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 \(\left( -\infty ,\frac { 63 }{ 10 } \right) \)

(-1, \(\infty\))

[3,\(\infty\))

Here 3x + 8 > 2
\(\Rightarrow\) 3x > 2 - 8 \(\Rightarrow\) 3x > - 6
Dividing both sides by 3, we have x > - 2
(i) When x is an integer then values of x that make the statement true are - 1, 0, 1, 2, 3, ... The solution set of inequality is {-1, 0,1,2, 3, ... }.
(ii) When x is a real number. The solution set of inequality is x \(\epsilon \) (-2, \(\infty\))

\(\frac { 73+67+72+x }{ 4 } \ge 75\)
88 marks

Rahul must score between 52 and 77 marks.

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

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].

On dividing both sides by 24, we get
= \(\frac { 24x }{ 24 } \) < \(\frac { 100 }{ 24 } \) \(\Rightarrow \) x < \(\frac { 25 }{ 6 } \)
When x is an integer.
In this case, solutions of given inequality are......, -4, -3, -2, -1,0,1,2,3,4
Hence , the solution  set of inequality is {........, -4,-3,-2,-1,0,1,2,3,4}.

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

Given inequality is  \(y+8\ge 2x.\)
In equation form, it can be written as y+8=2x or 2x-y=8
On putting y=0, we get 2x-0=8 \(\Rightarrow \) x=4
\(\therefore \) Line 2x-y=8 cuts the X-anix at A(4,0).
On putting x=0 we get,
\(2\times 0-y=8\quad \Rightarrow \quad y=-8\)
\(\therefore \) Line 2x-y=8 cuts the Y-axis at B(0,-8).
On joining the points A(4,0) and B(0,-8) by a dark line,we get the line AB.
[\(\therefore \) given inequality has sign \(\ge \) , so we draw dark line]
Let us check whether O(0,0) satisfy the inequality or not and shade the suitable region.
On putting x=y=0 in y+8\(\ge \)2x, we get
0+8\(\ge \)2.0
\(\Rightarrow \)8 \(\ge \) 0, which is true.
So shade the region containing (0,0) i.e. the region above the line AB.

(x - 3) (x - 5) > 0
case I When both are positive
 \(\therefore x>3\quad and\quad x>5 \Rightarrow x>5\)
case II When both are negative
\(x-3<0\quad and\quad x-5<0 \Rightarrow x<3\quad and\quad x<5 \Rightarrow x<3\)
Ans \((-\infty ,3) \cup (5,\infty )\)

Consider the equation of the line x+2y=8. We opbserve that the shaded region and the origin lie on the same side of the line.So, the corresponding inequality is \(x+2y<8.\)
Now, consider the equation of the line x-y=1.
We observe that the shaded region and orgin both lie on the same side. So, corresponding inequality is \(x-y<1.\)
Consider the equation of the line 2x+y=2.We observe that the shaded region and origin lie on opposite side of and is on the right side of Y-axis.So, \(x\ge 0\ and\ y\ge 0.\)
Thus, the linear inequalities corresponding to the given solution ser are
\(x+2y<8,2x+y>2\)
\(x-y>1\)
\(x\ge 0,y\ge 0\)

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°

Between 59^o F and 77^oF

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

2x + y≥ 6 … (1)
3x + 4y ≤ 12 … (2)
The graph of the lines, 2x + y= 6 and 3x + 4y = 12, are drawn in the figure below.
Inequality (1) represents the region above the line, 2x + y= 6 (including the line 2x + y= 6), and inequality (2) represents the region below the line, 3x + 4y =12 (including the line 3x + 4y =12).
Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the respective lines as follows.