Linear Inequations

Given 12x < 50

i. x ∈ R

When x is a real number, the solution of the given inequation is.

ii. x ∈ Z

As, when x is an integer, the maximum possible value of x is 4.

Thus, the solution of the given inequation is {…, –2, –1, 0, 1, 2, 3, 4}.

iii. x ∈ N

As, when x is a natural number, the maximum possible value of x is 4 and we know the natural numbers start from 1.

Thus, the solution of the given inequation is {1, 2, 3, 4}.

Given –4x > 30

i. x ∈ R

When x is a real number, the solution of the given inequation is.

ii. x ∈ Z

As, when x is an integer, the maximum possible value of x is –8.

Thus, the solution of the given inequation is {…, –11, –10, –9, –8}.

iii. x ∈ N

As natural numbers start from 1 and can never be negative, when x is a natural number, the solution of the given inequation is ∅.

Given 4x – 2 < 8

⇒ 4x – 2 + 2 < 8 + 2

⇒ 4x < 10

i. x ∈ R

When x is a real number, the solution of the given inequation is.

ii. x ∈ Z

As, when x is an integer, the maximum possible value of x is 2.

Thus, the solution of the given inequation is {…, –2, –1, 0, 1, 2}.

iii. x ∈ N

As, when x is a natural number, the maximum possible value of x is 2 and we know the natural numbers start from 1.

Thus, the solution of the given inequation is {1, 2}.

Given 3x – 7 > x + 1

⇒ 3x – 7 + 7 > x + 1 + 7

⇒ 3x > x + 8

⇒ 3x – x > x + 8 – x

⇒ 2x > 8

∴ x > 4

Thus, the solution of the given inequation is (4, ∞).

Given x + 5 > 4x – 10

⇒ x + 5 – 5 > 4x – 10 – 5

⇒ x > 4x – 15

⇒ 4x – 15 < x

⇒ 4x – 15 – x < x – x

⇒ 3x – 15 < 0

⇒ 3x – 15 + 15 < 0 + 15

⇒ 3x < 15

∴ x < 5

Thus, the solution of the given inequation is (–∞, 5).

Given 3x + 9 ≥ –x + 19

⇒ 3x + 9 – 9 ≥ –x + 19 – 9

⇒ 3x ≥ –x + 10

⇒ 3x + x ≥ –x + 10 + x

⇒ 4x ≥ 10

Thus, the solution of the given inequation is.

Given

⇒ 30 – 10x ≥ x + 20

⇒ x + 20 ≤ 30 – 10x

⇒ x + 20 – 20 ≤ 30 – 10x – 20

⇒ x ≤ 10 – 10x

⇒ x + 10x ≤ 10 – 10x + 10x

⇒ 11x ≤ 10

Thus, the solution of the given inequation is.

Given

⇒ 6x – 4 ≤ 20x – 15

⇒ 20x – 15 ≥ 6x – 4

⇒ 20x – 15 + 15 ≥ 6x – 4 + 15

⇒ 20x ≥ 6x + 11

⇒ 20x – 6x ≥ 6x + 11 – 6x

⇒ 14x ≥ 11

Thus, the solution of the given inequation is.

Given –(x – 3) + 4 < 5 – 2x

⇒ –x + 3 + 4 < 5 – 2x

⇒ –x + 7 < 5 – 2x

⇒ –x + 7 – 7 < 5 – 2x – 7

⇒ –x < –2x – 2

⇒ –x + 2x < –2x – 2 + 2x

∴ x < –2

Thus, the solution of the given inequation is (–∞, –2).

Given

⇒ 4x < 2 – 5x

⇒ 4x + 5x < 2 – 5x + 5x

⇒ 9x < 2

Thus, the solution of the given inequation is.

Given

⇒ 14x – 14 ≤ 30 + 15x

⇒ 14x – 14 + 14 ≤ 30 + 15x + 14

⇒ 14x ≤ 44 + 15x

⇒ 14x – 44 ≤ 44 + 15x – 44

⇒ 14x – 44 ≤ 15x

⇒ 15x ≥ 14x – 44

⇒ 15x – 14x ≥ 14x – 44 – 14x

∴ x ≥ –44

Thus, the solution of the given inequation is [–44, ∞).

Given

⇒ 13x ≥ 39

∴ x ≥ 3

Thus, the solution of the given inequation is [3, ∞).

Given

⇒ 5(x – 1) < 3(x – 35)

⇒ 5x – 5 < 3x – 105

⇒ 5x – 5 + 5 < 3x – 105 + 5

⇒ 5x < 3x – 100

⇒ 5x – 3x < 3x – 100 – 3x

⇒ 2x < –100

∴ x < –50

Thus, the solution of the given inequation is (–∞, –50).

Given

⇒ 3(2x + 3) < 4(x – 1)

⇒ 6x + 9 < 4x – 4

⇒ 6x + 9 – 9 < 4x – 4 – 9

⇒ 6x < 4x – 13

⇒ 6x – 4x < 4x – 13 – 4x

⇒ 2x < –13

Thus, the solution of the given inequation is.

Given

⇒ 2(5 – 2x) < x – 30

⇒ 10 – 4x < x – 30

⇒ 10 – 4x – 10 < x – 30 – 10

⇒ –4x < x – 40

⇒ x – 40 > –4x

⇒ x – 40 + 40 > –4x + 40

⇒ x > –4x + 40

⇒ x + 4x > –4x + 40 + 4x

⇒ 5x > 40

∴ x > 8

Thus, the solution of the given inequation is (8, ∞).

Given

⇒ 2(4 + 2x) ≥ 3(x – 6)

⇒ 8 + 4x ≥ 3x – 18

⇒ 8 + 4x – 8 ≥ 3x – 18 – 8

⇒ 4x ≥ 3x – 26

⇒ 4x – 3x ≥ 3x – 26 – 3x

∴ x ≥ –26

Thus, the solution of the given inequation is [–26, ∞).

Given

⇒ 2x – 7 < 3x – 6

⇒ 2x – 7 + 7 < 3x – 6 + 7

⇒ 2x < 3x + 1

⇒ 3x + 1 > 2x

⇒ 3x + 1 – 1 > 2x – 1

⇒ 3x > 2x – 1

⇒ 3x – 2x > 2x – 1 – 2x

∴ x > –1

Thus, the solution of the given inequation is (–1, ∞).

Given

⇒ 3(x – 2) ≤ 5x + 8

⇒ 3x – 6 ≤ 5x + 8

⇒ 3x – 6 + 6 ≤ 5x + 8 + 6

⇒ 3x ≤ 5x + 14

⇒ 5x + 14 ≥ 3x

⇒ 5x + 14 – 14 ≥ 3x – 14

⇒ 5x ≥ 3x – 14

⇒ 5x – 3x ≥ 3x – 14 – 3x

⇒ 2x ≥ –14

∴ x ≥ –7

Thus, the solution of the given inequation is [–7, ∞).

Given

For this inequation to be true, there are two possible cases.

i. 6x – 5 > 0 and 4x + 1 < 0

⇒ 6x – 5 + 5 > 0 + 5 and 4x + 1 – 1 < 0 – 1

⇒ 6x > 5 and 4x < –1

However,

Hence, this case is not possible.

ii. 6x – 5 < 0 and 4x + 1 > 0

⇒ 6x – 5 + 5 < 0 + 5 and 4x + 1 – 1 > 0 – 1

⇒ 6x < 5 and 4x > –1

However,

Thus, the solution of the given inequation is.

