Question

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

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Solution

(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:

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