Question

Show that the solution set of the following linear inequations is empty set:

(i) $x-2y\ge 0,2x-y\le -2,x\ge 0,y\ge 0$

(ii) $x+2y\le 3,3x+4y\ge 12,y\ge 1,x\ge 0andy\ge 0$

Answer 1

We have,

(i) $x-2y\ge 0,2x-y\le -2,x\ge 0andy\ge 0$

Converting the inequations into equations, we get

x-2y=0,2x-y=-2,x=0 and y=0

Region represented by $x-2y\ge 0$

Putting x=0 in x-2y=0,we get y=0.

Putting y=2 in x-2y=0,we get x=4.

$\therefore$ The line x-2y=0 meets the coordinates axes at (0,0).Joining these point (0,0) and (4,2) by a thick line.

Now,putting x=0 and y=0 in $x-2y\ge 0$,we get $0\ge 0$

Clearly,we find that (0,0) satisifes the inequation $x-2y\ge 0$.So,the portion containing the origin is represented by the given inequation.
Region represented by $2x-y\le 2$

putting x=0 and y=0 in 2x-y=-2,we get y=2

putting y=0 and y=0 in 2x-y=-2,we get x=$\frac{-2}{-2}=-1$

$\therefore$ The line 2x-y=-2 meets the coordinates axes at (0,2) and (-1,0).Joining these points by a thick line.

Now,putting x=0 and y=0 in $x-2y\ge 0$,we get $2x-y\le 2$,we get $0\le -2$

This is not possible.

Since,(0,0) does not satisfy the portion inequation $2x-y\le 2$

So,the portion containing the origin is represented by the given inequation $2x-y\le 2$

Region represented by $x\ge 0andy\ge 0$

Clearly,$x\ge 0andy\ge 0$ represented the first quadrant.

(ii) We have, $x+2y\le 3,3x+4y\ge 12,y\ge 1,x\ge 0andy\ge 0$

Converting the inequations into equations, we get

x+2y=3,3x+4y=12,y=1,x=0 and y=0

Region represented by $x+2y\le 3$

Putting x=0 in x+2y=3,we get y=$\frac{3}{2}$.

Putting y=2 in x+2y=3,we get x=3.

$\therefore$ The line x+2y=3 meets the coordinates axes at $(0,\frac{3}{2})$and (3,0).Join these point by a thick line.

Now,putting x=0 and y=0 in $x+2y\ge 3$,we get $0\ge 3$

Clearly,(0,0) satisifes the inequality $x+2y\ge 3$.So,the portion containing the origin represents the solution set of the inequation $x+2y\le 3$.

Region represented by $3x+4y\ge 12$

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

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

$\therefore$ The line 3x+4y=12 meets the coordinates axes at (0,3) and (4,0).Join these point by a thick line.

Now,putting x=0 and y=0 in $3x+4y\ge 12$,we get $0\le 12$

This is not possible.

Since,(0,0) does not satisifes the inequation $3x+4y\ge 12$.So,the portion not containing the origin represented by the inequation $3x+4y\ge 12$.

Region represented by $y\ge 1$.

Clearly,y=1 is a line parallel to x-axis at a distance of 1 units from the origin.Since (0,0) does not satisifes the inequation $y\ge 1$.

So,the portion containing the origin represented by the solution set of the inequation.

Region represented by $x\ge 0andy\ge 0$.

Clearly, $x\ge 0andy\ge 0$ represent the first quadrant.