Given

For this inequation to be true, there are two possible cases.

i. 2x – 3 > 0 and 3x – 7 > 0

⇒ 2x – 3 + 3 > 0 + 3 and 3x – 7 + 7 > 0 + 7

⇒ 2x > 3 and 3x > 7

However,

Hence,

ii. 2x – 3 < 0 and 3x – 7 < 0

⇒ 2x – 3 + 3 < 0 + 3 and 3x – 7 + 7 < 0 + 7

⇒ 2x < 3 and 3x < 7

However,

Hence,

Thus, the solution of the given inequation is.

Given

For this inequation to be true, there are two possible cases.

i. x – 5 > 0 and x – 2 > 0

⇒ x – 5 + 5 > 0 + 5 and x – 2 + 2 > 0 + 2

⇒ x > 5 and x > 2

∴ x ∈ (5, ∞) ∩ (2, ∞)

However, (5, ∞) ∩ (2, ∞) = (5, ∞)

Hence, x ∈ (5, ∞)

ii. x – 5 < 0 and x – 2 < 0

⇒ x – 5 + 5 < 0 + 5 and x – 2 + 2 < 0 + 2

⇒ x < 5 and x < 2

∴ x ∈ (–∞, 5) ∩ (–∞, 2)

However, (–∞, 5) ∩ (–∞, 2) = (–∞, 2)

Hence, x ∈ (–∞, 2)

Thus, the solution of the given inequation is (–∞, 2) ∪ (5, ∞).

Given

For this inequation to be true, there are two possible cases.

i. 2x – 3 ≥ 0 and x – 1 > 0

⇒ 2x – 3 + 3 ≥ 0 + 3 and x – 1 + 1 > 0 + 1

⇒ 2x ≥ 3 and x > 1

However,

Hence,

ii. 2x – 3 ≤ 0 and x – 1 < 0

⇒ 2x – 3 + 3 ≤ 0 + 3 and x – 1 + 1 < 0 + 1

⇒ 2x ≤ 3 and x < 1

However,

Hence, x ∈ (–∞, 1)

Thus, the solution of the given inequation is.

Given

For this inequation to be true, there are two possible cases.

i. 8x – 33 > 0 and 2x – 5 > 0

⇒ 8x – 33 + 33 > 0 + 33 and 2x – 5 + 5 > 0 + 5

⇒ 8x > 33 and 2x > 5

However,

Hence,

ii. 8x – 33 < 0 and 2x – 5 < 0

⇒ 8x – 33 + 33 < 0 + 33 and 2x – 5 + 5 < 0 + 5

⇒ 8x < 33 and 2x < 5

However,

Hence,

Thus, the solution of the given inequation is.

Given

For this inequation to be true, there are two possible cases.

i. x – 3 > 0 and x + 6 < 0

⇒ x – 3 + 3 > 0 + 3 and x + 6 – 6 < 0 – 6

⇒ x > 3 and x < –6

∴ x ∈ (3, ∞) ∩ (–∞, –6)

However, (3, ∞) ∩ (–∞, –6) = ∅

Hence, this case is not possible.

ii. x – 3 < 0 and x + 6 > 0

⇒ x – 3 + 3 < 0 + 3 and x + 6 – 6 > 0 – 6

⇒ x < 3 and x > –6

∴ x ∈ (–∞, 3) ∩ (–6, ∞)

However, (–∞, 3) ∩ (–6, ∞) = (–6, 3)

Hence, x ∈ (–6, 3)

Thus, the solution of the given inequation is (–6, 3).

Given

For this inequation to be true, there are two possible cases.

i. x > 0 and 4 – x < 0

⇒ x > 0 and 4 – x – 4 < 0 – 4

⇒ x > 0 and –x < –4

⇒ x > 0 and x > 4

∴ x ∈ (0, ∞) ∩ (4, ∞)

However, (0, ∞) ∩ (4, ∞) = (4, ∞)

Hence, x ∈ (4, ∞)

ii. x < 0 and 4 – x > 0

⇒ x < 0 and 4 – x – 4 > 0 – 4

⇒ x < 0 and –x > –4

⇒ x < 0 and x < 4

∴ x ∈ (–∞, 0) ∩ (–∞, 4)

However, (–∞, 0) ∩ (–∞, 4) = (–∞, 0)

Hence, x ∈ (–∞, 0)

Thus, the solution of the given inequation is (–∞, 0) ∪ (4, ∞).

Given

For this inequation to be true, there are two possible cases.

i. x + 7 > 0 and x + 3 < 0

⇒ x + 7 – 7 > 0 – 7 and x + 3 – 3 < 0 – 3

⇒ x > –7 and x < –3

∴ x ∈ (–7, ∞) ∩ (–∞, –3)

However, (–7, ∞) ∩ (–∞, –3) = (–7, –3)

Hence, x ∈ (–7, –3)

ii. x + 7 < 0 and x + 3 > 0

⇒ x + 7 – 7 < 0 – 7 and x + 3 – 3 > 0 – 3

⇒ x < –7 and x > –3

∴ x ∈ (–∞, –7) ∩ (–3, ∞)

However, (–∞, –7) ∩ (–3, ∞) = ∅

Hence, this case is not possible.

Thus, the solution of the given inequation is (–7, –3).

Given

For this inequation to be true, there are two possible cases.

i. 25x + 17 > 0 and 8x + 3 < 0

⇒ 25x + 17 – 17 > 0 – 17 and 8x + 3 – 3 < 0 – 3

⇒ 25x > –17 and 8x < –3

However,

Hence,

ii. 25x + 17 < 0 and 8x + 3 > 0

⇒ 25x + 17 – 17 < 0 – 17 and 8x + 3 – 3 > 0 – 3

⇒ 25x < –17 and 8x > –3

However,

Hence, this case is not possible.

Thus, the solution of the given inequation is.

Given

For this inequation to be true, there are two possible cases.

i. x + 5 > 0 and x – 5 > 0

⇒ x + 5 – 5 > 0 – 5 and x – 5 + 5 > 0 + 5

⇒ x > –5 and x > 5

∴ x ∈ (–5, ∞) ∩ (5, ∞)

However, (–5, ∞) ∩ (5, ∞) = (5, ∞)

Hence, x ∈ (5, ∞)

ii. x + 5 < 0 and x – 5 < 0

⇒ x + 5 – 5 < 0 – 5 and x – 5 + 5 < 0 + 5

⇒ x < –5 and x < 5

∴ x ∈ (–∞, –5) ∩ (–∞, 5)

However, (–∞, –5) ∩ (–∞, 5) = (–∞, –5)

Hence, x ∈ (–∞, –5)

Thus, the solution of the given inequation is (–∞, –5) ∪ (5, ∞).

Given x + 3 < 0 and 2x < 14

Let us consider the first inequality.

x + 3 < 0

⇒ x + 3 – 3 < 0 – 3

⇒ x < –3

∴ x ∈ ( –∞, –3) (1)

Now, let us consider the second inequality.

2x < 14

⇒ x < 7

∴ x ∈ (–∞, 7) (2)

From (1) and (2), we get

x ∈ (–∞, –3) ∩ (–∞, 7)

∴ x ∈ ( –∞, –3)

Thus, the solution of the given system of inequations is (–∞, –3).

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

⇒ x > 4

∴ x ∈ ( 4, ∞) (1)

Now, let us consider the second inequality.

11 – 5x ≤ 1

⇒ 11 – 5x – 11 ≤ 1 – 11

⇒ –5x ≤ –10

⇒ –x ≤ –2

⇒ x ≥ 2

∴ x ∈ (2, ∞) (2)

From (1) and (2), we get

x ∈ (4, ∞) ∩ (2, ∞)

∴ x ∈ (4, ∞)

Thus, the solution of the given system of inequations is (4, ∞).

