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Option : (B)

|x-2|/(x-2) ≥ 0

Case I : x > 2

\(\frac{x-2}{x-2}\) ≥ 0

1 ≥ 0 

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

Case II : x < 2

\(\frac{-(x-2)}{x-2}\) ≥ 0 

-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, ∞)

Option : (B)

3x + 17 < -13

-3x < -13 – 17 

-3x < -30 

3x > 30 

x > 10 

x ∈ (10, ∞)

Given,

–4x > 30

⇒ \(-\frac{4x}{4}\) > \(\frac{30}{4}\)

⇒ - x > \(\frac{15}{2}\)

∴ x < \(-\frac{15}{2}\) 

i. x ∈ R

When x is a real number, the solution of the given inequation is (-∞,\(-\frac{15}{2}\)).

ii. x ∈ Z

As - 8 < \(-\frac{15}{2}\) < -7,

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 ∅.

Let us consider as F₁ = 86^°F

And F₂ = 95°

As we know that, F = 9/5 C + 32

F₁ = 9/5 C₁ + 32

F₁ – 32 = 9/5 C₁

C₁ = 5/9 (F₁ – 32)

= 5/9 (86 – 32)

= 5/9 (54)

= 5 × 6

= 30° C

Then,

F₂ = 9/5 C₂ + 32

F₂ – 32 = 9/5 C₂

C₂ = 5/9 (F₂ – 32)

= 5/9 (95 – 32)

= 5/9 (63)

= 5 × 7

= 35° C

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

Given as

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

Suppose the marks in the third test be x.

Therefore let us find minimum x for which the average of all three papers would be at least 65 marks.

That is,

The average marks in three papers ≥ 65 …(i)

The average is given by:

Average = (sum of all numbers)/(Total number of items)

= (marks in 1^st two papers + marks in third test)/3

= (65 + 70 + x)/3

= (135 + x)/3

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

(135 + x)/3 ≥ 65

(135 + x) ≥ 65 × 3

(135 + x) ≥ 195

x ≥ 195 – 135

x ≥ 60

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

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

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

Let us consider as C₁ = 30°C

And the C₂ = 35°

As we know, F = 9/5 C + 32

F₁ = 9/5 C₁ + 32

= 9/5 × 30 + 32

= 9 × 6 + 32

= 54 + 32

= 86° F

Then,

F₂ = 9/5 C₂ + 32

= 9/5 × 35 + 32

= 9 × 7 + 32

= 63 + 32

= 95° F

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

Option : (B)

|x| < 5 

- 5 < x < 5

Given as

12x < 50

Therefore when we divide by 12, we get

12x/12 < 50/12

x < 25/6

(i) Given as x ∈ R

When x is a real number, the solution of the given inequation is (-∞, 25/6).

(ii) Given as x ∈ Z

When, 4 < 25/6 < 5

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

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

(iii) Given as x ∈ N

When, 4 < 25/6 < 5

Therefore when, when x is a natural number, the maximum possible value of x is 4. As we know that the natural numbers start from 1, the solution of the given inequation is {1, 2, 3, 4}.

Given as

-4x > 30

Therefore when we divide by 4, we get

-4x/4 > 30/4

-x > 15/2

x < – 15/2

(i) Given as x ∈ R

Whenever x is a real number, the solution of the given inequation is (-∞, -15/2).

(ii) Given as x ∈ Z

Whenever, -8 < -15/2 < -7

Therefore when, when x is an integer, the maximum possible value of x is -8.

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

(iii) Given as x ∈ N

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

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 

⇒ \(\frac{2x}{2}\) < \(\frac{14}{2}\)

⇒ 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,

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

⇒ \(\frac{-5x}{5}\) < \(\frac{10}{5}\)

⇒ –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 < x/6 - 5

⇒ \(\frac{5-2x}{3}\) < \(\frac{x-30}{6}\) 

⇒ 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

⇒ \(\frac{5x}{5}\) > \(\frac{40}{5}\)

∴ x > 8 

Thus,

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

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

⇒ \(\frac{3x}{3}\) < \(\frac{18}{3}\)

⇒ 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 as

4x – 2 < 8

4x – 2 + 2 < 8 + 2

4x < 10

Therefore divide by 4 on both sides we get,

4x/4 < 10/4

x < 5/2

(i) Given as x ∈ R

When x is a real number, the solution of the given inequation is (-∞, 5/2).

