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Given,

2(3-x) ≥ x/5 + 4

⇒ 6 - 2x ≥ \(\frac{x}{5}\)+ 4

⇒ 6 - 2x ≥ \(\frac{x+20}{5}\)

⇒ (6 - 2x) x 5 ≥ (\(\frac{x+20}{5}\)) x 5

⇒ 30 – 10x ≥ x + 20 

⇒ x + 20 ≤ 30 – 10x 

⇒ x + 20 – 20 ≤ 30 – 10x – 20 

⇒ x ≤ 10 – 10x 

⇒ x + 10x ≤ 10 – 10x + 10x 

⇒ 11x ≤ 10

⇒ \(\frac{11x}{11}\) ≤ \(\frac{10}{11}\)

∴ x ≤ \(\frac{10}{11}\)

Thus,

The solution of the given inequation is (-∞,\(\frac{10}{11}\)].

we have

2(3−x)≥5x​+4x∈R

10(3−x)≥x+20

30−10x≥x+20

10≥11x

x≤1110​

so x∈(−∞,1110​]

Therefore x∈(−∞,1110​]

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

⇒ \(\frac{3x}{3}\) > \(\frac{12}{3}\) 

⇒ x > 4 

∴ x ∈ ( 4, ∞) ...(1) 

Now, 

Let us consider the second inequality. 

11 – 5x ≤ 1 

⇒ 11 – 5x – 11 ≤ 1 – 11 

⇒ –5x ≤ –10 

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

⇒ –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,

(4+2x)/3 ≥ x/2 - 3

⇒ \(\frac{4+2x}{3}\) ≥ \(\frac{x-6}{2}\)

⇒ \((\frac{4+2x}{3})\) \(\times\) 3 \(\times\) 2 ≥ \((\frac{x-6}{2})\) \(\times\) 3 \(\times\) 2

⇒ 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 as 

[2(x - 1)]/5 ≤ [3(2 + x)]/7

(2x – 2)/5 ≤ (6 + 3x)/7

Multiplying both the sides by 5 we get,

(2x – 2)/5 × 5 ≤ (6 + 3x)/7 × 5

2x – 2 ≤ 5(6 + 3x)/7

7(2x – 2) ≤ 5(6 + 3x)

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

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

Given as

(x – 1)/3 + 4 < (x – 5)/5 – 2

Subtracting both sides by 4 we get,

(x – 1)/3 + 4 – 4 < (x – 5)/5 – 2 – 4

(x – 1)/3 < (x – 5)/5 – 6

(x – 1)/3 < (x – 5 – 30)/5

(x – 1)/3 < (x – 35)/5

On cross multiply we get,

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

Dividing both sides by 2, we get

2x/2 < -100/2

x < -50

Therefore, the solution of the given inequation is (-∞, -50).

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,

3x – 7 > x + 1 

⇒ 3x – 7 + 7 > x + 1 + 7 

⇒ 3x > x + 8

⇒ 3x – x > x + 8 – x 

⇒ 2x > 8

⇒ \(\frac{2x}{2}\) > \(\frac{8}{2}\) 

∴ x > 4 

Thus,

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

Given,

3x + 9 ≥ –x + 19 

⇒ 3x + 9 – 9 ≥ –x + 19 – 9 

⇒ 3x ≥ –x + 10 

⇒ 3x + x ≥ –x + 10 + x 

⇒ 4x ≥ 10

⇒ \(\frac{4x}{4}\) ≥ \(\frac{10}{4}\) 

∴ x ≥ \(\frac{5}{2}\)

Thus,

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

Given as

5x/2 + 3x/4 ≥ 39/4

On taking LCM

[2(5x) + 3x]/4 ≥ 39/4

13x/4 ≥ 39/4

Multiplying both the sides by 4 we get,

13x/4 × 4 ≥ 39/4 × 4

13x ≥ 39

Dividing both sides by 13, we get

13x/13 ≥ 39/13

x ≥ 39/13

x ≥ 3

Therefore, the solution of the given inequation is (3, ∞).

Given,

x - 2 ≤ (5x + 8)/3

⇒  (x - 2) \(\times\) ≤ (\(\frac{5x + 8}{3}\)) \(\times\) 3

⇒ 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

⇒ \(\frac{2x}{2}\) ≥ \(-\frac{14}{2}\)

∴ x ≥ –7 

Thus,

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

Given,

4x – 2 < 8 

⇒ 4x – 2 + 2 < 8 + 2 

⇒ 4x < 10

⇒ \(\frac{4x}{4}\) < \(\frac{10}{4}\)

∴ x < \(\frac{5}{2}\)

i. x ∈ R

When x is a real number, the solution of the given inequation is (-∞,\(\frac{5}{2}\)).

ii. x ∈ Z

As 2<\(\frac{5}{2}\)<3,

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 2<\(\frac{5}{2}\)<3, 

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,

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

⇒ \(\frac{3x}{3}\)<\(\frac{15}{3}\)

∴ x < 5 

Thus, 

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

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

⇒ \(\frac{2x}{2}\) > \(\frac{-2}{2}\) 

⇒ x > –1 

∴ x ∈ (–1, ∞) ....(1) 

Now,

Let us consider the second inequality. 

2x – 3 < 5 

⇒ 2x – 3 + 3 < 5 + 3 

⇒ 2x < 8

⇒ \(\frac{2x}{2}\) < \(\frac{8}{2}\)

⇒ 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).

(x+2)/(x²+1)>1/2

⇒ \(\frac{x+2}{x^2+1}\) - \(\frac{1}{2}\) > 0 

⇒ \(\frac{2(x+2)-(x^2+1)}{2(x^2+1)}\) > 0

⇒ \(\frac{-x^2+2x+3}{2(x^2+1)}\) > 0

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.

|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]

(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, ∞)

|1/x - 2| < 4

- 4 < (\(\frac{1}{x}\) - 2) < 4

- 2 < \(\frac{1}{x}\) < 6

Part I : \(\frac{1}{x}\) > - 2

⇒ x < \(\frac{-1}{2}\)

Part II : \(\frac{1}{x}\) < 6 

⇒ x > \(\frac{1}{6}\)

We have to take union of  x < \(\frac{-1}{2}\) and x > \(\frac{1}{6}\)

So, the final answer is x ∈ (- ∞,\(\frac{-1}{2}\)) ∪ (\(\frac{1}{6}\),∞).

x + 1/x ≥ 2

⇒  \(x\) + \(\frac{1}{x}\) - 2 ≥ 0

⇒ \(\frac{x^2-2x+1}{x}\) ≥ 0 

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, ∞).

|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, ∞)

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

⇒ 2x < 6 

⇒ x < 3

Part II : (x+2)/(x-1) < 4.

⇒ \(\frac{x+2}{x-1}\) - 4 < 0

⇒ \(\frac{(x+2)-4(x-1)}{x-1}\)< 0

⇒ \(\frac{-3x+6}{x-1}\) < 0

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)

|x + 1/x| > 2

⇒ - 2 < \(x\) + \(\frac{1}{x}\) < 2

Part I : \(x\) + \(\frac{1}{x}\) > - 2

⇒ \(\frac{x^2+2x+1}{x}\) > 0

⇒ \(\frac{(x+1)^2}{x}\) > 0

Square term is always positive and (x + 1)² ≠ 0. 

So, x > 0 and x ≠ -1

Part II :  \(x\) + \(\frac{1}{x}\)< 2

⇒ \(\frac{x^2-2x+1}{x}\) < 0

⇒ \(\frac{(x-1)^2}{x}\) < 0

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}

x²/(x-2)>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,∞)