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

⇒ x < 6

∴ x ∈ (–∞, 6) (2)

From (1) and (2), we get

x ∈ (2, ∞) ∩ (–∞, 6)

∴ x ∈ ( 2, 6)

Thus, the solution of the given system of inequations is (2, 6).

Given 2x + 6 ≥ 0 and 4x – 7 < 0

Let us consider the first inequality.

2x + 6 ≥ 0

⇒ 2x + 6 – 6 ≥ 0 – 6

⇒ 2x ≥ –6

⇒ x ≥ –3

∴ x ∈ [–3, ∞) (1)

Now, let us consider the second inequality.

4x – 7 < 0

⇒ 4x – 7 + 7 < 0 + 7

⇒ 4x < 7

(2)

From (1) and (2), we get

Thus, the solution of the given system of inequations is

Given 3x – 6 > 0 and 2x – 5 > 0

Let us consider the first inequality.

3x – 6 > 0

⇒ 3x – 6 + 6 > 0 + 6

⇒ 3x > 6

⇒ x > 2

∴ x ∈ ( 2, ∞) (1)

Now, let us consider the second inequality.

2x – 5 > 0

⇒ 2x – 5 + 5 > 0 + 5

⇒ 2x > 5

(2)

From (1) and (2), we get

Thus, the solution of the given system of inequations is

Given 2x – 3 < 7 and 2x > –4

Let us consider the first inequality.

2x – 3 < 7

⇒ 2x – 3 + 3 < 7 + 3

⇒ 2x < 10

⇒ x < 5

∴ x ∈ ( –∞, 5) (1)

Now, let us consider the second inequality.

2x > –4

⇒ x > –2

∴ x ∈ (–2, ∞) (2)

From (1) and (2), we get

x ∈ (–∞, 5) ∩ (–2, ∞)

∴ x ∈ (–2, 5)

Thus, the solution of the given system of inequations is (–2, 5).

Given 2x + 5 ≤ 0 and x – 3 ≤ 0

Let us consider the first inequality.

2x + 5 ≤ 0

⇒ 2x + 5 – 5 ≤ 0 – 5

⇒ 2x ≤ –5

(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

Thus, the solution of the given system of inequations is

Given 5x – 1 < 24 and 5x + 1 > –24

Let us consider the first inequality.

5x – 1 < 24

⇒ 5x – 1 + 1 < 24 + 1

⇒ 5x < 25

⇒ x < 5

∴ x ∈ (–∞, 5) (1)

Now, let us consider the second inequality.

5x + 1 > –24

⇒ 5x + 1 – 1 > –24 – 1

⇒ 5x > –25

⇒ x > –5

∴ x ∈ (–5, ∞) (2)

From (1) and (2), we get

x ∈ (–∞, 5) ∩ (–5, ∞)

∴ x ∈ (–5, 5)

Thus, the solution of the given system of inequations is (–5, 5).

Given 3x – 1 ≥ 5 and x + 2 > –1

Let us consider the first inequality.

3x – 1 ≥ 5

⇒ 3x – 1 + 1 ≥ 5 + 1

⇒ 3x ≥ 6

⇒ 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, ∞)

Thus, the solution of the given system of inequations is (2, ∞).

Given 11 – 5x > –4 and 4x + 13 ≤ –11

Let us consider the first inequality.

11 – 5x > –4

⇒ 11 – 5x – 11 > –4 – 11

⇒ –5x > –15

⇒ –x > –3

⇒ x < 3

∴ x ∈ (–∞, 3) (1)

Now, let us consider the second inequality.

4x + 13 ≤ –11

⇒ 4x + 13 – 13 ≤ –11 – 13

⇒ 4x ≤ –24

⇒ x ≤ –6

∴ x ∈ (–∞, –6] (2)

From (1) and (2), we get

x ∈ (–∞, 3) ∩ (–∞, –6]

∴ x ∈ (–∞, –6]

Thus, the solution of the given system of inequations is (–∞, –6].

Given 4x – 1 ≤ 0 and 3 – 4x < 0

Let us consider the first inequality.

4x – 1 ≤ 0

⇒ 4x – 1 + 1 ≤ 0 + 1

⇒ 4x ≤ 1

(1)

Now, let us consider the second inequality.

3 – 4x < 0

⇒ 3 – 4x – 3 < 0 – 3

⇒ –4x < –3

(2)

From (1) and (2), we get

∴ x ∈ ∅

Thus, there is no solution of the given system of inequations.

Given x + 5 > 2(x + 1) and 2 – x < 3(x + 2)

Let us consider the first inequality.

x + 5 > 2(x + 1)

⇒ x + 5 > 2x + 2

⇒ x + 5 – 5 > 2x + 2 – 5

⇒ x > 2x – 3

⇒ 2x – 3 < x

⇒ 2x – 3 + 3 < x + 3

⇒ 2x < x + 3

⇒ 2x – x < x + 3 – x

⇒ x < 3

∴ x ∈ (–∞, 3) (1)

Now, let us consider the second inequality.

2 – x < 3(x + 2)

⇒ 2 – x < 3x + 6

⇒ 2 – x – 2 < 3x + 6 – 2

⇒ –x < 3x + 4

⇒ 3x + 4 > –x

⇒ 3x + 4 – 4 > –x – 4

⇒ 3x > –x – 4

⇒ 3x + x > –x + x – 4

⇒ 4x > –4

⇒ x > –1

∴ x ∈ (–1, ∞) (2)

From (1) and (2), we get

x ∈ (–∞, 3) ∩ (–1, ∞)

∴ x ∈ (–1, 3)

Thus, the solution of the given system of inequations is (–1, 3).

Given 2(x – 6) < 3x – 7 and 11 – 2x < 6 – x

Let us consider the first inequality.

2(x – 6) < 3x – 7

⇒ 2x – 12 < 3x – 7

⇒ 2x – 12 + 12 < 3x – 7 + 12

⇒ 2x < 3x + 5

⇒ 3x + 5 > 2x

⇒ 3x + 5 – 5 > 2x – 5

⇒ 3x > 2x – 5

⇒ 3x – 2x > 2x – 5 – 2x

⇒ x > –5

∴ x ∈ (–5, ∞) (1)

Now, let us consider the second inequality.

11 – 2x < 6 – x

⇒ 11 – 2x – 11 < 6 – x – 11

⇒ –2x < –x – 5

⇒ –x – 5 > –2x

⇒ –x – 5 + 5 > –2x + 5

⇒ –x > –2x + 5

⇒ –x + 2x > –2x + 5 + 2x

⇒ x > 5

∴ x ∈ (5, ∞) (2)

From (1) and (2), we get

x ∈ (–5, ∞) ∩ (5, ∞)

∴ x ∈ (5, ∞)

Thus, the solution of the given system of inequations is (5, ∞).

Given 5x – 7 < 3(x + 3) and

Let us consider the first inequality.

5x – 7 < 3(x + 3)

⇒ 5x – 7 < 3x + 9

⇒ 5x – 7 + 7 < 3x + 9 + 7

⇒ 5x < 3x + 16

⇒ 5x – 3x < 3x + 16 – 3x

⇒ 2x < 16

⇒ x < 8

∴ x ∈ (–∞, 8) (1)

Now, let us consider the second inequality.

⇒ 2 – 3x ≥ 2(x – 4)

⇒ 2 – 3x ≥ 2x – 8

⇒ 2 – 3x – 2 ≥ 2x – 8 – 2

⇒ –3x ≥ 2x – 10

⇒ 2x – 10 ≤ –3x

⇒ 2x – 10 + 10 ≤ –3x + 10

⇒ 2x ≤ –3x + 10

⇒ 2x + 3x ≤ –6x + 10 + 6x

⇒ 5x ≤ 10

⇒ x ≤ 2

∴ x ∈ (–∞, 2] (2)

From (1) and (2), we get

x ∈ (–∞, 8) ∩ (–∞, 2]

∴ x ∈ (–∞, 2]

Thus, the solution of the given system of inequations is (–∞, 2].

