TOPICS

Unit-6(Examples)

Linear Inqualities

**Example-1 :-** Solve 30 x < 200 when

(i) x is a natural number,

(ii) x is an integer.

We have, 30x < 200 (Using Rule 2 of) 30x/30 < 200/30 (Divide By 30 Both sides) x < 20/3 (i) When x is a natural number, in this case the following values of x make the statement true. 1, 2, 3, 4, 5, 6. The solution set of the inequality is {1,2,3,4,5,6}. (ii) When x is an integer, the solutions of the given inequality are ..., – 3, –2, –1, 0, 1, 2, 3, 4, 5, 6 The solution set of the inequality is {...,–3, –2,–1, 0, 1, 2, 3, 4, 5, 6}

**Example-2 :-** Solve 5x – 3 < 3x + 1 when

(i) x is an integer,

(ii) x is a real number.

We have 5x - 3 < 3x + 1 5x - 3 + 3 < 3x + 1 + 3 (Rule-1 : Adding Both Sides by 3) 5x < 3x + 4 5x - 3x < 3x - 3x + 4 (Rule-2 : Substracting Both Sides by -3x) 2x < 4 2x/2 < 4/2 (Rule-3 : Dividing by 2 both sides) x < 2 (i) When x is an integer, the solutions of the given inequality are ..., – 4, – 3, – 2, – 1, 0, 1 (ii) When x is a real number, the solutions of the inequality are given by x < 2, i.e., all real numbers x which are less than 2. Therefore, the solution set of the inequality is x ∈ (– ∞, 2).

**Example-3 :-** Solve 4x + 3 < 6x + 7.

We have 4x + 3 < 6x + 7 4x + 3 - 3 < 6x + 7 - 3 (Rule-1 : Substracting Both Sides by 3) 4x < 6x + 4 4x - 6x < 6x - 6x + 4 (Rule-2 : Substracting Both Sides by -6x) -2x < 4 -2x/2 < 4/2 (Rule-3 : Dividing by 2 both sides) -x < 2 x > -2 i.e., all the real numbers which are greater than –2, are the solutions of the given inequality. Hence, the solution set is (–2, ∞).

**Example-4 :-** Solve (5-2x)/3 ≤ x/6 - 5

We have, (5-2x)/3 ≤ x/6 - 5 6(5-2x)/3 ≤ x - 30 2(5-2x) ≤ x - 30 10 - 4x ≤ x - 30 10 - 4x - 10 ≤ x - 30 - 10 -4x ≤ x - 40 -4x - x ≤ x - x - 40 -5x ≤ -40 5x ≥ 40 5x/5 ≥ 40/5 x ≥ 8 Thus, all real numbers x which are greater than or equal to 8 are the solutions of the given inequality, i.e., x ∈ [8, ∞).

**Example-5 :-** Solve 7x + 3 < 5x + 9. Show the graph of the solutions on number line.

We have 7x + 3 < 5x + 9 2x < 6 x < 3 The graphical representation of the solution is

**Example-6 :-** Solve (3x-4)/2 ≥ (x+1)/4 - 1. Show the graph of the solutions on number line.

We have, (3x-4)/2 ≥ (x+1)/4 - 1 4(3x-4)/2 ≥ x + 1 - 4 2(3x-4) ≥ x - 3 6x - 8 ≥ x - 3 6x - 8 + 8 ≥ x - 3 + 8 6x ≥ x + 5 6x - x ≥ x + 5 - x 5x ≥ 5 5x/5 ≥ 5/5 x ≥ 1 The graphical representation of solutions is

**Example-7 :-** The marks obtained by a student of Class XI in first and second terminal examination are 62 and 48, respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.

Let x be the marks obtained by student in the annual examination. Then (62+48+x)/3 ≥ 60 110 + x ≥ 180 110 - 110 + x ≥ 180 - 110 x ≥ 70 Thus, the student must obtain a minimum of 70 marks to get an average of at least 60 marks.

**Example-8 :-** Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.

Let x be the smaller of the two consecutive odd natural number, so that the other one is x + 2. Then, x > 10 ... (1) and x + ( x + 2) < 40 ... (2) x + x + 2 < 40 2x + 2 < 40 2x + 2 - 2 < 40 - 2 2x < 38 2x/2 < 38/2 x < 19 ... (3) From (1) and (3), we get 10 < x < 19 Since x is an odd number, x can take the values 11, 13, 15, and 17. So, the required possible pairs will be (11, 13), (13, 15), (15, 17), (17, 19).

