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 тАУ 3 > 0 тАУ 3
x > тАУ3
Now, let us consider the second inequality.
2x < 14
Divide both the sides by 2 we get,
2x/2 < 14/2
x < 7
тИ┤ The solution of the given system of inequation is (тАУ3, 7).
Question - 2 : - 2x тАУ 7 > 5 тАУ x, 11 тАУ 5x тЙд 1
Answer - 2 : -
Given:
2x тАУ 7 > 5 тАУ x and 11 тАУ 5x тЙд 1
Let us consider the first inequality.
2x тАУ 7 > 5 тАУ x
2x тАУ 7 + 7 > 5 тАУ x + 7
2x > 12 тАУ x
2x + x > 12 тАУ x + x
3x > 12
Divide both the sides by 3 we get,
3x/3 > 12/3
x > 4
тИ┤ x тИИ ( 4, тИЮ) тАж (1)
Now, let us consider the second inequality.
11 тАУ 5x тЙд 1
11 тАУ 5x тАУ 11 тЙд 1 тАУ 11
тАУ5x тЙд тАУ10
Divide both the sides by 5 we get,
-5x/5 тЙд -10/5
тАУx тЙд тАУ2
x тЙе 2
тИ┤ x тИИ [2, тИЮ) тАж (2)
From (1) and (2) we get
x тИИ (4, тИЮ) тИй [2, тИЮ)
x тИИ (4, тИЮ)
тИ┤ The solution of the given system of inequations is (4, тИЮ).
Question - 3 : - x тАУ 2 > 0, 3x < 18
Answer - 3 : -
Given:
x тАУ 2 > 0 and 3x < 18
Let us consider the first inequality.
x тАУ 2 > 0
x тАУ 2 + 2 > 0 + 2
x > 2
тИ┤ x тИИ ( 2, тИЮ) тАж(1)
Now, let us consider the second inequality.
3x < 18
Divide both the sides by 3 we get,
3x/3 < 18/3
x < 6
тИ┤ x тИИ (тАУтИЮ, 6) тАж(2)
From (1) and (2), we get
x тИИ (2, тИЮ) тИй (тАУтИЮ, 6)
x тИИ ( 2, 6)
тИ┤ The solution of the given system of inequations is (2, 6).
Question - 4 : - 2x + 6 тЙе 0, 4x тАУ 7 < 0
Answer - 4 : -
Given:
2x + 6 тЙе 0 and 4x тАУ 7 < 0
Let us consider the first inequality.
2x + 6 тЙе 0
2x + 6 тАУ 6 тЙе 0 тАУ 6
2x тЙе тАУ6
Divide both the sides by 2 we get,
2x/2 тЙе -6/2
x тЙе -3
тИ┤ x тИИ [тАУ3, тИЮ) тАж(1)
Now, let us consider the second inequality.
4x тАУ 7 < 0
4x тАУ 7 + 7 < 0 + 7
4x < 7
Divide both the sides by 4 we get,
4x/4 < 7/4
x < 7/4
тИ┤ x тИИ [тАУтИЮ, 7/4) тАж(2)
From (1) and (2), we get
x тИИ [-3, тИЮ) тИй (тАУтИЮ, 7/4)
x тИИ [-3, 7/4)
тИ┤ The solution of the given system of inequations is [-3, 7/4).
Question - 5 : - 3x тАУ 6 > 0, 2x тАУ 5 > 0
Answer - 5 : -
Given:
3x тАУ 6 > 0 and 2x тАУ 5 > 0
Let us consider the first inequality.
3x тАУ 6 > 0
3x тАУ 6 + 6 > 0 + 6
3x > 6
Divide both the sides by 3 we get,
3x/3 > 6/3
x > 2
тИ┤ x тИИ ( 2, тИЮ)тАж (1)
Now, let us consider the second inequality.
2x тАУ 5 > 0
2x тАУ 5 + 5 > 0 + 5
2x > 5
Divide both the sides by 2 we get,
2x/2 > 5/2
x > 5/2
тИ┤ x тИИ (5/2, тИЮ)тАж (2)
From (1) and (2), we get
x тИИ (2, тИЮ) тИй (5/2, тИЮ)
x тИИ (5/2, тИЮ)
тИ┤ The solution of the given system of inequations is (5/2, тИЮ).
Question - 6 : - 2x тАУ 3 < 7, 2x > тАУ4
Answer - 6 : -
Given:
2x тАУ 3 < 7 and 2x > тАУ4
Let us consider the first inequality.
2x тАУ 3 < 7
2x тАУ 3 + 3 < 7 + 3
2x < 10
Divide both the sides by 2 we get,
2x/2 < 10/2
x < 5
тИ┤ x тИИ ( тАУтИЮ, 5)тАж (1)
Now, let us consider the second inequality.