Given and 2(2x + 3) < 6(x – 2) + 10

Let us consider the first inequality.

⇒ 3(2x – 11) ≥ 4(4x – 18)

⇒ 6x – 33 ≥ 16x – 72

⇒ 6x – 33 + 33 ≥ 16x – 72 + 33

⇒ 6x ≥ 16x – 39

⇒ 16x – 39 ≤ 6x

⇒ 16x – 39 + 39 ≤ 6x + 39

⇒ 16x ≤ 6x + 39

⇒ 16x – 6x ≤ 6x + 39 – 6x

⇒ 10x ≤ 39

(1)

Now, let us consider the second inequality.

2(2x + 3) < 6(x – 2) + 10

⇒ 4x + 6 < 6x – 12 + 10

⇒ 4x + 6 < 6x – 2

⇒ 4x + 6 – 6 < 6x – 2 – 6

⇒ 4x < 6x – 8

⇒ 6x – 8 > 4x

⇒ 6x – 8 + 8 > 4x + 8

⇒ 6x > 4x + 8

⇒ 6x – 4x > 4x + 8 – 4x

⇒ 2x > 8

⇒ x > 4

∴ x ∈ (4, ∞) (2)

From (1) and (2), we get

∴ x ∈ ∅

Thus, there is no solution of the given system of inequations.

Given and

Let us consider the first inequality.

⇒ 7x – 1 < –6

⇒ 7x – 1 + 1 < –6 + 1

⇒ 7x < –5

(1)

Now, let us consider the second inequality.

⇒ 3x + 63 < 0

⇒ 3x + 63 – 63 < 0 – 63

⇒ 3x < –63

⇒ x < –21

∴ x ∈ (–∞, –21) (2)

From (1) and (2), we get

∴ x ∈ (–∞, –21)

Thus, the solution of the given system of inequations is (–∞, –21).

Given and

Let us consider the first inequality.

For this inequation to be true, there are two possible cases.

i. 11x – 2 > 0 and 7x – 1 < 0

⇒ 11x – 2 + 2 > 0 + 2 and 7x – 1 + 1 < 0 + 1

⇒ 11x > 2 and 7x < 1

However,

Hence, this case is not possible.

ii. 11x – 2 < 0 and 7x – 1 > 0

⇒ 11x – 2 + 2 < 0 + 2 and 7x – 1 + 1 > 0 + 1

⇒ 11x < 2 and 7x > 1

However,

Hence, (1)

Now, let us consider the second inequality.

For this inequation to be true, there are two possible cases.

i. x – 23 > 0 and x – 8 < 0

⇒ x – 23 + 23 > 0 + 23 and x – 8 + 8 < 0 + 8

⇒ x > 23 and x < 8

∴ x ∈ (23, ∞) ∩ (–∞, 5)

However, (23, ∞) ∩ (–∞, 5) = ∅

Hence, this case is not possible.

ii. x – 23 < 0 and x – 8 > 0

⇒ x – 23 + 23 < 0 + 23 and x – 8 + 8 > 0 + 8

⇒ x < 23 and x > 8

∴ x ∈ (–∞, 23) ∩ (8, ∞)

However, (–∞, 23) ∩ (8, ∞) = (8, 23)

Hence, x ∈ (8, 23) (2)

From (1) and (2), we get

∴ x ∈ ∅

Thus, there is no solution of the given system of inequations.

Given

The above inequality can be split into two inequalities.

and

Let us consider the first inequality.

⇒ 0 < –x

⇒ –x > 0

⇒ x < 0

∴ x ∈ (–∞, 0) (1)

Now, let us consider the second inequality.

⇒ –x < 6

⇒ x > –6

∴ x ∈ (–6, ∞) (2)

From (1) and (2), we get

x ∈ (–∞, 0) ∩ (–6, ∞)

∴ x ∈ (–6, 0)

Thus, the solution of the given system of inequations is (–6, 0).

Given 10 ≤ –5(x – 2) < 20

The above inequality can be split into two inequalities.

10 ≤ –5(x – 2) and –5(x – 2) < 20

Let us consider the first inequality.

10 ≤ –5(x – 2)

⇒ 10 ≤ –5x + 10

⇒ 10 – 10 ≤ –5x + 10 – 10

⇒ 0 ≤ –5x

⇒ 0 + 5x ≤ –5x + 5x

⇒ 5x ≤ 0

⇒ x ≤ 0

∴ x ∈ (–∞, 0] (1)

Now, let us consider the second inequality.

–5(x – 2) < 20

⇒ –5x + 10 < 20

⇒ –5x + 10 – 10 < 20 – 10

⇒ –5x < 10

⇒ –x < 2

⇒ x > –2

∴ x ∈ (–2, ∞) (2)

From (1) and (2), we get

x ∈ (–∞, 0] ∩ (–2, ∞)

∴ x ∈ (–2, 0]

Thus, the solution of the given system of inequations is (–2, 0].

Given –5 < 2x – 3 < 5

The above inequality can be split into two inequalities.

–5 < 2x – 3 and 2x – 3 < 5

Let us consider the first inequality.

–5 < 2x – 3

⇒ 2x – 3 > –5

⇒ 2x – 3 + 3 > –5 + 3

⇒ 2x > –2

⇒ x > –1

∴ x ∈ (–1, ∞) (1)

Now, let us consider the second inequality.

2x – 3 < 5

⇒ 2x – 3 + 3 < 5 + 3

⇒ 2x < 8

⇒ x < 4

⇒ x > –2

∴ x ∈ (–∞, 4) (2)

From (1) and (2), we get

x ∈ (–1, ∞) ∩ (–∞, 4)

∴ x ∈ (–1, 4)

Thus, the solution of the given system of inequations is (–1, 4).

Given

The above inequality can be split into two inequalities.

and

Let us consider the first inequality.

As x > 0, we have x + 1 > 0.

⇒ 4 ≤ 3(x + 1)

⇒ 4 ≤ 3x + 3

⇒ 3x + 3 ≥ 4

⇒ 3x + 3 – 3 ≥ 4 – 3

⇒ 3x ≥ 1

(1)

Now, let us consider the second inequality.

As x > 0, we have x + 1 > 0.

⇒ 3(x + 1) ≤ 6

⇒ 3x + 3 ≤ 6

⇒ 3x + 3 – 3 ≤ 6 – 3

⇒ 3x ≤ 3

⇒ x ≤ 1

∴ x ∈ (–∞, 1] (2)

From (1) and (2), we get

Thus, the solution of the given system of inequations is

We know that,

|x+a|>r ⟺ x>r–a or x<–(a+r)

Here,

We can verify the answers using the graph as well.

|4–x|+1<3

Subtracting 1 from both the sides.

⇒ |4–x|+1–1<3–1

⇒ |4–x|<2

We know that,

|a–x|<r ⟺ a–r<x<a+r

Here, a=4 and r=2

⇒ 4–2<x<4+2

⇒2<x<6

⇒xϵ(2, 6)

We can verify the answers using graph as well.

The equation can be re–written as

We know that,

|x–a|≤r ⟺ a–r ≤ x≤ a+r

Here,

Now, multiplying the whole inequality by 2 and dividing by 3, we get

We can verify the answers using graph as well.

Clearly, x≠2, as it will lead equation unmeaningful.

