    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

Open in App
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:   Suggest Corrections  0      Similar questions  Related Videos   Graphical Method of Solving LPP
MATHEMATICS
Watch in App  Explore more