\(8.2<\frac { 8.48+8.35+x }{ 3 } <8.5\)
Between 7.77 and 8.77

(11, 13), (13, 15), (15, 17)

(10, 12), (12, 14)

Let x litre of 2% boric acid solution be added to 640litres of 8%boric acid solution. Then
Total quantity of mixture = (640 + x) litres
Total boric acid in (640 + x) litres of mixture
= \(\frac { 2x }{ 100 } +\frac { 8 }{ 100 } \times 640=\frac { x }{ 50 } +\frac { 256 }{ 5 } \)
It is given that the resulting mixture must be more than 4% but less than 6% boric acid
\(\therefore\) \(\frac { 4 }{ 100 } (640+x)<\frac { x }{ 50 } +\frac { 256 }{ 5 } <\frac { 6 }{ 100 } (640+x)\)
\(\Rightarrow\)  \(\frac { 640+x }{ 25 } <\frac { x+2560 }{ 50 } <\frac { 1920+3x }{ 50 } \)
\(\Rightarrow\) 1280 + 2x \(\Rightarrow\) 1280 + 2x \(\Rightarrow\) x < 1280 and -2x < -640
\(\Rightarrow\) x < 1280 and x > 320
\(\Rightarrow\) 320 Thus 2% boric acid solution must be more than 320 litres but less than 1280 litres.

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 \(\frac { 2x-3 }{ 4 } +19\ge 13+\frac { 4x }{ 3 } \)
\(\frac { 2x-3 }{ 4 } -\frac { 4x }{ 3 } \)  \(\ge\) 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 \(\left( -\infty ,\frac { 63 }{ 10 } \right) \)

[3, \(\infty\))

The given inequality is y < -2
Draw the graph of line y = -2.

Putting (0, 0) in the given inequation, we have 0 < -2 which is false
\(\therefore\) half plane of y < -2 is away from origin

[-26,\(\infty\))

(-\(\infty\), -50)

(4,\(\infty\))

(3,\(\infty\)]

The given inequality is 2x +y \(\ge\) 4
Draw the graph of the line 2x + y = 4

Table of values satisfying the equation 2x +y = 4

x	2	1
y	0	2

Putting (0,0) in the given inequation, we have 2 x 0 + 0 \(\ge\) 0 \(\Rightarrow\) 0 \(\ge\)4, which is false
\(\therefore\) Half plane of 2x +y \(\ge\)4 is awy from origin
Also the given inequality is x + Y \(\le\)3.
Draw the graph of the line x + y = 3.
Table of values satisfying the equation x +y = 3

x	2	1
y	1	2

The putting (0,0) in the given inequation, we have 0 + 0 \(\le\) 3 \(\Rightarrow\) 0 \(\le\) 3, Which is true

\(\therefore\) Half plane inequality is 2x - 3y \(\le\)6
Table of values satisfyinh the equation 2x -3y = 6

x	0	3
y	-2	0

Putting (0, 0) in the given inequation, we have 2 x 0 -3 x 0 \(\le\) 6 \(\Rightarrow\) 0 \(\le\) 6, which is true
\(\therefore\) Half plane of 2x -3y \(\le\)6 is towards origin

The given inequality is 3x +4y \(\le\)60
Draw the graph of the line 3x +4y = 60
Table of values satisfying the equation 3x +4y = 60

x	8	12
y	9	6

Putting (0, 0) in the given inequation, we have 3 x 0 + 4 x 0 \(\le\) 60 \(\Rightarrow\) 0 \(\le\)60, which is true
\(\therefore\) Half plane of 3x+4y \(\le\) 60 is towards origin
Also the given inequality is x + 3y \(\le\) 30
Draw the graph of the line x +2y = 30
Table of values satisfying the equation x +3y = 30

x	0	9
y	10	7

Putting (0, 0) in the given inequation, we have 0 + 3 x 0 \(\le\) 30 \(\Rightarrow\) 0 \(\le\) 30, which is true
\(\therefore\) Half plane of c +3y \(\le\) 30 is towards origin