Now, two case arise:

Case1: x–2>0

⇒ x>2

In this case |x–2|=x–2

⇒ x ϵ (2, ∞)….(1)

Case 2: x–2<0

⇒ x<2

In this case, |x–2|=–(x–2)

Inequality doesn’t get satisfy

∴, this case gets nullified.

(from 1)

We can verify the answers using graph as well.

We know that, if we take reciprocal of any inequality we need to change the inequality as well.

Also,

|x|–3≠0

⇒ |x|>3 or |x|<3

For |x|<3

⇒ –3<x<3

⇒ xϵ (–3, 3) ….(1)

∴, The equation can be re–written as–

Adding 2 both the sides, we get–

|x|–3+3> 2+3

⇒ |x|>5

We know that,

|x |>a ⟺ x<–a or x>a

Here, a=5

⇒ x<–5 or x>5 ….(2)

⇒ x ϵ(–∞,–5 ) or x ϵ(5, ∞)

⇒xϵ(–∞,–5 )⋃(–3, 3)⋃(5, ∞) (from 1 and 2)

We can verify the answers using graph as well.

The equation can be re–written as

Adding 1 both the sides, we get,

Subtracting 3 both the sides

Clearly, x≠0, as it will lead equation unmeaningful.

Now, two case arise:

Case1: x+2>0

⇒ x>–2

In this case |x+2|=x+2

Considering Numerator,

2x–2>0

⇒ x>1

⇒ x ϵ (1, ∞) ….(1)

Case 2: x+2<0

⇒ x<–2

In this case, |x+2|=–(x+2)

Considering Numerator,

4x+2>0

ut x<–2

Now, from Denominator, we have–

⇒ x ϵ (–∞ , 0) …(2)

(1,∞) (from 1 and 2)

We can verify the answers using graph as well.

x≠1, as it will lead equation unmeaningful.

Now, on subtracting 2 from both the sides, we get–

Now, 3 case arises:

Case 1:1<x<∞

For this case, |2x–1|=2x–1 and |x–1|=x–1

⇒ x ϵ (1, ∞ ) …(1)

Case 2:

For this case: |2x–1|=2x–1 and |x–1|=–(x–1)

…(2)

Case 3:

For this case: |2x–1|=–(2x–1) and |x–1|=–(x–1)

Which is not possible, hence, this will give no solution.

(from 1 and 2)

We can verify the answers using graph as well.

Subtracting 6 from both the sides, we get–

|x–1|+|x–2|+|x–3|–6≥0

Here, we have 4 cases:

Case 1: –∞ <x<1

For this case, |x–1|=–(x–1), |x–2|=–(x–2) and |x–3|=–(x–3)

⇒ –(x–1+x–2+x–3+6)≥0

⇒ x–1+x–2+x–3+6<0

⇒ 3x<0

⇒ x<0

⇒ xϵ (–∞ , 0) …(1)

Case 2:1<x<2

For this case, |x–1|=x–1, |x–2|=–(x–2) and |x–3|=–(x–3)

⇒ x–1–x+2–x+3–6≥0

⇒ –x–2≥0

⇒ x+2<0

⇒ x<–2

Which doesn’t signify the interval

Case 3:2<x<3

For this case, |x–1|=x–1, |x–2|=x–2 and x–3=–(x–3)

⇒ x–1+x–2–x+3–6≥0

⇒ x–6≥0

⇒ x≥ 6

Which doesn’t signify the interval

Case 4:3<x<∞

For this case, |x–1|=x–1, |x–2|=x–2 and |x–3|=x–3

⇒ x–1+x–2+x–3–6≥0

⇒ x–1+x–2+x–3–6≥0

⇒ 3x–12>0

⇒ x>4

⇒ xϵ (4 , ∞ ) …(2)

⇒xϵ(–∞ , 0)⋃(4 , ∞ ) (from 1 and 2)

We can verify the answers using graph as well.

Clearly, |x–2|–2≠0

⇒ |x–2|≠2

⇒ x≠0 and x≠4

Now, 2 case arise:

Case 1:–∞ <x<2

For this, |x–2|=–(x–2)

⇒ x ϵ (0,1] …(1)

Case 2: 2<x<∞

For this, |x–2|=x–2

⇒ x ϵ [3,4) …(2)

⇒xϵ(0,1]⋃[3, 4) (from 1 and 2)

We can verify the answers using graph as well.

We know that, if we take reciprocal of any inequality we need to change the inequality as well.

Also,

|x|–3≠0

⇒ |x|>3 or |x|<3

For |x|<3

⇒ –3<x<3

⇒ x ϵ (–3, 3) …(1)

∴, The equation can be re–written as–

Adding 2 both the sides, we get–

|x|–3+3≥ 2+3

⇒ |x|≥5

We know that,

|x |≥a ⟺ x≤–a or x≥a

Here, a=5

⇒ x≤–5 or x≥5

⇒ x ϵ(–∞,–5 ] or x ϵ[5, ∞) …(2)

⇒xϵ(–∞,–5 ]⋃(–3, 3)⋃[5, ∞)(from 1 and 2)

We can verify the answers using graph as well.

Subtracting 3 from both sides, we get–

|x+1|+|x|–3>0

For this, we have 3 cases,

Case 1:–∞ <x<–1

For this, |x+1|=–(x+1) and |x|=–x

–(x+1)–x–3>0

–2x–1–3>0

2x+4<0

x<–2

⇒ x ϵ (–∞ , –2) …(1)

Case 2: –1<x<0

For this, |x+1|=x+1 and |x|=–x

x+1–x–3>0

–2>0

Which is absurd, thus it leads to no solution.

Case 3: 0<x<∞

For this, |x+1|=x+1 and |x|=x

x+1+x–3>0

2x–2>0

x>1

⇒ x ϵ (1 , ∞ ) …(2)

⇒xϵ(–∞ , –2)⋃(1 , ∞ ) (from 1 and 2)

We can verify the answers using graph as well.

We know that,

a≤ | x – c | ≤ b ⟺ x ϵ [–b+c, –a+c] ⋃ [a+c, b+c]

∴, 1≤|x–2|≤3 ⟺ xϵ [–3+2, –1+2] ⋃ [1+2, 3+2]

⟺xϵ[–1, 1]⋃[3, 5]

We can verify the answers using graph as well.

Subtracting 9 from both sides, we get–

|3–4x|–9≥0

For this, we have 2 cases,

Case 1:

For this, |3–4x|=3–4x

3–4x–9≥0

–4x–6≥0

4x≤–6

…(1)

Case 2: 0<x<∞

For this, |3–4x|=–(3–4x)

–3+4x–9≥0

4x–12≥0

4x≥12

⇒ x≥3

⇒ xϵ [3, ∞ ) …(2)

(from 1 and 2)

We can verify the answers using graph as well.

We need to assume two consecutive odd positive integers.

So, let the smaller odd positive integer be x.

Then, the other odd positive integer will be (x + 2).

Given: Both these numbers are smaller than 10. …(i)

And their sum is more than 11. …(ii)

So,

From given statement (i),

x < 10 …(iii)

x + 2 < 10

⇒ x < 10 – 2

⇒ x < 8 …(iv)

From given statement (ii),

Sum of these two consecutive odd positive integers > 11

⇒ (x) + (x + 2) > 11

⇒ x + x + 2 > 11

⇒ 2x + 2 > 11

⇒ 2(x + 1) > 11

…(v)

From inequalities (iv) & (v), we have

x < 8 &

It can be merged and written as

This means that x lies between 9/2 (or 4.5) and 8.

Note the odd positive integers lying between 4.5 and 8.

They are 5 and 7.

Also, consider inequality (iii), that is, x < 10.

