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

Let the length of the shortest piece be x cm, so that the lengths of second and third piece are(x+3) cm and 2x cm respectively.Then x+ (x+3)+2x \(\le \) 91
and 2x\(\ge \)(x+3)+5
From inequality (i), we get
\(4x+3\le 91\)
\( \Rightarrow 4x+3-3\le 91-3[subtarcting\quad 3\quad from\quad both\quad sides]\)
\(\Rightarrow 4x\le 88[dividing\quad both\quad sides\quad by\quad 4\)
\( \Rightarrow \frac { 4x }{ 4 } \le \frac { 88 }{ 4 } \)
\(\Rightarrow x\leq 22\)
From inequality (ii), we get
\(2x\ge x+8\)
\( \Rightarrow 2x-x\ge x+8-x[subtracting\quad x\quad from\quad both\quad sides]\)
\( \Rightarrow x\ge 8\)
From inequality (iii) and (iv), we get
\(8\le x\le 22\)
Hence, the shortest piece must be atleast 8 cm long but not more than 22cm long

Since the solution is to be kept between 68°F and 77°F, 68 < F < 77
\(\text { Putting } \mathrm{F}=\frac{9}{5} \mathrm{C}+32 \text { , we abtain }\)
\(68<\frac{9}{5} \mathrm{C}+32<77\)
\(\Rightarrow 68-32<\frac{9}{5} \mathrm{C}<77-32\)
\(\Rightarrow 36<\frac{9}{5} \mathrm{C}<45\)
\(\Rightarrow 36 \times \frac{5}{9}<\mathrm{C}<45 \times \frac{5}{9}\)
\(\Rightarrow 20<\mathrm{C}<25\)
Thus,the required range of temperature in degree celsius is between 20° and 25°

Let x be the smaller of the two consecutive odd natural number, so that the other one is x +2. Then, we should have
x > 10 ... (1)
and x + ( x + 2) < 40 ... (2)
Solving (2), we get
2x + 2 < 40
i.e., x < 19 ... (3)
From (1) and (3), we get
10 < x < 19
Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required possible pairs will be
(11, 13), (13, 15), (15, 17), (17, 19)

(10, 12), (12, 14)

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

At point (1, 0), given inequality 2(1) +3(0)>0 \(\Rightarrow\)2>0.which in true. So, shading is towards at that point

Write the given inequalities as \(|x-2|\ge 1\) and \(|x-2|\le 3\) Solve each inequality separately and then take the intersection of their solution sets
Ans. [-1, 1] \(\cup \) [3, 5]

Let x litres of 30% acid solution is required to be added. Then
Total mixture = (x + 600) litres
Therefore 30% x + 12% of 600 > 15% of (x + 600)
and 30% x + 12% of 600 < 18% of (x + 600)
or \(\frac{30 x}{100}+\frac{12}{100}(600)>\frac{15}{100}(x+600) \)
and \(\frac{30 x}{100}+\frac{12}{100}(600)<\frac{18}{100}(x+600)\)
or 30x + 7200 > 15x + 9000
and 30x + 7200 < 18x + 10800
or 15x > 1800 and 12x < 3600
or x > 120 and x < 300,
i.e. 120 < x < 300
Thus, the number of litres of the 30% solution of acid will have to be more than 120 litres but less than 300 litres.

We have 7x + 3 < 5x + 9 or
2x < 6 or x < 3
The graphical representation of the solutions are

We are given 30 x < 200
or \(\frac{30 x}{30}<\frac{200}{30}\) (Rule 2), i.e., x < 20 / 3.
(i) When x is a natural number, in this case the following values of x make the statement true.
1, 2, 3, 4, 5, 6.
The solution set of the inequality is {1,2,3,4,5,6}.
(ii) When x is an integer, the solutions of the given inequality are..., – 3, –2, –1, 0, 1, 2, 3, 4, 5, 6
The solution set of the inequality is {...,–3, –2,–1, 0, 1, 2, 3, 4, 5, 6}.

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

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.

Let x and x + 2 be two consecutive odd positive integers
Then x + 2 < 10 and  x + x 2 > 11
\(\Rightarrow\) x < 8  and  2x + 2 > 11
\(\Rightarrow\) x < 8 and  2x > -2 + 11
\(\Rightarrow\) x < 8 and  2x > 9
\(\Rightarrow\) x < 8 and  x >\(\frac { 9 }{ 2 } \)
\(\Rightarrow\)\(\frac { 9 }{ 2 } \) \(\Rightarrow\) x = 5 and 7
Thus required pairs of odd positive integers are (5, 7) and (7, 9).