Let the consecutive even positive integers be x and x + 2
\(\therefore\)  x > 5 . and x + 2 > 5
 \(\Rightarrow\) x > 5 and x > 3
 \(\Rightarrow\) x > 5 ....(i)
Also x + (x + 2) < 20
 \(\Rightarrow\) 2x < 20 - 2 => 2x < 18
 \(\Rightarrow\) x < 9 .....(ii)
From (i) and (ii), we have
5 Thus x = 6,7,8.
But x is even positive integer
\(\therefore\)  x = 6and8
Thus two pairs of even positive integers are 6, 8 and 8, 10

Here \(\frac { 2x+4 }{ x-3 } \le 4\),  \(x\neq 3\)
\(\Rightarrow\) \(\frac { 2x+4 }{ x-3 } -4\) \(\le\) 0 \(\Rightarrow\) \(\frac { 2x+4-4x+12 }{ x-3 } \le 0\)
\(\Rightarrow\) \(\frac { -2x+16 }{ x-3 } \le 0\) \(\Rightarrow\) -2x + 16\(\le\) 0
\(\Rightarrow\) -2x \(\le\) - 16
Dividing both sides by - 2
\(\therefore\)  \(\frac { -2x }{ -2 } \le \frac { -16 }{ -2 } \) \(\Rightarrow\) x \(\ge\) 8
Thus the Solution et of given in equation is [8, \(\infty\))

Here \(\frac { x+3 }{ x-2 } \ge 4\), \(x\neq 2\)
\(\Rightarrow\)  \(\frac { x+3 }{ x-2 } -4\) \(\ge\) \(\Rightarrow\) \(\frac { x+3-4x+8 }{ x-2 } \ge 0\)
\(\Rightarrow\) \(\frac { -3x+11 }{ x-2 } \ge 0\) \(\Rightarrow\) -3x +11 \(\ge\)0
\(\Rightarrow\) -3x\(\ge\)-11
Dividing both sides by -3
\(\therefore\) \(\frac { -3x }{ -3 } \le \frac { -11 }{ -3 } \) \(\Rightarrow\) \(x\le \frac { 11 }{ 3 } \)
Thus the solution set of given inequation is \(\left( -\infty ,\frac { 11 }{ 3 } \right) \) - {2}

Here \(\frac { 2x-3 }{ 4 } +19\ge 13+\frac { 4x }{ 3 } \)
\(\frac { 2x-3 }{ 4 } -\frac { 4x }{ 3 } \)  \(\ge\) 13 -19
\(\frac { 6x-9-16x }{ 12 } \) \(\ge\) -6  \(\Rightarrow\) \(\frac { -10x-9 }{ 12 } \) \(\ge\) -6
Multiplying both sides by 12
\(\therefore\) -10x - 9 \(\ge\) -6 x 12
\(\Rightarrow\) -10x -9 \(\ge\) -72
\(\Rightarrow\) -10x \(\ge\) -72 + 9
\(\Rightarrow\) -10x \(\ge\) -63
Dividing both sides by - 1 0
\(\therefore\)  \(\frac { -10x }{ 10 } \le \frac { -63 }{ -10 } \)
\(\therefore\)  \(x\le \frac { 63 }{ 10 } \)
Thus the solution set of given in equation is \(\left( -\infty ,\frac { 63 }{ 10 } \right) \)

We have, 5x –3 < 3x + 1
or 5x –3 + 3 < 3x +1 +3 (Rule 1)
or 5x < 3x +4
or 5x – 3x < 3x + 4 – 3x (Rule 1)
or 2x < 4 or x < 2 (Rule 2)
(i) When x is an integer, the solutions of the given inequality are ..., – 4, – 3, – 2, – 1, 0, 1
(ii) When x is a real number, the solutions of the inequality are given by x < 2, i.e., all real numbers x which are less than 2. Therefore, the solution set of the inequality is x ∈ (– ∞, 2).
We have considered solutions of inequalities in the set of natural numbers, set of integers and in the set of real numbers. Henceforth, unless stated otherwise, we shall solve the inequalities in this Chapter in the set of real numbers.

(i) Here 2x - 4 \(\le\) 0
\(\Rightarrow\) 2x \(\le\)4
Dividing both sides by 2
\(\therefore\) \(\frac { 2x }{ 2 } \le \frac { 4 }{ 2 } \) \(\Rightarrow\) x \(\le\) 2
Thus solution set of the given inequation is (-\(\infty\),2]
(ii) Here - 5x +15<0
\(\Rightarrow\) -5x<-15
Dividing both sides by -5
\(\therefore\)  \(\frac { -5x }{ -5 } >\frac { -15 }{ 5 } \) \(\Rightarrow\) x > 3
Thus solution set of the given inequation is (3, \(\infty\))

The given inequality is x > -3
Draw the graph of the line x = -3.
Putting (0, 0) in the given in equation, we have
0 > -3 which is true

\(\therefore\) Half plane of x > -3 is towards origin

Here \(\frac { 3(x-2) }{ 5 } \le \frac { 5(2-x) }{ 3 } \)
\(\Rightarrow \) \(\frac { 3x-6 }{ 5 } \le \frac { 10-5x }{ 3 } \) \(\Rightarrow\) \(\frac { 3x }{ 5 } -\frac { 6 }{ 5 } \le \frac { 10 }{ 3 } -\frac { -5x }{ 3 } \)
\(\Rightarrow \) \(\frac { 3x }{ 5 } +\frac { 5x }{ 3 } \le \frac { 10 }{ 3 } +\frac { 6 }{ 5 } \) \(\Rightarrow\) \(\frac { 9x+25 }{ 15 } \le \frac { 50+18 }{ 15 } \)
\(\Rightarrow \) \(\frac { 34x }{ 15 } \le \frac { 68 }{ 15 } \)
Multiplying both sides by 15, we have 34x\(\le\) 68
Dividing both sides by 34, we have x \(\le\) 2
Thus the solution set is (-\(\infty\), 2]