So, from inequalities (iii) & (v), we have

x < 10 &

It can be merged and written as

Note that, the upper limit here has shifted from 8 to 10. Now, x is odd integer from 4.5 to 10.

So, the odd integers from 4.5 to 10 are 5, 7 and 9.

Now, let us find pairs of consecutive odd integers.

Let x = 5, then (x + 2) = (5 + 2) = 7.

Let x = 7, then (x + 2) = (7 + 2) = 9.

Let x = 9, then (x + 2) = (9 + 2) = 11. But, 11 is greater than 10.

Hence, all such pairs of odd consecutive positive integers required are (5, 7) and (7, 9).

We need to assume two consecutive odd natural numbers.

So, let the smaller odd natural number be x.

Then, the other odd natural number will be (x + 2).

Given: Both these numbers are larger than 10. …(i)

And their sum is less than 40. …(ii)

So,

From given statement (i),

x > 10 …(iii)

x + 2 > 10

⇒ x > 10 – 2

⇒ x > 8

Since, the number must be greater than 10, x > 8 can be ignored.

From given statement (ii),

Sum of these two consecutive odd natural numbers < 40

⇒ (x) + (x + 2) < 40

⇒ x + x + 2 < 40

⇒ 2x + 2 < 40

⇒ 2(x + 1) < 40

⇒ x + 1 < 20

⇒ x < 20 – 1

⇒ x < 19 …(iv)

From inequalities (iii) & (iv), we have

x > 10 & x < 19

It can be merged and written as

10 < x < 19

From this inequality, we can say that x lies between 10 and 19.

So, the odd natural numbers lying between 10 and 19 are 11, 13, 15 and 17. (Excluding 19 as x < 19)

Now, let us find pairs of consecutive odd natural numbers.

Let x = 11, then (x + 2) = (11 + 2) = 13

Let x = 13, then (x + 2) = (13 + 2) = 15

Let x = 15, then (x + 2) = (15 + 2) = 17

Let x = 17, then (x + 2) = (17 + 2) = 19.

Hence, all such pairs of consecutive odd natural numbers required are (11, 13), (13, 15), (15, 17) and (17, 19).

We need to assume two consecutive even positive integers.

So, let the smaller even positive integer be x.

Then, the other even positive integer will be (x + 2).

Given: Both these numbers are larger than 5. …(i)

And their sum is less than 23. …(ii)

So,

From given statement (i),

x > 5 …(iii)

x + 2 > 5

⇒ x > 5 – 2

⇒ x > 3

Since, the number must be larger than 5, x > 3 can be ignored.

From given statement (ii),

Sum of these two consecutive even positive integers < 23

⇒ (x) + (x + 2) < 23

⇒ x + x + 2 < 23

⇒ 2x + 2 < 23

⇒ 2(x + 1) < 23

⇒ x + 1 < 11.5

⇒ x < 11.5 – 1

⇒ x < 10.5 …(iv)

From inequalities (iii) & (iv), we have

x > 5 & x < 10.5

It can be merged and written as

5 < x < 10.5

From this inequality, we can say that x lies between 5 and 10.5.

So, the even positive integers lying between 5 and 10.5 are 6, 8 and 10.

Now, let us find pairs of consecutive even positive integers.

Let x = 6, then (x + 2) = (6 + 2) = 8

Let x = 8, then (x + 2) = (8 + 2) = 10

Let x = 10, then (x + 2) = (10 + 2) = 12.

Hence, all such pairs of consecutive even positive integers required are (6, 8), (8, 10) and (10, 12).

Given: Marks scored by Rohit in two tests are 65 and 70.

To find Minimum marks in the third test to make an average of at least 65 marks.

Let marks in the third test be x.

According to the question, we need to find minimum x for which the average of all three papers would be at least 65 marks.

That is,

Average marks in three papers ≥ 65 …(i)

Now, we know that average is given as

So, an average of the marks in these three tests is given by

Substituting this value of average in the inequality (i), we get

⇒ (135 + x) ≥ 65 × 3

⇒ (135 + x) ≥ 195

⇒ x ≥ 195 – 135

⇒ x ≥ 60

This inequality means that Rohit should score at least 60 marks in his third test to have an average of at least 65 marks.

So, the minimum marks to get an average of 65 marks is 60.

Thus, the minimum marks required in the third test is 60.

We have been given that, a solution is kept between 86° F and 95° F.

Let F₁ = 86° F

And let F₂ = 95°

The conversion formula is

We need to convert Fahrenheit into degree Celcius.

So, take F₁ = 86° F

Using the formula, we have

⇒ C₁ = 5 × 6

⇒ C₁ = 30° C

Now, take F₂ = 95° F

Using the formula, we have

⇒ C₂ = 5 × 7

⇒ C₂ = 35° C

Therefore, the range of temperature of the solution in degree Celsius is 30° C and 35° C.

We have been given that, a solution is kept between 30° C and 35° C.

Let C₁ = 30° C

And let C₂ = 35° C

The conversion formula is given by

We need to convert degree Celsius to degree Fahrenheit.

So, take C₁ = 30° C

Using the formula, we have

⇒ F₁ = 9 × 6 + 32

⇒ F₁ = 54 + 32

⇒ F₁ = 86° F

Now, take C₂ = 35° C

Using the formula, we have

⇒ F₂ = 9 × 7 + 32

⇒ F₂ = 63 + 32

⇒ F₂ = 95° F

Therefore, the range of temperature of the solution in degree Fahrenheit is 86° F and 95° F.

Given that, there is total of five papers that Shikha has attended.

The score in the first four papers is 87, 95, 92 and 94.

To receive grade ‘A,’ the average marks in the five papers must be 90 or more.

Let marks in the fourth paper be x.

According to the question, we need to find minimum x for which the average of all five papers would be at least 90 marks.

That is,

Average marks in five papers ≥ 90 …(i)

Let us find the average marks in five papers. It is given by

So,

Substituting this value of average in the inequality (i), we get

⇒ (368 + x) ≥ 90 × 5

⇒ (368 + x) ≥ 450

⇒ x ≥ 450 – 368

⇒ x ≥ 82

This inequality means that Shikha should score at least 82 marks in her fifth test to have an average of at least 90 marks.

So, the minimum marks to get an average of 90 marks is 82.

Thus, the minimum marks required in the fifth test is 82.

We have been a week’s data,

Revenue, R = 2x

Where x = number of cassettes produced and sold in a week.

We know that profit is given by,

Profit = Revenue – Cost …(i)

Revenue is the income that a business has from its normal business activities, usually from the sale of goods and services to customers.

A cost is the value of money that has been used up to produce something or deliver a service and hence is not available for use anymore.

And Profit is the gain in the business.

So, it is justified that profit in any business would be measured by the difference in the capital generated by the business and the capital used up in the business.

Profit generated by the company manufacturing cassettes is given by,

Profit = R – C (from (i))

Where, R = Revenue

C = Cost of cassette

Here,

If R < C, then

Profit < 0

⇒ There is a loss.

If R = C, then

Profit = 0

⇒ There is no profit no loss.

If R > C, then

Profit > 0

⇒ There is a profit.

We need to find the number of cassettes sold to make a profit. That is, we need to find x.

So, R > C (to realize a profit)

Substituting values of R and C. We get

⇒ x > 300 × 2

⇒ x > 600

This means that x must be greater than 600.

Thus, the company must sell more than 600 cassettes to realize a profit.

