Linear Inequalities Solved Problems

Page # 2 LINEAR INEQUALITIES SOLVED PROBLEMS Ex.2 Solve the following inequations: Sol. The given inequation is x 3 0 x 5 x 3 0 x 5 ....(1) 4 5 3x  4 5  2 12    5 3x  4 5   12 2 12  –5  6(3x–4)  5 –5+24  18x  5+24 19  18x  29 19 29 x  S = 18 18  Solution set S on numbr lines  X' 19 18 29 18 19 29  18 , 18    is represented as X Ex.4 Solve the following system of inequations: 2x – 7 > 5 – x, 11 – 5x  1 Sol. The given inequations are : 2x – 7 > 5 – x ...(1) and 11 – 5x1 ...(2) Ex.5 Solve the following system of inequations 5x 3x 39 2x  1 x  11 3x  1     and 4 8 8 12 3 4 Sol. The given inequations are : 5x 3x 39   ...(1) 4 8 8 2x  1 x  11 3x  1   and ...(2) 12 3 4 10x  3x 39  From (1)  13x > 39 8 8  x > 3 Thus the solution set of (1) is (3, ) – + 0 8 8 8 3 3x  4 5  Ex.3 Solve . 2 12 Sol. We have 2 Clearly the intersection of these solution sets is the sets (4, ) Hence the solution set of the given system of inequations is (4, ) 8 x 3 is + ve. x 5  either x–3>0, x–5>0 or x–3<0, x–5<0 i.e., either x>3, x>5 or x<3, x<5 i.e., either x> 5 or x<3 [ x>5  x>3 and x<3  x<5] Hence ths solution set is (–, 3),  (5, ), which on the number line as below: – + Since 0 8 6.3 4 8 x 2 0 From (2), 11 – 5x  11  –5x  1 –11  –5x  –10  x  2 Thus the solution set of (2) is [2, ) – + 8 10x 63   x  6.3 10 10  Solution set is the set of all numbers which are less than or equal to 6.3.  S = (–, 6.3] Solution set on number line is represented as  From (1), 2x+x>7+5  3x > 12  x > 4 Thus the solution set of (1) is (4, ) – + 8 2x  3 4x + 8  2 + 4 3 Represent the solution set on number line. Sol. The given inequation is 2x  3 4x + 8  2 + 4 3  3(2x–3)+9624+16x (Multiplying both sides by 12)  6x–9+9624+16x  6x–16x24+9–96 Ex.1 Solve 3 From (2), (2x  1)  4(x  11) 3x  1 2x  43 3x  1    12 4 12 4  –2x + 43 < 3 (3x+1) [Multiplying both sides by 12]  –2x + 43 < 9x +3 40  –11x < –40  x > 11  40  , Thus the solution set of (2) is   11  40 11 Clearly, the intersection of these solution sets  40  , is the set  11   Hence, the solution set of the given system  40  , of inequations is   11  Ex.6 Find the region enclosed by the following inequations : x+y–2  0, 2x+y–3  0, x  0, y  0. Also find the ordered pairs of the vertices of the regions. Sol. We have x + y – 2  0 Page # 3 LINEAR INEQUALITIES 2x + y – 3  0 and x  0, The correspondin equation of (1) is x + y – 2 = 0  y = 2 – x To draw the graph : Table : x Y .... .... 1 1 2 0 y  0 .... .... 0 3 1 1 .... .... Plot the points (0, 3) (1, 1) and join them, we get the graph of (6) 3  ,0 2  This st. line meets x-axis at  Put (0, 0) in x + y – 2  0, we get 0 + 0 – 2  0  –2  0, true So we shade the portion containing (0, 0) Put (0,0) in 2x+y–3  0, we get –3  0, true So we shade the portion containing (0,0) Also x  0 is the potion on and right of y-axis and y  0 is the portion on and above x-axis. Thus the required region is as shown shaded in the fig. X = 0 (0, 3) 3 · C 2· (0, 2) 1· O · 1 Y .... .... Plot the points (1, 1), (2, 0) join them, we get the graph of (5) This straight line meet y-axis at (0, 2) The corresponding equation of (2) is 2x + y – 3 = 0      y = 3 – 2x. To draw the graph : Table : x Y Let us draw the graph of 2x + y = 24 For x = 0, y = 24 For y = 0, 2x = 24 i.e., x = 12  line 2x + y = 24 meets OX in A (12, 0) and OY in L (0, 24). · B(1, 1) D(2, 0) · · A 3 ( , 0) 2 Hence the solution set of the given system of inequation is the quadrilateral OABC whose vertices are 3  , 0  , B (1, 1), C(0, 2). 2  O(0, 0), A  Ex.7 Solve the following system of inequations graphically. 2x+y24, x+y11, 2x+5y40, x  0, y  0 Sol. The given inequations are 2x + y  24 ...(i) x + y  11 ...(ii) 2x + 5y  40 ...(iii) x  0, y  0 ...(iv) Consider a set of rectangular cartesian axes OXY in the plane. It is clear that any point which satisfies x  0, y  0 lies in the first quadrant · L(9, 24) · ·11 M(0, 11) N(0, 8) O · 8D · · B (11, 0) A · (12, 0)· C(20, 0) X Again we draw the graph of x + y = 11 For x = 0, y = 11 For y = 0, x = 11  line x + y = 11 meets OX in B (11, 0) and OY in M (0, 11). Now, we draw the graph of 2x + 5y = 40 For x = 0, 5y = 40 i.e., y = 8 For y = 0, 2x = 40 i.e., x = 20  line 2x + 5y = 40 meets OX in C (20 , 0) and OY on N (0, 8) Put (0, 0) in (i), (ii) and (iii) we get 0  24, 0  11, 0  40 respectively which are all true. So in all the three cases we shade the portin containing (0, 0). Since feasible region is the region which satisfies all the given constraints.  OBDN is the feasible region, which represents the solution set of the given system of inequations. Ex.8 A manufacturer has 600 litres of a 12% solution of acid. How many litres of a 30% acid solution must be added to it so that acid content in the resulting mixture will be more than 15% but less than 18% ? Sol. Let number of litres of 30% acid solution be x. From given condition 30% of x+12% of 600>15% of (x+600) and 30% of x+12% of 600<18% of (x+600) 30 12 15 x  600  (x  600)  100 100 100 30 12 15 x  600  (x  600) and 100 100 100  30x + 7200 > 15x + 9000 and 30x + 7200 < 18x + 10800  15x > 1800 and 12x < 3600  x > 120 and x < 300  120 < x < 300  number of lines of the 30% solution of acid will have to be more than 120 but less than 300.