**Example-9 :-** Solve 3x + 2y > 6 graphically.

Graph of 3x + 2y = 6 is given as dotted line in the Figure. This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, We note that 3 (0) + 2 (0) > 6 or 0 > 6, which is false. Hence, half plane I is not the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality. In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.

**Example-10 :-** Solve 3x – 6 ≥ 0 graphically in two dimensional plane.

Graph of 3x – 6 = 0 is given in the Figure. We select a point, say (0, 0) and substituting it in given inequality, We note that: 3 (0) – 6 ≥ 0 or – 6 ≥ 0, which is false. Thus, the solution region is the shaded region on the right hand side of the line x = 2.

**Example-11 :-** Solve y < 2 graphically.

Graph of y = 2 is given in the Figure. Let us select a point, (0, 0) in lower half plane I and putting y = 0 in the given inequality, We note that 1 × 0 < 2 or 0 < 2, which is true. Thus, the solution region is the shaded region below the line y = 2. Hence, every point below the line determines the solution of the given inequality.

**Example-12 :-** Solve the following system of linear inequalities graphically.

x + y ≥ 5 ... (1)

x – y ≤ 3 ... (2)

The graph of linear equation x + y = 5 is drawn in Figure Put x = 0, then we get 0 + y = 5 or y = 5 i.e., (0,5) Put y = 0, then we get x + 0 = 5 or x = 5 i.e., (5,0) Similarly, The graph of linear equation x - y = 3 is drawn in Figure Put x = 0, then we get 0 - y = 3 or y = -3 i.e., (0,-3) Put y = 0, then we get x - 0 = 3 or x = 3 i.e., (3,0) Then we note that solution of inequality, (1) is represented by the shaded region above the line x + y = 5, including the points on the line. (2) represents the shaded region abovethe line x – y = 3, including the points on the line. Clearly, the double shaded region, common to the above two shaded regions is the required solution region of the given system of inequalities.

**Example-13 :-** Solve the following system of inequalities graphically

5x + 4y ≤ 40 ... (1)

x ≥ 2 ... (2)

y ≥ 3 ... (3)

The graph of linear equation 5x + 4y = 40, x = 2 and y = 3 is drawn in Figure Put x = 0, then we get 0 + 4y = 40 or y = 10 i.e., (0,10) Put y = 0, then we get 5x + 0 = 40 or x = 8 i.e., (8,0) Then we note that the inequality, (1) represents shaded region below the line 5x + 4y = 40 and inequality, (2) represents the shaded region right of line x = 2 but inequality, (3) represents the shaded region above the line y = 3. Hence, shaded region including all the point on the lines are also the solution of the given system of the linear inequalities.

**Example-14 :-** Solve the following system of inequalities

8x + 3y ≤ 100 ... (1)

x ≥ 0 ... (2)

y ≥ 0 ... (3)

The graph of linear equation 8x + 3y = 100, x = 0 and y = 0 is drawn in Figure Put x = 0, then we get 0 + 3y = 100 or y = 100/3 i.e., (0,33.4) Put y = 0, then we get 8x + 0 = 100 or x = 100/8 i.e., (12.5,0) The inequality 8x + 3y ≤ 100 represents the shaded region below the line, including the points on the line 8x + 3y =100. Since x ≥ 0, y ≥ 0, every point in the shaded region in the first quadrant, including the points on the line and the axes, represents the solution of the given system of inequalities.

**Example-15 :-** Solve the following system of inequalities graphically

x + 2y ≤ 8 ... (1)

2x + y ≤ 8 ... (2)

x > 0 ... (3)

y > 0 ... (4)

The graph of linear equation x + 2y = 8 is drawn in Figure Put x = 0, then we get 0 + 2y = 8 or y = 4 i.e., (0,4) Put y = 0, then we get x + 0 = 8 or x = 8 i.e., (8,0) Similarly, The graph of linear equation 2x + y = 8 is drawn in Figure Put x = 0, then we get 0 + y = 8 or y = 8 i.e., (0,8) Put y = 0, then we get 2x + 0 = 8 or x = 4 i.e., (4,0) The inequality (1) and (2) represent the region below the two lines, including the point on the respective lines. Since x ≥ 0, y ≥ 0, every point in the shaded region in the first quadrant represent a solution of the given system of inequalities.

CLASSES