We are given with a triangle,

The longest side of this triangle = 3 × Shortest side …(i)

The third side of this triangle = Longest side – 2 cm …(ii)

The perimeter of the triangle ≥ 61 cm …(iii)

Let

Shortest side of the triangle = a

The longest side of the triangle = b

The third side of the triangle = c

So

From (i),

b = 3 × a

⇒ b = 3a …(iv)

From (ii),

c = b – 2

⇒ c = 3a – 2 (∵ b = 3a) …(v)

Then, perimeter is given by

Perimeter of the triangle = a + b + c

Substituting the values of b and c from equation (iv) and (v) respectively, we get

Perimeter of the triangle = a + (3a) + (3a – 2)

⇒ Perimeter of the triangle = 7a – 2 …(vi)

Putting the value of perimeter of the triangle from (v) in inequality (iii), we get

7a – 2 ≥ 61

⇒ 7a ≥ 61 + 2

⇒ 7a ≥ 63

⇒ a ≥ 9

This means, ‘a’ which is the shortest side of the triangle is 9 or more than 9.

Thus, the minimum length of the shortest side of the triangle is 9 cm.

Given: Volume of the existing solution = 1125 liters

Amount of acid in the existing solution = 45% of 1125 …(i)

And the rest 55% of 1125 liters is the amount of water in it, which need not be computed.

Let the water added (in liters) be x in 1125 liters of solution.

According to the question,

x liters of water has to be added to 1125 liters of the 45% solution.

We can say that, even if x liters of water is added to the 1125 liters of solution, acid content will not change. Only water content and the whole volume of the solution will get affected.

So, the resulted solution will have acid content as follows:

The acid content in the solution after adding x liters of water = 45% of 1125 …(ii)

[∵ we know that the amount of acid content will not change after adding water to the whole solution. So, from equation (i), we have this conclusion]

Also, according to the question,

This resulting mixture will contain more than 25% acid content.

So, we have

Acid content in the solution after adding x litres of water > 25% of new mixture

⇒ 45% of 1125 > 25% of (1125 + x) [∵ from equation (ii)]

⇒ 45 × 1125 > 25(1125 + x)

⇒ 9 × 1125 > 5(1125 + x)

⇒ 9 × 225 > 1125 + x

⇒ 2025 > 1125 + x

⇒ x < 2025 – 1125

⇒ x < 900

Also,

This resulting mixture will contain less than 30% acid content.

So, we have

Acid content in the solution after adding x litres of water < 30% of new mixture

⇒ 45% of 1125 < 30% of (1125 + x) [∵ from equation (ii)]

⇒ 45 × 1125 < 30(1125 + x)

⇒ 9 × 1125 < 6(1125 + x)

⇒ 3 × 1125 < 2(1125 + x)

⇒ 3375 < 2250 + 2x

⇒ 2x + 2250 > 3375

⇒ 2x > 3375 – 2250

⇒ 2x > 1125

⇒ x > 562.5

Thus, we have got x < 900 and x > 562.5.

⇒ 562.5 < x < 900

Hence, the required liters of water to be added to 1125 liters of solution is between 562.5 liters and 900 liters.

Volume of the 8% solution = 640 litres

Boric acid present in the 8% solution = 8% of 640 …(i)

And the rest 92% of 640 litres is water in the 8% solution.

Let volume of 2% solution added to 640 liters be x.

Boric acid present in 2% solution = 2% of x …(ii)

New volume of 8% solution = 640 + x …(iii)

Boric acid present in the new solution (that is, after adding x litres of 2% solution to 8% solution) = Boric acid present in the 8% solution + Boric acid present in the 2% solution [from (i) & (ii)]

⇒ Boric acid present in the new solution = 8% of 640 + 2% of x

⇒Boric acid present in the new solution =

⇒Boric acid present in the new solution = …(iv)

According to the question,

The resulting mixture is to be more than 4% but less than 6% boric acid.

That is, the boric acid content in the resulting mixture must be more than 4% but less than 6% boric acid.

So, first let us take boric acid content in the resulting mixture to be more than 4%.

⇒ Boric acid present in the new solution > 4% of the new volume of 8% solution

[from (iii) & (iv)]

⇒ 2x + 5120 > 2560 + 4x

⇒ 5120 – 2560 > 4x – 2x

⇒ 2560 > 2x

⇒ 2x < 2560

⇒ x < 1280

Now, let us the take boric acid in the resulting mixture to be less than 6%.

⇒ Boric acid present in the new solution < 6% of the new volume of 8% solution

[from (iii) & (iv)]

⇒ 2x + 5120 < 3840 + 6x

⇒ 5120 – 3840 < 6x – 2x

⇒ 1280 < 4x

⇒ 4x > 1280

⇒ x > 320

We have

x < 1280 & x > 320

⇒ 320 < x < 1280

Hence, the required liters of 2% solution to be added to 8% of the solution is between 320 liters and 1280 liters.

Given that,

pH value of the first reading = 7.48

pH value of the second reading = 7.85

We need to find the range of the pH value for the third reading so that the acidity level in the pool is normal.

But the acidity level in the pool is considered normal when the average pH reading of the three measurements is between 7.2 and 7.8.

That is, 7.2 < average pH reading of the three measurements < 7.8 …(i)

Let us find the average pH reading of the three measurements.

For this, let the pH value of the third reading be x.

Then, the average is given by

Substituting this value of average in inequality (i), we get

Multiply 3 throughout the inequality, we have

⇒ 22.6 < 15.33 + x < 23.4

Now, subtract 15.33 throughout the inequality,

⇒ 22.6 – 15.33 < 15.33 + x – 15.33 < 23.4 – 15.33

⇒ 7.27 < x < 8.07

This means, x lies between values 7.27 and 8.07.

Thus, the pool’s acidity level would be normal when the range of pH value in the third measurement would be between 7.27 and 8.07.

First, we will find the solutions of the given equation by hit and trial method and afterward we will plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + 2y – 4 ≤ 0

First, we will find the solutions of the given equation by hit and trial method and afterward we will plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + 2y ≥ 6

First, we will find the solutions of the given equation by hit and trial method and afterward we will plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + 2 ≥ 0

x ≥ –2

As there is only one variable ‘x,’ and y = 0, which means that x has only one value when considered as an equation. Therefore no table is required in this problem.

First, we will find the solutions of the given equation by hit and trial method and afterward we will plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x – 2y < 0

x < 2y

First, we will find the solutions of the given equation by hit and trial method and afterward we will plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

– 3x + 2y ≤ 6

First, we will find the solutions of the given equation by hit and trial method and afterward we will plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x ≤ 8 – 4y

x + 4y ≤ 8

First, we will find the solutions of the given equation by hit and trial method and afterward we will plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

0 ≤ 2x – 5y + 10

First, we will find the solutions of the given equation by hit and trial method and afterward we will plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

3y > 6 – 2x

2x + 3y > 6

First, we will find the solutions of the given equation by hit and trial method and afterward we will plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

y> 2x – 8

First, we will find the solutions of the given equation by hit and trial method and afterward we will plot the graph of the equation and shade the side containing solutions of the inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

3x – 2y ≤ x + y – 8

2x – 3y ≤ –8

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + 3y ≤ 6

3x + 2y ≤ 6

x ≥ 0, y ≥ 0

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + 3y≤ 6

x + 4y ≤ 4

x ≥ 0, y ≥ 0

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x – y ≤ 1

x + 2y≤ 8

2x + y ≥ 2

x ≥ 0, y ≥ 0

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + y ≥ 1

7x + 9y ≤ 63

x ≤6, y ≤ 5, x ≥ 0, y ≥ 0

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + 3y≤35

y ≥ 3, x ≥ 2, x ≥ 0, y ≥ 0

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality.

