Question

Solve the following systems of linear inequations graphically :

$\left(i\right)2x+3y\le 6,3x+2y\le 6,x\ge 0,y\ge 0\left(ii\right)2x+3y\le 6,x+4y\le 4,x\ge 0,y\ge 0\left(iii\right)x+y\ge 1,x+2y\le 8,2x+y\ge 2,x\ge 0,y\ge 0\left(iv\right)x+y\ge 1,7x+9y\le 63,y\le 5,x\ge 0,y\ge 0\left(v\right)2x+3y\le 35,y\ge 3,x\ge 2,x\ge 0,y\ge 0$

Answer 1

$\left(i\right)2x+3y\le 6,3x+2y\le 6,x\ge 0,y\ge 0Wehave,2x+3y\le 6,3x+2y\le 6,x\ge 0,y\ge 0$

Converting the given in equation into equations, the in equations reduce to 2x+3y=6.

3x+2y=6,x=0 and y=0

Region represented by $2x+3y\le 6$.

Putting x=0 in the equation 2x+3y=6,

we get y=$\frac{6}{3}=2.$

Putting y=0 in the equation 2x+3y=6,

we get x=$\frac{6}{3}=2.$

$\therefore$ This Putting 2x+3y=6 meets the coordinate axes at (0.2) and (3,0).Draw a thick line joining these points,we find that (0,0) satisfies inequation $2x+3y\le 6.$

Region represented by $3x+2y\le 6:$

Putting x =0 in the equation

3x+2y=6,we get y=$\frac{6}{2}=3$

Putting y=0 in the equation

3x+2y=6,we get y=$\frac{6}{2}=3$

$\therefore$ This Putting 3x+2y=6 meets the coordinate axes at (0,3) and (2,0).Draw a thick line joining these points,we find that (0,0) satisfies inequation $3x+2y\le 6.$

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

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

(ii) we have,

$2x+3y\le 6,x+4y\le 4,x\ge 0,y\ge 0$

Converting the given inequation into equations, the inequations reduce to 2x+3y=6,

x+4y=4,x=0 and y=0

Region represented by $2x+3y\le 6$

Putting x=0 inequation 2x+3y=6,

we get y=$\frac{6}{3}=2.$

Putting y=$\frac{6}{3}=2.$

Putting y=0 in 2x+3y=6,

we get x=$\frac{6}{2}=3.$

$\therefore$ This Putting 2x+3y=6 meets the coordinates axes at (0,2) and (3,0).Draw a thick line joining these points.

Now,putting x=0 and y=0 inequation $2x+3y\le 6.$

Clearly,we find that (0,0) satisifes inequation $2x+3y\le 6.$

Region represented by $x+4y\le 4.$

Putting x=0 in x +4y=0

we get,$y=\frac{4}{4}=1$

Putting y=0 in x+4y=4,

we get x=4

$\therefore$ This x+4y=4 meets the coordinate axes at (0,1) and (4,0).Draw a thick line joining these points.

Now,putting x=0,y=0

$x+4y\le 4$,we get $0\le 4$

Clearly,we find that (0,0) satisfies inequation $x+4y\le 4$.

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

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

(iii)

we have,

$x+y\ge 1,x+2y\le 8,2x+y\ge 2,x\ge 0,y\ge 0$

Converting the inequations into equations, we obtain x-y=1,x+2y=8,2x+y=2,x=0 and y=0

Region represented by $x-y\le 1:$

Putting x=0 inequation x-y=1,

we get y=-1

Putting y=0 inequation x-y=1,

we get x=1

$\therefore$ The line x-y=1 meets the co-ordinate axes at (0,-1) and (1,0).Draw a thick line joining these points.

Now,putting x=0 and y=0 inequation $x-y\le 1$ in x-1,we get,$0\le 1$

Clearly,we find that (0,0) satisifes inequation $x-y\le 1$

Region represented by $x+2y\le 8.$

Putting x=0 in x +2y=8,

we get,$y=\frac{8}{2}=4$

Putting y=0 in x+2y=8,

we get x=8

$\therefore$ The line x+2y=8 meets the coordinate axes at (8,0) and (0,4).Draw a thick line joining these points.

Now,putting x=0 and y=0 in $x+2y\le 8.$,we get,$0\le 8$

Clearly,we find that (0,0) satisfies inequation $x+2y\le 8$.

Region represented by $2x+2y\ge 2.$

Putting x=0,in 2x+y=2,we get y=2

Putting y=0,in 2x+y=2,we get x=$\frac{2}{2}=1$

The line 2x+y=2 meets the coordinate axes at (0,2) and (1,0).Draw a thick line joining these points.

(iv) We have,

$x+y\ge 1,7x+9y\le 63,x\le 6,$

Converting the inequations into equations,we obtain

x+y=1,7x+9y=63,x=6,y=5,x=0 and y=0

Region represented by $x+y\ge 1:$

Putting x=0 in x+y=1,we get y=1

Putting y=0 in x+y=1,we get x=1

$\therefore$ The line x+y=1 meets the coordinate axes at (0,1) and (1,0).join these point by a thick line.

Now,putting x=0 and y=0 inequation $x+y\ge 1$, we get $0\ge 1$

This is not possible

$\therefore$ (0,0) is not satisfies the inequality $x+y\ge 1$.So,the portion not containing the origin is represented by the inequation $x+y\ge 1$.

Region represented by $7x+9y\le 63.$

Putting x=0 in 7x+9y=63,we get,$y=\frac{63}{9}=7$.

Putting y=0 in 7x+9y=63,we get,$y=\frac{63}{7}=9$.

$\therefore$ The line 7x+9y=63 meets the coordinate axes at (0,7) and (9,0).Join these points by a thick line.

(v) $2x+3y\le 35,y\ge 3,x\ge 2,x\ge 0,y\ge 0$

we have $2x+3y\le 35,y\ge 3,x\ge 2,x\ge 0,y\ge 0$

Region represented by $2x+3y\le 35$

Putting x=0 inequation $2x+3y=35,wegety=\frac{35}{3}$

Putting y=0 inequation $2x+3y=35,wegetx=\frac{35}{2}$

$\therefore$ The line 2x+3y=35 meets the coordinate axes at (0,35) and $(\frac{35}{2},0)$.join these point by a thick line.

Now,putting x=0 and y=0 inequation $2x+3y\le 35$, we get $0\le 35$

Clearly,(0,0) satisfies the inequality $2x+3y\le 35$,so,the portion containing the origin represents the solution $2x+3y\le 35$.

Region represented by $y\ge 3.$

Clearly,y=3 is a line parallel to x axis at a distance 3 units from the origin,since (0,0) does not satisfies the inequation $y\ge 3$.

So the portion not containing the origin is represented by the inequation $y\ge 3$.

Region represented by $x\ge 2$.

Clearly,x=2 is a line parallel to y-axis at a distance 2 units from the origin.Since (0,0) does not satisfies the inequation $x\ge 2$ the portion not containing the origin is represented by the given inequations.

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

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

The common region of the above five regions represents the solution set of the given inequations as shown below: