – 7 < 0
Let us consider the first inequality.
2x + 6 ≥ 0
2x + 6 – 6 ≥ 0 – 6
2x ≥ –6
Divide both the sides by 2 we get,
2x/2 ≥ -6/2
x ≥ -3
∴ x ∈ [–3, ∞) …(1)
Now, let us consider the second inequality.
4x – 7 < 0
4x – 7 + 7 < 0 + 7
4x < 7
Divide both the sides by 4 we get,
4x/4 < 7/4
x < 7/4
∴ x ∈ [–∞, 7/4) …(2)
From (1) and (2), we get
x ∈ [-3, ∞) ∩ (–∞, 7/4)
x ∈ [-3, 7/4)
∴ The solution of the given system of inequations is [-3, 7/4).
Question - 5 : - 3x – 6 > 0, 2x – 5 > 0
Answer - 5 : -
Given:
3x – 6 > 0 and 2x – 5 > 0
Let us consider the first inequality.
3x – 6 > 0
3x – 6 + 6 > 0 + 6
3x > 6
Divide both the sides by 3 we get,
3x/3 > 6/3
x > 2
∴ x ∈ ( 2, ∞)… (1)
Now, let us consider the second inequality.
2x – 5 > 0
2x – 5 + 5 > 0 + 5
2x > 5
Divide both the sides by 2 we get,
2x/2 > 5/2
x > 5/2
∴ x ∈ (5/2, ∞)… (2)
From (1) and (2), we get
x ∈ (2, ∞) ∩ (5/2, ∞)
x ∈ (5/2, ∞)
∴ The solution of the given system of inequations is (5/2, ∞).
Question - 6 : - 2x – 3 < 7, 2x > –4
Answer - 6 : -
Given:
2x – 3 < 7 and 2x > –4
Let us consider the first inequality.
2x – 3 < 7
2x – 3 + 3 < 7 + 3
2x < 10
Divide both the sides by 2 we get,
2x/2 < 10/2
x < 5
∴ x ∈ ( –∞, 5)… (1)
Now, let us consider the second inequality.
2x > –4
Divide both the sides by 2 we get,
2x/2 > -4/2
x > –2
∴ x ∈ (–2, ∞)… (2)
From (1) and (2), we get
x ∈ (–∞, 5) ∩ (–2, ∞)
x ∈ (–2, 5)
∴ The solution of the given system of inequations is (–2, 5).
Question - 7 : - 2x + 5 ≤ 0, x – 3 ≤ 0
Answer - 7 : -
Given:
2x + 5 ≤ 0 and x – 3 ≤ 0
Let us consider the first inequality.
2x + 5 ≤ 0
2x + 5 – 5 ≤ 0 – 5
2x ≤ –5
Divide both the sides by 2 we get,
2x/2 ≤ –5/2
x ≤ – 5/2
∴ x ∈ (–∞, -5/2]… (1)
Now, let us consider the second inequality.
x – 3 ≤ 0
x – 3 + 3 ≤ 0 + 3
x ≤ 3
∴ x ∈ (–∞, 3]… (2)
From (1) and (2), we get
x ∈ (–∞, -5/2] ∩ (–∞, 3)
x ∈ (–∞, -5/2]
∴ The solution of the given system of inequations is (–∞, -5/2].
Question - 8 : - 5x – 1 < 24, 5x + 1 > –24
Answer - 8 : -
Given:
5x – 1 < 24 and 5x + 1 > –24
Let us consider the first inequality.
5x – 1 < 24
5x – 1 + 1 < 24 + 1
5x < 25
Divide both the sides by 5 we get,
5x/5 < 25/5
x < 5
∴ x ∈ (–∞, 5)… (1)
Now, let us consider the second inequality.
5x + 1 > –24
5x + 1 – 1 > –24 – 1
5x > –25
Divide both the sides by 5 we get,
5x/5 > -25/5
x > -5
∴ x ∈ (–5, ∞)… (2)
From (1) and (2), we get
x ∈ (–∞, 5) ∩ (–5, ∞)
x ∈ (–5, 5)
∴ The solution of the given system of inequations is (–5, 5).
Question - 9 : - 3x – 1 ≥ 5, x + 2 > -1
Answer - 9 : -
Given:
3x – 1 ≥ 5 and x + 2 > –1
Let us consider the first inequality.
3x – 1 ≥ 5
3x – 1 + 1 ≥ 5 + 1
3x ≥ 6
Divide both the sides by 3 we get,
3x/3 ≥ 6/3
x ≥ 2
∴ x ∈ [2, ∞)… (1)
Now, let us consider the second inequality.
x + 2 > –1
x + 2 – 2 > –1 – 2
x > –3
∴ x ∈ (–3, ∞)… (2)
From (1) and (2), we get
x ∈ [2, ∞) ∩ (–3, ∞)
x ∈ [2, ∞)
∴ The solution of the given system of inequations is [2, ∞).
Question - 10 : - 11 – 5x > –4, 4x + 13 ≤ –11
Answer - 10 : -
Given:
11 – 5x > –4 and 4x + 13 ≤ –11
Let us consider the first inequality.
11 – 5x > –4
11 – 5x – 11 > –4 – 11
–5x > –15
Divide both the sides by 5 we get,
-5x/5 > -15/5
–x > –3
x < 3
∴ x ∈ (–∞, 3) (1)
Now, let us consider the second inequality.
4x + 13 ≤ –11
4x + 13 – 13 ≤ –11 – 13
4x ≤ –24
Divide both the sides by 4 we get,
4x/4 ≤ –24/4
x ≤ –6
∴ x ∈ (–∞, –6] (2)
From (1) and (2), we get
x ∈ (–∞, 3) ∩ (–∞, –6]
x ∈ (–∞, –6]
∴ The solution of the given system of inequations is (–∞, –6].