(ii) Given as x ∈ Z

When, 2 < 5/2 < 3

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

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

(iii) Given as x ∈ N

When, 2 < 5/2 < 3

Therefore when, when x is a natural number, the maximum possible value of x is 2. As we know that the natural numbers start from 1, the solution of the given inequation is {1, 2}.

Given,

2x – 3 < 7 and 2x > –4 

Let us consider the first inequality. 

2x – 3 < 7 

⇒ 2x – 3 + 3 < 7 + 3 

⇒ 2x < 10

⇒ \(\frac{2x}{2}\) < \(\frac{10}{2}\)

⇒ x < 5 

∴ x ∈ ( –∞, 5) ...(1) 

Now, 

Let us consider the second inequality. 

2x > –4

⇒ \(\frac{2x}{2}\) > \(\frac{-4}{2}\)

⇒ 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,

5x – 1 < 24 and 5x + 1 > –24 

Let us consider the first inequality. 

5x – 1 < 24

⇒ 5x – 1 + 1 < 24 + 1 

⇒ 5x < 25

⇒ \(\frac{5x}{5}\) < \(\frac{25}{5}\)

⇒ x < 5 

∴ x ∈ (–∞, 5) ...(1) 

Now,

Let us consider the second inequality. 

5x + 1 > –24 

⇒ 5x + 1 – 1 > –24 – 1 

⇒ 5x > –25

⇒ \(\frac{5x}{5}\) > \(\frac{-25}{5}\) 

⇒ 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 as 

3x – 7 > x + 1

3x – 7 + 7 > x + 1 + 7

3x > x + 8

3x – x > x + 8 – x

2x > 8

On dividing both sides by 2, we get

2x/2 > 8/2

x > 4

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

Given as 

3x – 6 > 0 and 2x – 5 > 0

Now, let us consider the first inequality.

3x – 6 > 0

3x – 6 + 6 > 0 + 6

3x > 6

Dividing both the sides by 3 we get,

3x/3 > 6/3

x > 2

∴ x ∈ ( 2, ∞)… (1)

Then, let us consider the second inequality.

2x – 5 > 0

2x – 5 + 5 > 0 + 5

2x > 5

Dividing 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, ∞)

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

Given,

2x + 6 ≥ 0 and 4x – 7 < 0 

Let us consider the first inequality. 

2x + 6 ≥ 0 

⇒ 2x + 6 – 6 ≥ 0 – 6 

⇒ 2x ≥ –6

⇒ \(\frac{2x}{2}\) ≥ \(\frac{-6}{2}\)

⇒ x ≥ –3 

∴ x ∈ [–3, ∞) ...(1) 

Now,

Let us consider the second inequality. 

4x – 7 < 0 

⇒ 4x – 7 + 7 < 0 + 7 

⇒ 4x < 7

⇒ \(\frac{4x}{4}\) < \(\frac{7}{4}\)

⇒ x < \(\frac{7}{4}\)

∴ x ∈ (- ∞,\(\frac{7}{4}\)) ....(2)

From (1) and (2), we get

x ∈ [- 3,∞) ∩ (- ∞,\(\frac{7}{4}\))

∴ x ∈ [-3,\(\frac{7}{4}\))

Thus, 

The solution of the given system of inequations is  [-3,\(\frac{7}{4}\)).

Given as

2x – 3 < 7 and 2x > –4

Now, let us consider the first inequality.

2x – 3 < 7

2x – 3 + 3 < 7 + 3

2x < 10

Dividing both the sides by 2 we get,

2x/2 < 10/2

x < 5

∴ x ∈ ( –∞, 5)… (1)

Then, let us consider the second inequality.

