We have \(7\le \frac { \left( 3x+11 \right) }{ 2 } \le 11\)
\(\Rightarrow\) 14 \(\le\) 3x +11 \(\le\) 22  \(\Rightarrow\)  3 \(\le\) 3x \(\le\) 11
\(\Rightarrow\) 1 \(\le\) x \(\le\) \(\frac { 11 }{ 3 } \)

More than 230 litres but less than 920 litres]

(-7,-3)

(-1, \(\infty\))

(i) 1,2,3
(ii) ( , -3, -2, -1, 0, 1, 2, 3)

Here 5x - 3 \(\ge\) 3x - 5
\(\Rightarrow\) 5x - 3x \(\ge\) -5 + 3 \(\Rightarrow\) 2x \(\ge\) -2
Dividing both sides by 2, we have x \(\ge\) -1
The solution set is [-1, \(\infty\))
The representation of the solution set on the number line is

Here 3x - 2 < 2x + 1
\(\Rightarrow\) 3x - 2x < 1 + 2 \(\Rightarrow\) x < 3
The solution set is (-\(\infty\), 3)
The representation ofthe solution set on the number line is

Here 3(2 - x)\(\ge\) 2 (1-x)
\(\Rightarrow\) 6- 3x \(\ge\) 2 - 2x \(\Rightarrow\) - 3x + 2x \(\ge\) 2 - 6
\(\Rightarrow\) -x \(\le\) - 4
Dividing both sides by -1, we have
 \(\frac { -x }{ -1 } <\frac { -4 }{ -1 } \) \(\Rightarrow\) \(x\le 4\)
Thus the solution set is (-\(\infty\), 4]

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]

Here 3x - 7 > 5x - 1
\(\Rightarrow\) 3x - 5x > - 1 + 7 \(\Rightarrow\) - 2x > 6
Dividing both sides by -2, we have
\(\frac { -2x }{ -2 } >\frac { 6 }{ -2 } \) \(\Rightarrow\) x < -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.

2(2x+3)−10<6(x−2)
⇒4x+6−10<6x−12
⇒4x−4<6x−12
⇒−4+12<6x−4x
⇒8<2x
⇒4 Thus, all real numbers x, which are greater than 4, are the solution of the given inequality.
Hence, the solution set of the given inequality is (4, \(\infty \))

{-1, 0, 1, 2, 3,...}

We have 5(2x-7) -3 (2x + 3) \(\le\)0 and 2x +19 \(\le\) 6x +47
From inequality (i),we get
5(2x-7) -3 (2x + 3) \(\le\)0
\(\Rightarrow\)  10x - 35 - 6x - 9 \(\le\) 0 and -4x \(\le\) 28
\(\Rightarrow\) -4x -44 \(\le\) 0  and x \(\ge\) -7 [adding 44 on both sides]
\(\Rightarrow\) 4x\(\le\)44 and x \(\ge\) -7
\(\Rightarrow\) x\(\le\)11  and x\(\ge\) -7 [dividing both sides by 4]
\(\therefore\) The solution set is (- \(\infty\), 11].
From inequality (ii), we get
\(2 x+19 \leq 6 x+47\)
\(\Rightarrow \quad 2 x+19-2 x \leq 6 x+47-2 x\)  [subtracting 2x from both sides
\(\Rightarrow 19 \leq 4 x+47 \)
\(\Rightarrow 19-47 \leq 4 x+47-47\)  [subtracting 47 from both sides]
\(\Rightarrow \quad-28 \leq 4 x \text { or } 4 x \geq-28\)
\(\Rightarrow \frac{4 x}{4} \geq \frac{-28}{4}\) [dividing both sides by 4]
\(\Rightarrow x \geq-7\)
\(\therefore \text { The solution set is }[-7, \infty) \text { . }\)
Now, let us draw the graphs of the solutions of both inequalities on number line.

It can be seen that the values of x, which are common to both are lying in the interval [-7,11].
Hence, the solution set of given system of inequations is [- 7, 11] and this can be represented graphically on the number line as