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Question

Solve the following systems of inequations graphically:

(i) 2x+y8,x+2y8,x+y6

(ii) 12x+12y840,3x+6y300,8x+4y480 x0,y0

(iii) x+2y40,3x+y30,4x+3y60,x0,y0

(iv) 5x+y10,2x+2y12,x+4y12,x0,y0

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Solution

(i) We have,

2x+y8,x+2y8,x+y6

Converting the given inequation into equations,we obtain,

2x+y=8,x+2y=8,and x+y=6

Region represented by 2x+y8

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+y8,we get 08

This is not possible.

We find that (0,0) is not satisifes the inequation 2x+y8

So,the portion not containing the origin is represented by the given inequation

Region represented by x+2y8

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+2y8,we get 08

This is not possible.

We find that (0,0) is not satisifes the inequation x+2y8

So,the portion not containing the origin is represented by the given inequation.

Region represented by x+y6

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+y6,we get 06

Therefore,(0,0) satisfies x+y6,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+12y840,3x+6y300,8x+4y480 x0,y0

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+12y840

Putting x=0 in 12x+12y= 840,we get y=84012=70

Putting y=0 in by 12x+12y840,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+12y840,we get 0840

Therefore,(0,0) satisfies the inequality 12x+12y840,

So,the portion not containing the origin is represented by the given inequation 12x+12y840

Region represented by 3x+6y300

Putting x=0 in 3x+6y300,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+6y300,we get 0300

Therefore ,(0,0) satisfies the inequality 8x+4y480

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 0480

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 x0 and y0

Clearly x0 and y0 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+2y40,3x+y30,4x+3y60,x0,y0

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+2y40

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+2y40,we get 040

therefore,(0,0) satisfies the inequality x+2y40,

So,the portion not containing the origin is represented by the given inequation x+2y40

Region represented by 3x+y30

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+y30,we get 030

This is not possible.

Therefore ,(0,0) satisfies the inequality 3x+y30.

So,the portion not containing the origin is represented by the given inequation 3x+y30.

Region represented by 4x+3y60

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+3y60

This is not possible.

Therefore,(0,0) satisfies the inequality 4x+3y60

So, the portion containing the origin is represented the solution set of the inequation \4x+3y60.

Region represented by x0 and y0

Clearly x0 and y0 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+y10,2x+2y12,x+4y12,x0,y0

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+y10

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+y10,we get 010

therefore,(0,0) satisfies the inequality 5x+y10,

So,the portion not containing the origin is represented by the given inequation 5x+y10

Region represented by 2x+2y12

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 012,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+4y12

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+4y12

So, the portion containing the origin is represented the solution set of the inequation x+4y12

Region represented by x0 and y0

Clearly x0 and y0 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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