2x > –4

Dividing 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)

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

Given as 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

Divide both sides by 3, we get

3x/3 < 15/3

x < 5

Therefore, the solution of the given inequation is (-∞, 5).

Given as

5x – 1 < 24 and 5x + 1 > –24

Now, let us consider the first inequality.

5x – 1 < 24

5x – 1 + 1 < 24 + 1

5x < 25

Dividing both the sides by 5 we get,

5x/5 < 25/5

x < 5

∴ x ∈ (–∞, 5)… (1)

Then, let us consider the second inequality.

5x + 1 > –24

5x + 1 – 1 > –24 – 1

5x > –25

Dividing 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)

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

Given as

3x – 1 ≥ 5 and x + 2 > –1

Now, let us consider the first inequality.

3x – 1 ≥ 5

3x – 1 + 1 ≥ 5 + 1

3x ≥ 6

Dividing both the sides by 3 we get,

3x/3 ≥ 6/3

x ≥ 2

∴ x ∈ (2, ∞)… (1)

Then, 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, ∞)

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

Given as

(|x + 2| – x)/x < 2

Now, let us rewrite the equation as

|x + 2|/x – x/x < 2

|x + 2|/x – 1 < 2

On adding 1 on both sides, we get

|x + 2|/x – 1 + 1 < 2 + 1

|x + 2|/x < 3

On subtracting 3 on both sides, we get

|x + 2|/x – 3 < 3 – 3

|x + 2|/x – 3 < 0

So, clearly it states, x ≠ 2 therefore two case arise:

Case 1: x + 2 > 0

x > –2

Now, in this case |x + 2| = x + 2

x + 2/x – 3 < 0

(x + 2 – 3x)/x < 0

– (2x – 2)/x < 0

(2x – 2)/x < 0

Now, let us consider only the numerators, we get

2x – 2 > 0

x > 1

x ϵ (1, ∞) ….(1)

Case 2: x + 2 < 0

x < –2

In this case, |x + 2| = – (x + 2)

-(x + 2)/x – 3 < 0

(-x – 2 – 3x)/x < 0

– (4x + 2)/x < 0

(4x + 2)/x < 0

Then, let us consider only the numerators, we get

4x + 2 > 0

x > – 1/2

But x < -2

From the denominator we have,

x ∈ (–∞ , 0) …(2)

From (1) and (2)

∴ x ∈ (–∞ , 0) ⋃ (1, ∞)

Given as

2x + 6 ≥ 0 and 4x – 7 < 0

Now, let us consider the first inequality.

2x + 6 ≥ 0

2x + 6 – 6 ≥ 0 – 6

2x ≥ –6

Dividing both the sides by 2 we get,

2x/2 ≥ -6/2

x ≥ -3

∴ x ∈ [–3, ∞) …(1)

Then, let us consider the second inequality.

4x – 7 < 0

4x – 7 + 7 < 0 + 7

4x < 7

Dividing 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).

Given as 3x + 9 ≥ – x + 19

3x + 9 – 9 ≥ – x + 19 – 9

3x ≥ – x + 10

3x + x ≥ – x + 10 + x

4x ≥ 10

Divide both sides by 4, we get

4x/4 ≥ 10/4

x ≥ 5/2

Therefore, the solution of the given inequation is (5/2, ∞).

Given,

3x – 6 > 0 and 2x – 5 > 0 

Let us consider the first inequality. 

3x – 6 > 0 

⇒ 3x – 6 + 6 > 0 + 6 

⇒ 3x > 6 

⇒ \(\frac{3x}{3}\) > \(\frac{6}{3}\)

⇒ x > 2 

∴ x ∈ ( 2, ∞) ...(1) 

Now, 

Let us consider the second inequality.

2x – 5 > 0 

⇒ 2x – 5 + 5 > 0 + 5 

⇒ 2x > 5

⇒ \(\frac{2x}{2}\) > \(\frac{5}{2}\)

⇒ x > \(\frac{5}{2}\)

∴ x ∈ (\(\frac{5}{2}\),∞) ...(2)

From (1) and (2), we get

 x ∈ (2,∞) ∩ (\(\frac{5}{2}\),∞)

∴ x ∈ (\(\frac{5}{2}\),∞)

Thus, 

The solution of the given system of inequations is  (\(\frac{5}{2}\),∞).

