Question

Solve the following systems of linear inequations graphically :

(i) 2x+3y≤6,3x+2y≤6,x≥0,y≥0(ii) 2x+3y≤6,x+4y≤4,x≥0,y≥0(iii) x+y≥1,x+2y≤8,2x+y≥2,x≥0,y≥0(iv) x+y≥1,7x+9y≤63,y≤5,x≥0,y≥0(v) 2x+3y≤35,y≥3,x≥2,x≥0,y≥0

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Solution

(i) 2x+3y≤6,3x+2y≤6,x≥0,y≥0We have, 2x+3y≤6,3x+2y≤6,x≥0,y≥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≤6.

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

we get y=63=2.

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

we get x=63=2.

∴ 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≤6.

Region represented by 3x+2y≤6:

Putting x =0 in the equation

3x+2y=6,we get y=62=3

Putting y=0 in the equation

3x+2y=6,we get y=62=3

∴ 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≤6.

Region represented by x≥0 and y≥0.

Clearly x≥0 and y≥0. represent the first quadrant.

(ii) we have,

2x+3y≤6,x+4y≤4,x≥0,y≥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≤6

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

we get y=63=2.

Putting y=63=2.

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

we get x=62=3.

∴ 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≤6.

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

Region represented by x+4y≤4.

Putting x=0 in x +4y=0

we get,y=44=1

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

we get x=4

∴ 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≤4,we get 0≤4

Clearly,we find that (0,0) satisfies inequation x+4y≤4.

Region represented by x≥0 and y≥0.

Clearly x≥0 and y≥0 represent the first quadrant.

(iii)

we have,

x+y≥1,x+2y≤8,2x+y≥2,x≥0,y≥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≤1:

Putting x=0 inequation x-y=1,

we get y=-1

Putting y=0 inequation x-y=1,

we get x=1

∴ 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≤1 in x-1,we get,0≤1

Clearly,we find that (0,0) satisifes inequation x−y≤1

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 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≤8.,we get,0≤8

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

Region represented by 2x+2y≥2.

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

Putting y=0,in 2x+y=2,we get x=22=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≥1,7x+9y≤63,x≤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≥1:

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

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

∴ 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≥1, we get 0≥1

This is not possible

∴ (0,0) is not satisfies the inequality x+y≥1.So,the portion not containing the origin is represented by the inequation x+y≥1.

Region represented by 7x+9y≤63.

Putting x=0 in 7x+9y=63,we get,y=639=7.

Putting y=0 in 7x+9y=63,we get,y=637=9.

∴ 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≤35,y≥3,x≥2,x≥0,y≥0

we have 2x+3y≤35,y≥3,x≥2,x≥0,y≥0

Region represented by 2x+3y≤35

Putting x=0 inequation 2x+3y=35,we get y=353

Putting y=0 inequation 2x+3y=35,we get x=352

∴ The line 2x+3y=35 meets the coordinate axes at (0,35) and (352,0).join these point by a thick line.

Now,putting x=0 and y=0 inequation 2x+3y≤35, we get 0≤35

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

Region represented by y≥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≥3.

So the portion not containing the origin is represented by the inequation y≥3.

Region represented by x≥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≥2 the portion not containing the origin is represented by the given inequations.

Region represented by x≥0 and y≥0.

Clearly,x≥0 and y≥0. represent thr first quadrant.

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

0

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