We have, 4x-y>0
Its equation form i9s 4x-y=0
On putting y=0 in Eq.(i) we get 
4x-0=0\(\Rightarrow\)x=0
On putting y=0 in Eq. (i) we get 
4(1)-y=0\(\Rightarrow\)y=4

x	1	0
y	4	0

Thus, the 4x-y=0 passes throughy the points A(0, 0) and B(1, 4) by adotted line.

Now , the point, say (1, 2) not lying on the line, to check whether satisfies the given linear or not.
At(1, 2) 4(1)-2>0
\(\Rightarrow\)2>0, which is true.
(1, 2) satisfies the given inequality, so we shade the region which contains (1, 2). Thus shaded region (excluding all pints on the line )gives the solution region.

Given quality 5x+2y\(\le\)10
Corresponding equation of line is
5x+2y=10 
On putting x=0, we get 
5(0) +2y = 10 \(\Rightarrow\)y=5 
Thus, the line intersect the Y-axis at point (0, 5)
On putting y=0, we get 
5x+2(0) =10\(\Rightarrow\)x=2

x	0	2
y	5	0

THus, the line untersect the X-axis at poit(2, 0) Join the points A(0,5) and B (2, 0) with a dark line
[\(\because\) given inequality has sign'\(\le\) ']
 

On putting (0,  0) in given inequality, we get 
5\(\times\)0+2\(\times\)0\(\le\)10
\(\Rightarrow\)0 \(\le\)10, which is true
So, the half plane of 5x+2y\(\le\)10 contains origin.Now, shade the half plane which contain (0,0)
Here, the shaded region including the points on the line represent the solution region of given inequality

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 x + 2y \(\le\)10.
Draw the graph of the line x + 2y = 10.
Table of values satisfying the equation x+ 2y = 10

x	2	4
y	4	3

Putting (0, 0) in the given inequation, we have 0 + 2 x 0 \(\le\) 10 \(\Rightarrow\) 0 \(\le\) 10, which is true

\(\therefore\) Half plane ofx + 2y \(\le\)10is towards origin.
Also the given inequality is x + y \(\ge\) 1.
Draw the graph of the line x + y = 1.
Table of values satisfying the equation x +y = 1

x	0	1
y	1	0

Putting (0, 0) in the given inequation, we have 0+ 0 \(\ge\) 1\(\Rightarrow\) 0 \(\ge\) 1, which is false.
\(\therefore\) Half plane of x+ y \(\ge\) 1is away from origin.
Also the given inequality is x - y \(\le\) O.
Draw the graph of the line x - y = O.
Table of values satisfying the equation x-y =0

x	1	2
y	1	2

Putting (2, 0) in the given inequation, we have
2 - 0 \(\le\) 0 \(\Rightarrow\) 2 \(\le\) 0 which is false.
\(\therefore\) Half plane ofx - y \(\le\) 0 is away from origin.

The given inequality is 2x +y \(\ge\) 4
Draw the graph of the line 2x + y = 4

Table of values satisfying the equation 2x +y = 4

x	2	1
y	0	2

Putting (0,0) in the given inequation, we have 2 x 0 + 0 \(\ge\) 0 \(\Rightarrow\) 0 \(\ge\)4, which is false
\(\therefore\) Half plane of 2x +y \(\ge\)4 is awy from origin
Also the given inequality is x + Y \(\le\)3.
Draw the graph of the line x + y = 3.
Table of values satisfying the equation x +y = 3

x	2	1
y	1	2

The putting (0,0) in the given inequation, we have 0 + 0 \(\le\) 3 \(\Rightarrow\) 0 \(\le\) 3, Which is true

\(\therefore\) Half plane inequality is 2x - 3y \(\le\)6
Table of values satisfyinh the equation 2x -3y = 6

x	0	3
y	-2	0

Putting (0, 0) in the given inequation, we have 2 x 0 -3 x 0 \(\le\) 6 \(\Rightarrow\) 0 \(\le\) 6, which is true
\(\therefore\) Half plane of 2x -3y \(\le\)6 is towards origin

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

x	8	12
y	9	6

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

x	0	9
y	10	7

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 x + y \(\le\) 6
Draw the graph of the line x + y = 6.

Table of value satisfying the equation x + y = 6

x	3	4
y	3	2

Putting (0, 0) in the given inequation, we have
0 + 0 \(\le\) 6 \(\Rightarrow\) 0 \(\le\) 6, which is true
Half plane of x +y \(\le\) 6 is towards origin
Also the given unequality is x +y \(\ge\)4
Draw the graph of the line x + y = 4
Table of values satisfying the equation x + y = 4

x	2	1
y	2	3

Putting (0, 0) in the given inequation, we have 0 + 0 \(\ge\) 4 \(\Rightarrow\) 0 \(\ge\) 4, which is false
Hald plane of x +y \(\ge\) 4 is awy from 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.