Given,

5x – 7 < 3(x + 3) and 1 - 3x/2 ≥ x - 4

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

⇒ \(\frac{2x}{2}\) < \(\frac{16}{2}\) 

⇒ x < 8 

∴ x ∈ (–∞, 8) ...(1) 

Now,

Let us consider the second inequality.

1 - \(\frac{3x}{2}\) ≥ x - 4

⇒ \(\frac{2-3x}{2}\) ≥ x - 4

⇒ \((\frac{2-3x}{2})\) \(\times\) 2 ≥ (x - 4) \(\times\) 2

⇒ 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

⇒ \(\frac{5x}{5}\) ≤ \(\frac{10}{5}\) 

⇒ 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].

As 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 ∈ (–3, 3) …. (1)

Now, the equation can be re–written as

|x| – 3 > 2

Then, let us add 3 on both the sides, we get

|x| – 3 + 3 > 2 + 3

|x| > 5

Suppose ‘a’ be a fixed real number. Now,

|x| > a ⟺ x < –a or x > a

Here, a = 5

x < –5 or x > 5 …. (2)

From (1) and (2)

x ∈ (–∞,–5 ) or x ∈ (5, ∞)

∴ x ∈ (–∞,–5 ) ⋃ (–3, 3) ⋃ (5, ∞)

Given,

0 < -x/2 < 3

The above inequality can be split into two inequalities.

0 <\(-\frac{x}{2}\) and \(-\frac{x}{2}\) < 3

Let us consider the first inequality.

 0 <\(-\frac{x}{2}\) 

⇒ 0\(\times\)2 < \(-\frac{x}{2}\)\(\times\)2

⇒ 0 < –x 

⇒ –x > 0 

⇒ x < 0 

∴ x ∈ (–∞, 0) ....(1) 

Now,

Let us consider the second inequality.

\(-\frac{x}{2}\) < 3

⇒ \(-\frac{x}{2}\) \(\times\) 2 < 3 \(\times\)2

⇒ –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).

Suppose ‘r’ be a positive real number and ‘a’ be a fixed real number. Now,

|x + a| > r ⟺ x > r – a or x < – (a + r)

Here, a = 1/3 and r = 8/3

x > 8/3 – 1/3 or x < – (8/3 + 1/3)

x > (8 - 1)/3 or x < – (8 + 1)/3

x > 7/3 or x < – 9/3

x > 7/3 or x < – 3

x ∈ (7/3, ∞) or x ∈ (–∞, -3)  

∴ x ∈ (–∞, -3) ∪ (7/3, ∞)

Given as 2(3 – x) ≥ x/5 + 4

6 – 2x ≥ x/5 + 4

6 – 2x ≥ (x + 20)/5

5(6 – 2x) ≥ (x + 20)

30 – 10x ≥ x + 20

30 – 20 ≥ x + 10x

10 ≥ 11x

11x ≤ 10

Divide both sides by 11, we get

11x/11 ≤ 10/11

x ≤ 10/11

Therefore, the solution of the given inequation is (-∞, 10/11).

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 as |4 – x| + 1 < 3

Now, let us subtract 1 from both the sides, we get

|4 – x| + 1 – 1 < 3 – 1

|4 – x| < 2

Suppose ‘r’ be a positive real number and ‘a’ be a fixed real number. Now,

|a – x| < r ⟺ a – r < x < a + r

Here, a = 4 and r = 2

4 – 2 < x < 4 + 2

2 < x < 6

∴ x ∈ (2, 6)

Given,

3x – 1 ≥ 5 and x + 2 > –1 

Let us consider the first inequality. 

