Show that the solution set of the following linear in equations is an unbounded set :

$x + y \geq 9, 3 x + y \geq 12, x \geq 0, y \geq 0$

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Solution

We have,

Converting the inequations into equations,we get x+y=9,3x+y=12,x=0 and y=0

Region represented by $x + y \geq 9$

Putting x=0 inequation x+y=9,we get y=9

Putting y=0 inequation x+y=9,we get x=9

$∴$ The line x+y=9 meets the coordinate axes at (0,9) and (9,0).Join these points by a thick line.

Now,putting x=0 and y=0 in $x + y \geq 9, w e g e t 0 \geq 9$

This is not possible.

$∴$ We find that (0,0) is not satisfies the inequation $x + y \geq 9$

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

Region represented by $3 x + y \geq 12$

Putting x=0 inequation 3x+y=12,we get y=12

Putting y=0 inequation 3x+y=9,we get x= $\frac{12}{3} = 4$

$∴$ The line 3x+y=12 meets the coordinates axes at (0,12) and (4,0).Joining these points by a thick line.

Now,Putting x=0 and y=0 in $3 x + y \geq 12, w e g e t 0 \geq 12$

This is not possible.

$∴$ We find that (0,0) is not satisfies the inequation $3 x + y \geq 12$

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

Region represented by $x \geq 0 a n d y \geq 0$ :

Clearly, $x \geq 0 a n d y \geq 0$ represent the first quadrant.

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