2x-1 >\(\frac { 2x+7 }{ 3 } \) > 2 \(\Rightarrow \) -22 \(\le \) -3 > 2x + 7 < 6
\(\Rightarrow \)  6x - 3 > 2x + 7 and 2x + 7> 6
\(\Rightarrow \)  4x > 10 and 2x > -1 \(\Rightarrow \) x > \(\frac { 5 }{ 2 } \) and x > -\(\frac { 1 }{ 2 } \)
[\(\frac { 5 }{ 2 } \), \(\infty \)].

(4, 9)

Given that, 30 < C < 35 ......(i)
and F = \(\frac { 9 }{ 5 } \) C + 32 \(\Rightarrow \)  F - 32 = \(\frac { 9 }{ 5 } \) C \(\Rightarrow \) C = \(\frac { 5}{ 9 } \)(F - 32)
On putting the value of C inequality (i), we get
30 < \(\frac { 5}{ 9 } \)(F - 32)< 35 \(\Rightarrow \) 30\(\times \frac { 9 }{ 5 } \)<\(\frac { 5}{ 9 } \)(F - 32 ) \(\times \) \(\frac { 9 }{ 5 } \)< 35 \(\times \) \(\frac { 9 }{ 5 } \) [multiplying by \(\frac { 9 }{ 5 } \) on each term]
\(\Rightarrow \) 6 \(\times \)9\(\times \)9 \(\Rightarrow \)  54 < F - 32 < 63
\(\Rightarrow \) 54 + 32  [adding 32 in each term]
\(\Rightarrow \) 86 < F < 95
Thus, the required range of temperature is between 86^oF and 95^oF.
His temperature is not normal. Normal temperature of body is 98 ^o6 F. So, he need medical attention.

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

Also, IQ = \(\frac { MA }{ CA } \) \(\times \) 100
\(\therefore \) 80 \(\le \) \(\frac { MA }{ CA } \)\(\times \) 100 \(\le \)140
\(\Rightarrow \) 80 \(\le \)\(\frac { MA }{ 12 } \) \(\times \) 100 \(\le \) 140
9.6 \(\le \) MA \(\le \) 16.8
range of mental age is[9.6,16.8]

It is given that, 68 < F < 77
On putting F = \(\frac { 9 }{ 5 } \)C + 32, we get
68 < \(\frac { 9 }{ 5 } \)C + 32 < 77
\(\Rightarrow \) 68 - 32 < \(\frac { 9 }{ 5 } \)C < 77 -32 [subtracting 32 from each term]
\(\Rightarrow \) 36 < \(\frac { 9 }{ 5 } \)C < 45
\(\Rightarrow \) \(36 \times \frac{5}{9}<C<45 \times \frac{5}{9}\)  [multiplying \(\frac { 5 }{ 9 } \) in each term]
\(\Rightarrow \)  4\(\times\)5 < C < 5 x 5  \(\Rightarrow \)   20 < C < 25
Hence, the required range of temperature is between 20^o C and 25^o C.

Use \(|x|\le a\Longrightarrow -a\le x\le a\).
[-1,3]

We have, \(\frac { 5-3x }{ 3 } \) \(\le \)  \(\frac { x }{ 6 } \) - 5.    
\(\Rightarrow \) \(\frac { 5-3x }{ 3 } \) \(\le \) \(\frac { x - 30}{ 6 } \)
\(\Rightarrow \) \(\frac { 5-3x }{ 3 } \)  \(\times \) 6 \(\le \) \(\frac { x - 30}{ 6 } \) \(\times \)6 [multiplying both sides by 6 ]
\(\Rightarrow \) 2(5-3x) \(\le \) x -30
\(\Rightarrow \)  10 - 6x \(\le \) x -30
\(\Rightarrow \)  10 - 6x - 10  \(\le \) x -30 - 10 [subtracting 10 from both sides]
\(\Rightarrow \)  -6x \(\le \) x -40
\(\Rightarrow \) -6x - x \(\le \) x - 40 - x [subtracting x from both sides]
\(\Rightarrow \)   -7x \(\le \) -40
\(\Rightarrow \)   -7x  \(\ge \) 40  [multiplying by -1 on both sides] 
\(\Rightarrow \) \(\frac {7 x }{ 7 } \) \(\ge \)\(\frac { 40}{ 7 } \) [dividing by 7 on both sides]
\(\Rightarrow \) x  \(\ge \) \(\frac { 40}{ 7 } \)
i.e   x \(\varepsilon\) [\(\frac { 40}{ 7 } \), \(\infty \))
Hence, the required solution set is [\(\frac { 40}{ 7 } \), \(\infty \)).

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].

For the equation x+y=20, the shaded area and origin both lies on the same side of the live, therefore the compounding inequality is \(x+y\le 20\).
Similarly, \(3x+2y\le 48\)
Also,the shaded portion lines in Ist quadrant.
\(\therefore \quad x\ge 0\quad and\quad y\ge 0\)
Ans. \(x+y\le 20,3x+2y\le 48,x\ge 0,y\ge 0\)

Consider 2x+3y=3. We observe that, the shaded region and the origin lie on opposite side of this line.Therefore,\(2x+3y\ge 3\) is the linear inequality correspondingto the line 2x+3y=3.
Consider  3x+4y=18. We observe that, the shaded region and the origin lie on the same side of this line.Therefore, \(3x+4y\le 18\) is the linear inequality corresponding to the line 3x+4y=18.
Consider -7x+4y=14.It is clear from the figure that the shaded region and the origin lie on the same side of this line.Therefore, \(-7x+4y\le 14\) is the linear inequality correspondiong to the line -7x+4y=14.
Consider x-6y=3.It is clear form the figure that the shaded region and origin lie on the same side of this line. Therefore, \(x-6y\le 3\)is the linear inequality corresponding to the line x-6y=3.Also, the shaded region line is the first quadrant only.Therefore,\(x\ge 0,y\ge 0.\)
Thus, the linear inequlities corresponding to the given solution set are
\(2x+3y\ge 3,3x+4y\le 18,\quad -7x+4y\le 14,\quad x-6y\le 3\quad ,x\ge 0,y\ge 0.\)

\(\frac { 2x+1 }{ 7x-1 } -5>0,\frac { x+7 }{ x-8 } -2>0 \Rightarrow \frac { -33x+6 }{ 7x-1 } >0\frac { -x+23 }{ (x-8) } >0\)
\(\Rightarrow (-33x+6)(7x-1)>0,(-x+23)(x-8)>0\)
\(\\ \Rightarrow x>7,x<23\)

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. \(\left[ \frac { 1 }{ 3 } ,1 \right] \)