3x – 1 ≥ 5 

⇒ 3x – 1 + 1 ≥ 5 + 1 

⇒ 3x ≥ 6

⇒ \(\frac{3x}{3}\) ≥ \(\frac{6}{3}\)

⇒ 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, ∞) 

Thus, 

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

Given as

|x – 2|/(x – 2) > 0

Clearly it states, x ≠ 2 therefore two case arise:

Case 1: x – 2 > 0

x > 2

Now, 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)

– (x – 2)/(x – 2) > 0

– 1 > 0

Here, Inequality doesn’t get satisfy

This case gets nullified.

∴ x ∈ (2, ∞) from (1)

Given as

x – 2 > 0 and 3x < 18

Now, let us consider the first inequality.

x – 2 < 0

x – 2 + 2 < 0 + 2

x < 2

∴ x ∈ ( 2, ∞) …(1)

Then, let us consider the second inequality.

3x < 18

Dividing 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).

Given as

(3x – 2)/5 ≤ (4x – 3)/2

Multiplying both the sides by 5 we get,

(3x – 2)/5 × 5 ≤ (4x – 3)/2 × 5

(3x – 2) ≤ 5(4x – 3)/2

3x – 2 ≤ (20x – 15)/2

Multiplying both the sides by 2 we get,

(3x – 2) × 2 ≤ (20x – 15)/2 × 2

6x – 4 ≤ 20x – 15

20x – 15 ≥ 6x – 4

20x – 15 + 15 ≥ 6x – 4 + 15

20x ≥ 6x + 11

20x – 6x ≥ 6x + 11 – 6x

14x ≥ 11

Dividing both sides by 14, we get

14x/14 ≥ 11/14

x ≥ 11/14

Therefore, the solution of the given inequation is (11/14, ∞).

Given,

11 – 5x > –4 and 4x + 13 ≤ –11 

Let us consider the first inequality. 

11 – 5x > –4 

⇒ 11 – 5x – 11 > –4 – 11 

⇒ –5x > –15

⇒ \(\frac{-5x}{5}\) > \(\frac{-15}{5}\)

⇒ –x > –3 

⇒ x < 3 

∴ x ∈ (–∞, 3) ...(1) 

Now,

Let us consider the second inequality. 

4x + 13 ≤ –11 

⇒ 4x + 13 – 13 ≤ –11 – 13 

⇒ 4x ≤ –24

⇒ \(\frac{4x}{4}\) ≤ \(\frac{-24}{4}\)

⇒ 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,

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

⇒ \(\frac{4x}{4}\) > \(\frac{-4}{4}\)

⇒ 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 as

|(3x – 4)/2| ≤ 5/12

Now, we can rewrite it as

|3x/2 – 4/2| ≤ 5/12

|3x/2 – 2| ≤ 5/12

Suppose ‘r’ be a positive real number and ‘a’ be a fixed real number. Now,

|x – a| ≤ r ⟺ a – r ≤ x ≤ a + r

Here, a = 2 and r = 5/12

2 – 5/12 ≤ 3x/2 ≤ 2 + 5/12

(24 - 5)/12 ≤ 3x/2 ≤ (24 + 5)/12

19/12 ≤ 3x/2 ≤ 29/12

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

19/18 ≤ x ≤ 29/18

∴ x ∈ [19/18, 29/18]

Given as

11 – 5x > –4 and 4x + 13 ≤ –11

Now, let us consider the first inequality.

11 – 5x > –4

11 – 5x – 11 > –4 – 11

–5x > –15

Dividing both the sides by 5 we get,

-5x/5 > -15/5
–x > –3

x < 3

∴ x ∈ (–∞, 3) (1)

Then, let us consider the second inequality.

4x + 13 ≤ –11

4x + 13 – 13 ≤ –11 – 13

4x ≤ –24

Dividing 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]

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

Given as

2x + 5 ≤ 0 and x – 3 ≤ 0

Now, let us consider the first inequality.

2x + 5 ≤ 0

2x + 5 – 5 ≤ 0 – 5

2x ≤ –5

Dividing both the sides by 2 we get,

2x/2 ≤ –5/2

x ≤ – 5/2

∴ x ∈ (–∞, -5/2]… (1)

Then, let us consider the second inequality.

x – 3 ≤ 0

x – 3 + 3 ≤ 0 + 3

x ≤ 3

∴ x ∈ (–∞, 3]… (2)

From the equation (1) and (2), we get

x ∈ (–∞, -5/2) ∩ (–∞, 3)

x ∈ (–∞, -5/2) 

Hence, the solution of the given system of inequations is (–∞, -5/2).