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x – 2y ≥ 0

2x – y ≤ –2

x ≥ 0, y ≥ 0

The lines do not intersect each other for x ≥ 0, y ≥ 0

Hence, there is no solution for the given inequations.

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + 2y≤ 3

3x + 4y≥ 12

y ≥ 1, x ≥ 0, y ≥ 0

In this question, we will apply the concept of a common solution area to find the signs of inequality by using their given equations and the given common solution area(shaded part).

If a line is in the form ax+by = c and c is positive constant(in case of negative c the rule becomes opposite), so there will be two cases to discuss, which are,

If a line is above origin :–

(i). If the shaded area is below the line then ax+by<c

(ii). If the shaded area is above the line then ax+by>c

If a line is below origin then the rule becomes opposite.

According to the rules the answer will be,

In this question we will apply the concept of common solution area to find the signs of inequality by using their given equations and the given common solution area(shaded part).

If a line is in the form ax+by = c and c is positive constant(in case of negative c the rule becomes opposite), so there will be two cases to discuss, which are,

If a line is above origin :–

(i). If the shaded area is below the line then ax+by<c

(ii). If the shaded area is above the line then ax+by>c

If a line is below origin, then the rule becomes opposite.

According to the rules, the answer will be,

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + y ≥ 9

3x + y ≥ 12

x ≥ 0, y ≥ 0

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

2x + y ≥ 8

x + 2y ≥ 8

x + y ≤ 6

First we will find the solutions of the given equations by hit and trial method and afterwards we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

12 + 12y ≤ 840

x + y ≤ 70

3x + 6y ≤ 300

x + 2y ≤ 100

8x + 4y ≤ 480

2x + y ≤ 120

x ≥ 0, y ≥ 0

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + 2y≤ 40

3x + y ≥ 30

4x + 3y≥ 60

x ≥ 0, y ≥ 0

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

5x + y ≥ 10

2x + 2y ≥ 12

x + y ≥ 6

x + 4y ≥ 12

x ≥ 0, y ≥ 0

First, we will find the solutions of the given equations by hit and trial method and afterward we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of the solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e., x and y–intercepts always,

x + 2y ≥ 3

3x + 4y ≥ 12

x ≥ 0, y ≥ 1

First we will find the solutions of the given equations by hit and trial method and afterwards we will plot the graph of the equations and shade the side with grey color containing common solutions or intersection of solution set of each inequality,

You can choose any value but find the two mandatory values which are at x = 0 and y = 0, i.e. x and y intercepts always,

2x + y≥ 8

x + 2y≥ 10

x ≥ 0, y ≥ 0.

⇒ x² > 0 and x – 2 > 0

⇒ x > 0 and x > 2

We have to take intersection of x > 0 and x > 2

So, answer should be x ∈ (2, ∞)

⇒

⇒

Case I : x² – 2x + 1 ≥ 0 and x > 0

(x – 1)² ≥ 0 and x > 0

So, by taking intersection x > 0

Case II : : x² – 2x + 1 ≤ 0 and x < 0

(x – 1)² ≤ 0 and x < 0

Square term is always positive so case II is irrelevant.

Then, the final answer of question is x ∈ (0, ∞)

(x² – 2x + 1)(x – 4) ≥ 0

(x – 1)²(x – 4) ≥ 0

⇒ (x – 1)² ≥ 0 and (x – 4) ≥ 0

⇒ Square term is always positive and x ≥ 4

So, answer is x ∈ [4, ∞)

|2 – x| = x – 2

⇒ |x – 2| = x – 2

We know that mode is always positive or zero. So, x – 2 is also positive or zero.

⇒ x – 2 ≥ 0

⇒ x ≥ 2

So, final answer is x ∈ [2, ∞)

|x – 1| ≤ 3

⇒ -3 ≤ x – 1 ≤ 3

⇒ -2 ≤ x ≤ 4

|x – 1 | ≤ 1

⇒ -1 ≤ x – 1 ≤ 1

⇒ 0 ≤ x ≤ 2

We have to take intersection of x ∈ [-2, 4] and x ∈ [0, 2]

So, final answer is x ∈ [0, 2]

Part I :

⇒

Part II:

⇒

We have to take union of and

So, the final answer is .

Here, denominator i.e., x² + 1 is always positive and not equal to zero. So, neglect it.

⇒ -x² + 2x + 3 > 0

⇒ x² – 2x – 3 < 0

⇒ (x – 3)(x + 1) < 0

Case I : (x – 3) < 0 and (x + 1) > 0

⇒ x < 3 and x > -1

By takin intersection x ∈ (-1, 3)

Case II : (x – 3) > 0 and (x + 1) < 0

⇒ x > 3 and x < -1

By taking intersection x ∈ ∅. So, case II is irrelevant.

So, the complete solution is x ∈ (-1, 3)

The integral solution is 0, 1 and 2. So, number of integral

solution is 3.

Part I : 5x + 2 < 3x + 8

⇒ 2x < 6

⇒ x < 3

Part II :

Case I : -3x + 6 < 0 and x – 1 > 0

⇒ x > 2 and x > 1

By taking intersection x ∈ (2, ∞)

Case II : -3x + 6 > 0 and x – 1 < 0

⇒ x < 2 and x < 1

By takin intersection x ∈ (-∞, 1)

Taking union of case I and case II, x ∈ (-∞, 1) (2, ∞)

We have to take intersection of part I and part II, we have

final answer i.e., x ∈ (-∞, 1) (2, 3)

Part I :

Square term is always positive and (x + 1)² ≠ 0. So, x > 0 and x ≠ -1

Part II :

Square term is always positive and (x – 1)² ≠ 0. So, x < 0 and x ≠ 1

So, by taking union of part I and part II we have final solution i.e., x ∈ R – {-1, 0, 1}

|x – 1| ≥ |x – 3|

Squaring both sides

|x – 1|² ≥ |x – 3|²

(x – 1)² ≥ (x – 3)²

x² – 2x + 1 ≥ x² – 6x + 9

4x – 8 ≥ 0

x ≥ 2

x ∈ [2, ∞)

We know that when we change the sign of inequalities then greater tan changes to less than and vice versa also true.

So, -x > -7

-3x + 17 < -13

-3x < -13 – 17

-3x < -30

3x > 30

x > 10

x ∈ (10, ∞)

x < y and b > 0

because b is greater than zero.

|x| < 5

-5 < x < 5

|x| > a

x < -a and x > a

x ∈ (-∞, -a) (a, ∞)

|x – 1| > 5

x – 1 < -5 and x – 1 > 5

x < -4 and x > 6

x ∈ (-∞, -4) (6, ∞)

|x + 2| ≤ 9

-9 ≤ x + 2 ≤ 9

-11 ≤ x ≤ 7

x ∈ [-11, 7]

The given figure is shaded between -3 to 3 on x-axis.

x ∈ [-3, 3]

-3 ≤ x ≤ 3

|x| ≤ 3

The given figure is highlighted between -∞ to 5 and 5 to ∞

So, x ∈ (-∞, 5] [5, ∞)

x ≤ -5 and x ≥ 5

|x| ≥ 5

|x + 2| ≤ 5

-5 ≤ x + 2 ≤ 5

-7 ≤ x ≤ 3

x ∈ [-7, -3]

Case I : x > 2

1 ≥ 0

It is true that 1 is always greater than 0 so case I is also true x > 2

Case II : x < 2

-1 ≥ 0

It is false that -1 is not greater than 0 so case II is also false.

So, the final solution is x > 2 i.e., x ∈ (2, ∞)

|x + 3| ≥ 10

x + 3 ≤ -10 and x + 3 ≥ 10

x ≤ -13 and x ≥ 7

x ∈ (-∞, -13] [7, ∞)