We have 6 \(\le\) -3 (2x-4) <12
-2 \(\ge\) (2x -4) > -4  \(\Rightarrow\) 2 \(\ge\) 2x >0
1\(\ge\)c > 0 \(\Rightarrow\) 0 < x \(\le\) 1

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

More than 230 litres but less than 920 litres]

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

More than or equal to 52 but less than 77

[3,\(\infty\))

Here \(\frac { x }{ 4 } <\frac { (5x-2) }{ 3 } -\frac { (7x-3) }{ 5 } \)
\(\Rightarrow\)  \(\frac { x }{ 4 } <\frac { 5x }{ 3 } -\frac { 2 }{ 3 } -\frac { 7x }{ 5 } +\frac { 3 }{ 5 } \)
\(\Rightarrow\) \(\frac { x }{ 4 } <\frac { 5x }{ 3 } +\frac { 7x }{ 5 } <\frac { -2 }{ 3 } +\frac { 3 }{ 5 } \)
\(\Rightarrow\) \(\frac { 15x-100x+84x }{ 60 } \) < \(\frac { -10+9 }{ 15 } \)
\(\Rightarrow\) \(\frac { -x }{ 60 } <\frac { -1 }{ 15 } \)
Multiplying both sides by 60, we have -x < - 4
Dividing both sides by -1, we have x > 4
Thus the solution (4,\(\infty\))

Here 3(x-1) \(\le\) 2(x-3)
\(\Rightarrow\) 3x-3 \(\le\) 2x - 6 \(\Rightarrow\) 3x - 2x \(\le\)-6 +3
\(\Rightarrow\) x \(\le\) - 3
Thus the solution set is (-\(\infty\) ,-3]

Let at s km below the Earth's surface, the temperature is between 155^oC and 205^oC.
\(\therefore \)  155 < T(s) < 205
\(\Rightarrow \) 155 < 30 + 25(s - 3)< 205
Ans. 8< s < 10

Let x be the marks obtained by Ravi in the third unit test.
Since the student should have an average of at least 60 marks.
\( \frac{70+75+x}{3} \geq 60 \)
\( \Rightarrow 145+x \geq 180\)
\( \Rightarrow x \geq 180-145 \)
\( \Rightarrow x \geq 35 \)
Thus the student must obtain a minimum of 35 marks to have an average of at least 60 marks.

Use \(|x|\ge a\Longrightarrow x\ge a\) or \(x\le -a\)
(\(-\infty\) \(\frac { -3 }{ 2 } \)] \(\cup \) [3, \(\infty \))

Use \(|x|\ge a\Longrightarrow x\ge a\) or  \(x\le -a\)
(-\(\infty \), -13] \(\cup \) [7,\(\infty \))

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].

(-2, \(\infty \) )

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 false statement.Therefore, the given half plane does not contain the origin.

For the equation 7x+10y=70, shaded area and origin both lie on the same side of the line, therefore corresponding linear inequality is 7x+10y\(\le 70\). For the equation x+y=4, shaded area and origin lie on the opposite side of the line, therefore corresponding linear inequality is \(x+y\ge 4.\) Also, the solution region is above the X-axis and on the right side of Y-axis, therefore we have \(x\ge 0 \ and \ y\ge 0\) .Hence, linear inequalities will be
\(7x+10y\le 70\)
\(x+y\ge 4\)
\(x,y\ge 0\)

More than 230 liters but less than 920 liters.

We have, \(\frac { 4 }{ x+1 } \le 3\le \frac { 6 }{ x+1 } \)
\(\Longrightarrow \)\(4\le 3\left( x+1 \right) \le 6\quad \left[ \because x+1\neq 0\Longrightarrow x\neq -1 \right] \)
\(\Longrightarrow \)\(\frac { 4 }{ 3 } \le x+1\le 2\)
\(\Longrightarrow \)\(\frac { 4 }{ 3 } -1\le x\le 2-1\Longrightarrow \frac { 1 }{ 3 } \le x\le 1\)
Ans. \(\left[ \frac { 1 }{ 3 } ,1 \right] \)

Let x litres ofwater be added to 1125 litres of 45% acid solution.
Then total quantity of mixture = (1125 + x) litres
\(\frac { 45 }{ 100 } \times 1125+0\times \frac { x }{ 100 } >\frac { 25 }{ 100 } \times \left( 1125+x \right) \) and \(\frac { 45 }{ 100 } \times 1125+0\times \frac { x }{ 100 } <\frac { 30 }{ 100 } \times \left( 1125+x \right) \)
Combining the above inequations, we get
\(\frac { 25 }{ 100 } \times 100\le \frac { 2025\times 100 }{ 4(1125+x) } \le \frac { 30 }{ 100 } \times 100\)
\(\Rightarrow\) \(25\le \frac { 50625 }{ 1125+x } \le 30\)
\(\Rightarrow\) \(25\le \frac { 50625 }{ 1125+x } \) and \(\frac { 50625 }{ 1125+x } \le 30\)
\(\Rightarrow\) 28125 + 25x \(\le\) 50625 and 50625 \(\le\)33750 + 30x
\(\Rightarrow\) 25x \(\le\) 22500 and 30x \(\ge\) 1687.5
\(\Rightarrow\) x \(\le\) 900 and x \(\ge\) 562.5
\(\Rightarrow\) 562.5 \(\le\) x \(\le\) 900

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

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

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

The given inequality is 2x + y \(\ge\) 8
Draw the graph of the line 2x +y = 8

Table of values satisfying the eqaution we have
2x +y = 8

Putting (0,0) in the given inequation, we have
2 x 0 + 0 \(\ge\) 8 \(\Rightarrow\) 0 \(\ge\)8, which is false
\(\therefore\) Half plane of 2x +y  \(\ge\) 8 is away from origin
Also the given inequality is x + 2y 
Draw the graph of the line x +2y = 10
Table of the values satisfying the equation x + 2y = 10

Putting (0,0) in the given inequation , we have 0 + 2 x 0 \(\ge\) 10 \(\Rightarrow\) 0 \(\le\) 10, Which is false
\(\therefore\) half plane of x +2y \(\ge\) 10 is away from origin 

The given inequality ix x + y \(\ge\) 4
Draw the gtaph of the line x + y = 4

Total of value satisfying the equation x +y = 4

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

Putting (3, 0) in the given inequation, we have
2 x 3 - 0 > 0 => 6 > 0, which is true.
\(\therefore\) Half plane of 2x - y > 0 containing (3,0)