Given as

2x – 7 > 5 – x and 11 – 5x ≤ 1

Now, 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

Dividing both the sides by 3 we get,

3x/3 > 12/3

x > 4
∴ x ∈ (4, ∞) … (1)

Then, let us consider the second inequality.

11 – 5x ≤ 1

11 – 5x – 11 ≤ 1 – 11

–5x ≤ –10

Dividing 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, ∞)

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

Given as x + 3 < 0 and 2x < 14

Now, let us consider the first inequality.

x + 3 < 0

x + 3 – 3 < 0 – 3

x < –3

Then, let us consider the second inequality.

2x < 14

Dividing both the sides by 2 we get,

2x/2 < 14/2

x < 7

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

Given as –(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

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

Given,

4x – 1 ≤ 0 and 3 – 4x < 0 

Let us consider the first inequality. 

4x – 1 ≤ 0 

⇒ 4x – 1 + 1 ≤ 0 + 1 

⇒ 4x ≤ 1

⇒ \(\frac{4x}{4}\) ≤ \(\frac{1}{4}\) 

⇒ x ≤ \(\frac{1}{4}\)

∴ x ∈ ( -∞,\(\frac{1}{4}\)] ... (1)

Now,

Let us consider the second inequality. 

3 – 4x < 0 

⇒ 3 – 4x – 3 < 0 – 3 

⇒ –4x < –3

⇒ \(\frac{-4x}{4}\) < \(\frac{-3}{4}\) 

⇒ -x < \(-\frac{3}{4}\)

⇒ x > \(\frac{3}{4}\)

∴ x ∈ (\(\frac{3}{4}\),∞) ... (2)

From (1) and (2), we get

 x ∈ ( -∞,\(\frac{1}{4}\)] ∩  (\(\frac{3}{4}\),∞)

∴ x ∈ ∅

Thus,

There is no solution of the given system of inequations.

Given,

2x + 5 ≤ 0 and x – 3 ≤ 0 

Let us consider the first inequality. 

2x + 5 ≤ 0 

⇒ 2x + 5 – 5 ≤ 0 – 5 

⇒ 2x ≤ –5

⇒ \(\frac{2x}{2}\) ≤ \(\frac{-5}{2}\) 

⇒ x ≤ \(-\frac{5}{2}\)

∴ x ∈ (- ∞,\(-\frac{5}{2}\)] ...(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 ∈  (-∞,\(-\frac{5}{2}\)] ∩ (-∞, 3]

∴ x ∈  (-∞,\(-\frac{5}{2}\)]

Thus,

The solution of the given system of inequations is  (-∞,\(-\frac{5}{2}\)].

Given as x/5 < (3x - 2)/4 – (5x - 3)/5

x/5 < [5(3x - 2) – 4(5x - 3)]/4(5)

x/5 < [15x – 10 – 20x + 12]/20

x/5 < [2 – 5x]/20

Multiplying both the sides by 20 we get,

x/5 × 20 < [2 – 5x]/20 × 20

4x < 2 – 5x

4x + 5x < 2 – 5x + 5x

9x < 2

Dividing both sides by 9, we get

9x/9 < 2/9

x < 2/9

Therefore, the solution of the given inequation is (-∞, 2/9).

Given as

(2x + 3)/4 – 3 < (x – 4)/3 – 2

Adding 3 on both sides we get,

(2x + 3)/4 – 3 + 3 < (x – 4)/3 – 2 + 3

(2x + 3)/4 < (x – 4)/3 + 1

(2x + 3)/4 < (x – 4 + 3)/3

(2x + 3)/4 < (x – 1)/3

On cross multiply we get,

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

Dividing both sides by 2, we get

2x/2 < -13/2

x < -13/2

Therefore, the solution of the given inequation is (-∞, -13/2).