2x > тАУ4
Divide both the sides by 2 we get,
2x/2 > -4/2
x > тАУ2
тИ┤ x тИИ (тАУ2, тИЮ)тАж (2)
From (1) and (2), we get
x тИИ (тАУтИЮ, 5) тИй (тАУ2, тИЮ)
x тИИ (тАУ2, 5)
тИ┤ The solution of the given system of inequations is (тАУ2, 5).
Question - 7 : - 2x + 5 тЙд 0, x тАУ 3 тЙд 0
Answer - 7 : -
Given:
2x + 5 тЙд 0 and x тАУ 3 тЙд 0
Let us consider the first inequality.
2x + 5 тЙд 0
2x + 5 тАУ 5 тЙд 0 тАУ 5
2x тЙд тАУ5
Divide both the sides by 2 we get,
2x/2 тЙд тАУ5/2
x тЙд тАУ 5/2
тИ┤ x тИИ (тАУтИЮ, -5/2]тАж (1)
Now, let us consider the second inequality.
x тАУ 3 тЙд 0
x тАУ 3 + 3 тЙд 0 + 3
x тЙд 3
тИ┤ x тИИ (тАУтИЮ, 3]тАж (2)
From (1) and (2), we get
x тИИ (тАУтИЮ, -5/2] тИй (тАУтИЮ, 3)
x тИИ (тАУтИЮ, -5/2]
тИ┤ The solution of the given system of inequations is (тАУтИЮ, -5/2].
Question - 8 : - 5x тАУ 1 < 24, 5x + 1 > тАУ24
Answer - 8 : -
Given:
5x тАУ 1 < 24 and 5x + 1 > тАУ24
Let us consider the first inequality.
5x тАУ 1 < 24
5x тАУ 1 + 1 < 24 + 1
5x < 25
Divide both the sides by 5 we get,
5x/5 < 25/5
x < 5
тИ┤ x тИИ (тАУтИЮ, 5)тАж (1)
Now, let us consider the second inequality.
5x + 1 > тАУ24
5x + 1 тАУ 1 > тАУ24 тАУ 1
5x > тАУ25
Divide both the sides by 5 we get,
5x/5 > -25/5
x > -5
тИ┤ x тИИ (тАУ5, тИЮ)тАж (2)
From (1) and (2), we get
x тИИ (тАУтИЮ, 5) тИй (тАУ5, тИЮ)
x тИИ (тАУ5, 5)
тИ┤ The solution of the given system of inequations is (тАУ5, 5).
Question - 9 : - 3x тАУ 1 тЙе 5, x + 2 > -1
Answer - 9 : -
Given:
3x тАУ 1 тЙе 5 and x + 2 > тАУ1
Let us consider the first inequality.
3x тАУ 1 тЙе 5
3x тАУ 1 + 1 тЙе 5 + 1
3x тЙе 6
Divide both the sides by 3 we get,
3x/3 тЙе 6/3
x тЙе 2
тИ┤ x тИИ [2, тИЮ)тАж (1)
Now, let us consider the second inequality.
x + 2 > тАУ1
x + 2 тАУ 2 > тАУ1 тАУ 2
x > тАУ3
тИ┤ x тИИ (тАУ3, тИЮ)тАж (2)
From (1) and (2), we get
x тИИ [2, тИЮ) тИй (тАУ3, тИЮ)
x тИИ [2, тИЮ)
тИ┤ The solution of the given system of inequations is [2, тИЮ).
Question - 10 : - 11 тАУ 5x > тАУ4, 4x + 13 тЙд тАУ11
Answer - 10 : -
Given:
11 тАУ 5x > тАУ4 and 4x + 13 тЙд тАУ11
Let us consider the first inequality.
11 тАУ 5x > тАУ4
11 тАУ 5x тАУ 11 > тАУ4 тАУ 11
тАУ5x > тАУ15
Divide both the sides by 5 we get,
-5x/5 > -15/5
тАУx > тАУ3
x < 3
тИ┤ x тИИ (тАУтИЮ, 3) (1)
Now, let us consider the second inequality.
4x + 13 тЙд тАУ11
4x + 13 тАУ 13 тЙд тАУ11 тАУ 13
4x тЙд тАУ24
Divide both the sides by 4 we get,
4x/4 тЙд тАУ24/4
x тЙд тАУ6
тИ┤ x тИИ (тАУтИЮ, тАУ6] (2)
From (1) and (2), we get
x тИИ (тАУтИЮ, 3) тИй (тАУтИЮ, тАУ6]
x тИИ (тАУтИЮ, тАУ6]
тИ┤ The solution of the given system of inequations is (тАУтИЮ, тАУ6].