Here 5x - 3 < 7
5x < 7 + 3  \(\Rightarrow\) 5x < 10
Dividing both sides by 5, we have x < 2
(i) When x is an integer then values of x that make the statement true are ..., -3, -2, -1, 0, 1
The solution set of inequality is {..., -3, -2, -1,0, I}.
(ii) When x is a real number. The solution set of inequality is x \(\epsilon \) (-\(\infty\),2)

We have, x + 2y \(\ge \) 4
On putting x = 0 and y = 0, we get
0 + 2(0) \(\ge \) 4 \(\Rightarrow \) 0 \(\ge \) 4
This is a false statement. Therefore, the given half plane does not contain the origin.

Given, \(|3x-2|\le \frac { 1 }{ 2 } \)
\(\therefore \) \(\Rightarrow\) \(\frac { -1 }{ 2 } \le \left( 3x-2 \right) \le \frac { 1 }{ 2 } \quad \left[ \therefore |x|\le a\Longrightarrow -a\le x\le a \right]\)
\(\Longrightarrow\)     \(-\frac { 1 }{ 2 } +2\le 3x-2+2\le \frac { 1 }{ 2 } +2\)  [adding 2 on each term]
\(\Longrightarrow\) \(\frac { 3 }{ 2 } \le 3x\le \frac { 5 }{ 2 } \)
\(\Longrightarrow\) \(\frac { 3 }{ 2 } \times \frac { 1 }{ 3 } \le \frac { 3x }{ 3 } \le \frac { 5 }{ 2 } \times \frac { 1 }{ 3 } \)     [dividing each term by 3]
\(\Longrightarrow\) \(\frac { 1 }{ 2 } \le x\le \frac { 5 }{ 6 } \)  i.e.  \(x\in \left[ \frac { 1 }{ 2 } ,\frac { 5 }{ 6 } \right] \) 

Hence, the required solution set is \(\left[ \frac { 1 }{ 2 } ,\frac { 5 }{ 6 } \right] \)

We have, - 3 \(\le \) 4 - \(\frac { 7x }{ 2 } \) \(\le \) 18.
On subtracting 4 from each term, we get
- 3 - 4 \(\le \) 4 - \(\frac { 7x }{ 2 } \) - 4 \(\le \) 18 - 4 \(\Longrightarrow \) - 7 \(\le \) - \(\frac { 7x }{ 2 } \) \(\le \) 14
On multiplting each term by (\(\frac { -2 }{ 7 } \)), we get
- 7 (\(\frac { -2 }{ 7 } \)) \(\ge \) \(\frac { -7 }{ 2 } \) x \(\times \) (\(\frac { -2 }{ 7 } \)) \(\ge \) 14 \(\times \) (\(\frac { -2 }{ 7 } \))
[while multiplying each term by the same negative number, then the sign of inequalities will get change]
\(\Longrightarrow \) 2 \(\ge \) x \(\ge \) - 4 or  - 4 \(\le \) x \(\le \) 2  or  x \(\in \) [- 4, 2]
Hence, solution set of given system of inequations is [- 4, 2].

We have, 3x - 5 < x + 7
\(\Rightarrow \) 3x -5 + 5 < x + 7 + 5  [adding 5 on both sides]
\(\Rightarrow \) 3x < x + 12
\(\Rightarrow \) 3x -x < x + 12 - x [Subtracting x from both sides]
\(\Rightarrow \) 2x < 12
\(\Rightarrow \) \(\frac { 2x }{ 2 } \) < \(\frac { 12 }{ 2 } \)
\(\Rightarrow \) x < 6
Now if  x is a whole number, then the solution set {0, 1, 2, 3, 4, 5}

We have,   6 \(\le \) - 3(2x - 4) < 12 or  6 \(\le \) - 6x + 12 < 12
On subtracting 12 from each term, we get
6 - 12 \(\le \) - 6x + 12 - 12 < 12 - 12
\(\Longrightarrow \) - 6 \(\le \) -6x < 0
On dividing each term by - 6, we get \(\frac { -6 }{ -6 } \) \(\ge \) \(\frac { -6x }{ -6 } >\frac { 0 }{ -6 } \)
[ while dividing each term by the same negative number, then sign of inequalities will get change]
\(\Longrightarrow \)  1 \(\ge \) x > 0
Which can be written as 0 < x \(\le \) 1.
Hence, solution set of given system of inequalities is (0, 1].

We have,3x - 5 < x + 7
\(\Rightarrow \) 3x -5 + 5 < x + 7 + 5 [adding 5 on both sides]
\(\Rightarrow \) 3x < x + 12
\(\Rightarrow \) 3x -x < x + 12 - x [Subtracting x from both sides]
\(\Rightarrow \) 2x < 12
\(\Rightarrow \) \(\frac { 2x }{ 2 } \) < \(\frac { 12 }{ 2 } \)
\(\Rightarrow \) x < 6
Now if x is natural number, then the solution set is {1, 2, 3, 4, 5}