Solve the following systems of inequations graphically:
(i) 2x+y≥8,x+2y≥8,x+y≤6
(ii) 12x+12y≤840,3x+6y≤300,8x+4y≤480 x≥0,y≥0
(iii) x+2y≤40,3x+y≥30,4x+3y≥60,x≥0,y≥0
(iv) 5x+y≥10,2x+2y≥12,x+4y≥12,x≥0,y≥0
(i) We have,
2x+y≥8,x+2y≥8,x+y≤6
Converting the given inequation into equations,we obtain,
2x+y=8,x+2y=8,and x+y=6
Region represented by 2x+y≥8
Putting x=0 in2x+y=8,we get y=8
Putting y=0 in 2x+y=8,we get x=82=4
∴ The line 2x+y=8 meets the coordinates axes at (0,8) and (4,0).Join these points by a thick line.
Now ,putting x=0 and y=0 in 2x+y≥8,we get 0≥8
This is not possible.
∴ We find that (0,0) is not satisifes the inequation 2x+y≥8
So,the portion not containing the origin is represented by the given inequation
Region represented by x+2y≥8
Putting x=0 in x+2y=8,we get y=82=4
Putting y=0 in x+2y=8,we get x=8
∴ The line x+2y=8 meets the coordinates axes at (0,4) and (8,0).Joining these points by a thick line.
Now ,putting x=0 and y=0 in x+2y≥8,we get 0≥8
This is not possible.
∴ We find that (0,0) is not satisifes the inequation x+2y≥8
So,the portion not containing the origin is represented by the given inequation.
Region represented by x+y≤6
Putting x=0 in x+y=6,we get y=6
Putting y=0 in x+y=6,we get x=6
∴ The line x+y=6 meets the coordinates axes at (0,6) and (6,0).
Joining these points by a thick line.
Now ,putting x=0 and y=0 in x+y≤6,we get 0≥6
Therefore,(0,0) satisfies x+y≤6,so the portion containing the origin is represented by the given inequation.The common region of the above three regions represents the solution set of the given inequations as shown below:
(ii) We have,
12x+12y≤840,3x+6y≤300,8x+4y≤480 x≥0,y≥0
Converting the given inequation into equations,we obtain,
12x+12y= 840, 3x+6y= 300,8x+4y=480, x = 0 and y=0
Region represented by 12x+12y≤840
Putting x=0 in 12x+12y= 840,we get y=84012=70
Putting y=0 in by 12x+12y≤840,we get x=84012=70
∴ The line 12x+12y= 840meets the coordinates axes at (0,70) and (70,0).Join these points by a thick line.,
Now ,putting x=0 and y=0 in 12x+12y≤840,we get 0≤840
Therefore,(0,0) satisfies the inequality 12x+12y≤840,
So,the portion not containing the origin is represented by the given inequation 12x+12y≤840
Region represented by 3x+6y≤300
Putting x=0 in 3x+6y≤300,we get y=3006=50
Putting y=0 in x=3003=100
∴ The line 3x+6y= 300meets the coordinates axes at (0,50) and (100,0).Joining these points by a thick line.
Now ,putting x=0 and y=0 in 3x+6y≤300,we get 0≤300
Therefore ,(0,0) satisfies the inequality 8x+4y≤480
Putting x=0 in 8x+4y=480,we get, y=4804=120
Putting y=0 in 8x+4y=480,we get, x=4808=60
∴ The line 8x+4y=480 meets the coordinates axes at (0,120) and (60,0).Join these points by a thick line.
Now ,putting x=0 and y=0 in 8x+4y=480,we get 0≤480
Therefore,(0,0) satisfies the inequality 8x+4y=480
So, the portion containing the origin is represented the solution set of the inequation 8x+4y=480
Region represented by x≥0 and y≥0
Clearly x≥0 and y≥0 represents the first quadrant.
The common region of the above five regions represents the solution set of the given inequations as shown below:
(iii) We have,
x+2y≤40,3x+y≥30,4x+3y≥60,x≥0,y≥0
Converting the given inequation into equations,we obtain,
x+2y=40,3x+y=30,4x+3y=60,x=0 and y=0
Region represented by x+2y≤40
Putting x=0 in x+2y=40,we get y=402=20
Putting y=0 in by x+2y=40,we get x=40
∴ The line x+2y=40,meets the coordinates axes at (0,20) and (40,0).Join these points by a thick line.
Now ,putting x=0 and y=0 in x+2y≤40,we get 0≤40
therefore,(0,0) satisfies the inequality x+2y≤40,
So,the portion not containing the origin is represented by the given inequation x+2y≤40
Region represented by 3x+y≥30
Putting x=0 in 3x+y=30,we get y=30
Putting y=0 in 3x+y=30,we get x=303=10
∴ The line 3x+y=30,meets the coordinates axes at (0,30) and (10,0).Join these points by a thick line.
Now,putting x=0 and y=0 in 3x+y≥30,we get 0≥30
This is not possible.
Therefore ,(0,0) satisfies the inequality 3x+y≥30.
So,the portion not containing the origin is represented by the given inequation 3x+y≥30.
Region represented by 4x+3y≥60
Putting x=0 in 4x+3y= 60,we get, y=603=20
Putting y=0 in 4x+3y= 60,we get, x=604=15
∴ The line 4x+3y= 60,meets the coordinates axes at (0,20) and (15,0).Join these points by a thick line.
Now ,putting x=0 and y=0 in 4x+3y≥60
This is not possible.
Therefore,(0,0) satisfies the inequality 4x+3y≥60
So, the portion containing the origin is represented the solution set of the inequation \4x+3y≥60.
Region represented by x≥0 and y≥0
Clearly x≥0 and y≥0 represents the first quadrant.
The common region of the above five regions represents the solution set of the given inequations as shown below:
(iv) We have,
5x+y≥10,2x+2y≥12,x+4y≥12,x≥0,y≥0
Converting the given inequation into equations,we obtain,
5x+y=10,2x+2y=1,x+4y=12,x=0 and y=0
Region represented by 5x+y≥10
Putting x=0 in 5x+y=10,we get y=10
Putting y=0 in by 5x+y=10,we get x=105=2
∴ The line 5x+y=10,meets the coordinates axes at (0,10) and (2,0).Join these points by a thick line.
Now ,putting x=0 and y=0 in 5x+y≥10,we get 0≥10
therefore,(0,0) satisfies the inequality 5x+y≥10,
So,the portion not containing the origin is represented by the given inequation 5x+y≥10
Region represented by 2x+2y≥12
Putting x=0 in 2x+2y=12,we get y=122=6
Putting y=0 in 2x+2y=12,we get x=122=6
∴ The line 2x+2y=12,meets the coordinates axes at (0,6) and (6,0).Join these points by a thick line.
Now,putting x=0 and y=0 in 2x+2y=12,we get 0≥12,which is not possible.
Therefore ,(0,0) satisfies the inequality 2x+2y=12.
So,the portion not containing the origin is represented by the given inequation 2x+2y=12.
Region represented by x+4y≥12
Putting x=0 in x+4y= 12,we get, y=124=3
Putting y=0 in x+4y= 12,we get, x=12
∴ The line x+4y= 12,meets the coordinates axes at (0,3) and (12,0).Join these points by a thick line.
Now ,putting x=0 and y=0 in x+4y= 12,which is not possible.
Therefore,(0,0) satisfies the inequality x+4y≥12
So, the portion containing the origin is represented the solution set of the inequation x+4y≥12
Region represented by x≥0 and y≥0
Clearly x≥0 and y≥0 represents the first quadrant.
The common region of the above five regions represents the solution set of the given inequations as shown below: