Chapter 6 – Linear Inequalities

> 3, x ≤ 4, x + y ≥ 9. (i) Inequalities which do not involve variables are called numerical inequalities. For

< 8, 5 ≥ 2. (ii) Inequalities which involve variables are called literal inequalities. For example, x > 3, y ≤ 5, x – y ≥ 0. (iii) An inequality may contain more than one variable and it can be linear, quadratic or cubic etc. For eaxmple, 3x – 2 < 0 is a linear inequality in one variable, 2x + 3y ≥ 4 is a linear inequality in two variables and x2 + 3x + 2 < 0 is a quadratic inequality in one variable. (iv) Inequalities involving the symbol ‘>’ or ‘<’ are called strict inequalities. For example, 3x – y > 5, x < 3. (v) Inequalities involving the symbol ‘≥’ or ‘≤’ are called slack inequalities. For example, 3x – y ≥ 5, x ≤ 5. 6.1.2 Solution of an inequality (i) The value(s) of the variable(s) which makes the inequality a true statement is called its solutions. The set of all solutions of an inequality is called the solution set of the inequality. For example, x – 1 ≥ 0, has infinite number of solutions as all real values greater than or equal to one make it a true statement. The inequality x2 + 1 < 0 has no solution in R as no real value of x makes it a true statement. To solve an inequality we can (i) Add (or subtract) the same quantity to (from) both sides without changing the sign of inequality. (ii) Multiply (or divide) both sides by the same positive quantity without changing the sign of inequality. However, if both sides of inequality are multiplied (or divided) by the same negative quantity the sign of inequality is reversed, i.e., ‘>’ changes into ‘<’ and vice versa. LINEAR INEQUALITIES 99 6.1.3 Representation of solution of linear inequality in one variable on a number line To represent the solution of a linear inequality in one variable on a number line, we use the following conventions: (i) If the inequality involves ‘≥’ or ‘≤’, we draw filled circle (•) on the number line to indicate that the number corresponding to the filled circle is included in the solution set. (ii) If the inequality involves ‘>’ or ‘<’, we draw an open circle (O) on the number line to indicate that the number corresponding to the open circle is excluded from the solution set. 6.1.4 Graphical representation of the solution of a linear inequality

• (a)To represent the solution of a linear inequality in one or two variables graphically in a plane, we proceed as follows: (i) If the inequality involves ‘≥’ or ‘≤’, we draw the graph of the line as a thick line to indicate that the points on this line are included in the solution set. (ii) If the inequality involves ‘>’ or ‘<’, we draw the graph of the line as dotted line to indicate that the points on the line are excluded from the solution set.
• (b)Solution of a linear inequality in one variable can be represented on number line as well as in the plane but the solution of a linear inequality in two variables of the type ax + by > c, ax + by ≥ c, ax + by < c or ax + by ≤ c (a ≠ 0, b ≠ 0) can be represented in the plane only.
• (c)Two or more inequalities taken together comprise a system of inequalities and the solutions of the system of inequalities are the solutions common to all the inequalities comprising the system. 6.1.5 Two important results (a) If a, b ∈ R and b ≠ 0, then (i) ab > 0 or > 0 ⇒ a and b are of the same sign. (ii) ab < 0 or < 0 ⇒ a and b are of opposite sign. (b) If a is any positive real number, i.e., a > 0, then (i) | x | < a ⇔ – a < x < a |x| ≤ a ⇔ – a ≤ x ≤ a (ii) | x | > a ⇔ x < – a or x > a | x | ≥ a ⇔ x ≤ – a or x ≥ a

Solve the inequality, 3x – 5 < x + 7, when

• (i)x is a natural number
• (ii)x is a whole number
• (iii)x is an integer
• (iv)x is a real number Solution We have 3x – 5 < x + 7 ⇒ 3x < x + 12 (Adding 5 to both sides) ⇒ 2x < 12 (Subtracting x from both sides) ⇒ x<6 (Dividing by 2 on both sides) (i) Solution set is {1, 2, 3, 4, 5} (ii) Solution set is {0, 1, 2, 3, 4, 5} (iii) Solution set is {....– 3, – 2, –1, 0, 1, 2, 3, 4, 5} (iv) Solution set is {x : x ∈ R and x < 6}, i.e., any real number less than 6. x−2

Solve >2 x+5 x−2 Solution We have >2 x+5 x−2 ⇒ – 2>0 [Subtracting 2 from each side] x+5 – ( x +12) ⇒ >0 x+5 x +12 ⇒ <0 (Multiplying both sides by – 1) x+5 ⇒ x + 12 > 0 and x + 5 < 0 [Since < 0 ⇒ a and b are of opposite signs] x + 12 < 0 and x + 5 > 0 ⇒ x > – 12 and x < – 5 x < – 12 and x > – 5 (Not possible) Therefore, – 12 < x < –5, i.e. x ∈ (–12, –5) LINEAR INEQUALITIES 101

Solve 3 – 4 x ≥ 9 . Solution We have 3 – 4 x ≥ 9 . ⇒ 3 – 4x ≤ – 9 or 3 – 4x ≥ 9 (Since x ≥ a ⇒ x ≤ – a or x ≥ a) ⇒ – 4x ≤ – 12 or – 4x ≥ 6 −3 ⇒ x≥3 or x≤ (Dividing both sides by – 4) −3 ⇒ x∈(− ∞, ] ∪ [3, ∞ )

Solve 1 ≤ |x –2 | ≤ 3. Solution We have 1 ≤ |x – 2| ≤ 3 ⇒ x − 2 ≥1 and x − 2 ≤3 ⇒ (x – 2 ≤ – 1 or x – 2 ≥ 1) and (– 3 ≤ x – 2 ≤ 3) ⇒ (x ≤ 1 or x ≥ 3) and (– 1 ≤ x ≤ 5) ⇒ x ∈ (– ∞, 1] ∪ [3, ∞) and x ∈ [ –1, 5] Combining the solutions of two inequalities, we have x ∈ [–1, 1] ∪ [3, 5]

The cost and revenue functions of a product are given by C(x) = 20 x + 4000 and R(x) = 60x + 2000, respectively, where x is the number of items produced and sold. How many items must be sold to realise some profit? Solution We have, profit = Revenue – Cost = (60x + 2000) – (20x + 4000) = 40x – 2000 To earn some profit, 40x – 2000 > 0 ⇒ x > 50 Hence, the manufacturer must sell more than 50 items to realise some profit.

Solve for x, x +1 + x > 3 . Solution On LHS of the given inequality, we have two terms both containing modulus. By equating the expression within the modulus to zero, we get x = – 1, 0 as critical points. These critical points divide the real line in three parts as (– ∞, – 1), [–1, 0), [0, ∞). Case I When – ∞ < x < – 1 x +1 + x > 3 ⇒ – x – 1 – x > 3 ⇒ x < – 2. Case II When – 1 ≤ x < 0, x +1 + x > 3 ⇒ x + 1 – x > 3 ⇒ 1 > 3 (not possible) Case III When 0 ≤ x < ∞, x +1 + x > 3 ⇒ x + 1 + x > 3 ⇒ x > 1. Combining the results of cases (I), (II) and (III), we get x ∈ (– ∞ , – 2) ∪ (1, ∞)

Solve for x, >1 x+2 x+3 + x Solution We have >1 x+2 x+3 + x ⇒ –1> 0 x+2 x+3 − 2 ⇒ >0 x+2 Now two cases arise: Case I When x + 3 ≥ 0, i.e., x ≥ – 3. Then x+3 − 2 x+3 − 2 >0 ⇒ >0 x+2 x+2 x +1 ⇒ >0 x+2 ⇒ {(x + 1) > 0 and x + 2 > 0} or {x + 1 < 0 and x + 2 < 0} ⇒ {x > –1 and x > –2} or {x < – 1 and x < – 2} ⇒ x > –1 or x < – 2 ⇒ x ∈ (–1, ∞) or x ∈ (– ∞, – 2) ⇒ x ∈ (–3, –2) ∪ ( – 1, ∞) [Since x ≥ – 3] ... (1) LINEAR INEQUALITIES 103 Case II When x + 3 < 0, i.e., x < – 3 x+3 − 2 − x − 3− 2 >0 ⇒ >0 x+2 x+2 − ( x + 5) x+5 ⇒ >0 ⇒ <0 x+2 x+2 ⇒ (x + 5 < 0 and x + 2 > 0) or (x + 5 > 0 and x + 2 < 0) ⇒ (x < – 5 and x > – 2) or (x > – 5 and x < –2) it is not possible. ⇒ x ∈ (– 5 , – 2) ... (2) Combining (I) and (II), the required solution is x ∈ (– 5 , – 2) ∪ (– 1, ∞)

Solve the following system of inequalities : x 1 6x 1 ≥ , <

Find the linear inequalities for which the shaded region in the given figure is the solution set. Solution

• (i)Consider 2x + 3y = 3. We observe that the shaded region and the origin lie on opposite side of this line and (0, 0) satisfies 2x + 3y ≤ 3. Therefore, we must have 2x + 3y ≥ 3 as linear inequality corresponding to the line 2x + 3y = 3.
• (ii)Consider 3x + 4y = 18. We observe that the shaded region and the origin lie on the same side of this line and (0, 0) satisfies 3x + 4y ≤ 18. Therefore, 3x + 4y ≤ 18 is the linear inequality Fig 6.1 corresponding to the line 3x + 4y = 18.
• (iii)Consider –7x + 4y = 14. It is clear from the figure that the shaded region and the origin lie on the same side of this line and (0, 0) satisfies the inequality – 7x + 4y ≤ 14. Therefore, –7x + 4y ≤ 14 is the inequality corresponding to the line –7x + 4y = 14.
• (iv)Consider x – 6y = 3. It may be noted that the shaded portion and origin lie on the same side of this line and (0, 0) satisfies x – 6y ≤ 3. Therefore, x – 6y ≤ 3 is the inequality corresponding to the line x – 6y = 3. (v) Also the shaded region lies in the first quadrant only. Therefore, x ≥ 0, y ≥ 0. Hence, in view of (i), (ii), (iii), (iv) and (v) above, the linear inequalities corresponding to the given solution set are : 2x + 3y ≥ 3,3x + 4y ≤ 18 –7x + 4y ≤14, x – 6y ≤ 3, x ≥ 0, y ≥ 0.

If ≥ 0 , then x−2

• (A)x ∈ [2, ∞)
• (B)x ∈ (2, ∞)
• (C)x ∈ (– ∞, 2)
• (D)x ∈ (– ∞, 2] LINEAR INEQUALITIES 105 x−2 Solution (B) is the correct choice. Since ≥ 0 , for x − 2 ≥ 0, and x – 2 ≠ 0. x−2

The length of a rectangle is three times the breadth. If the minimum perimeter of the rectangle is 160 cm, then

• (A)breadth > 20 cm
• (B)length < 20 cm
• (C)breadth x ≥ 20 cm
• (D)length ≤ 20 cm Solution (C) is the correct choice. If x cm is the breadth, then 2 (3x + x) ≥ 160 ⇒ x ≥ 20

Solutions of the inequalities comprising a system in variable x are represented on number lines as given below, then Fig 6.2

• (A)x ∈ (– ∞, – 4] ∪ [3, ∞)
• (B)x ∈ [– 3, 1]
• (C)x ∈ (– ∞, – 4) ∪ [3, ∞)
• (D)x ∈ [– 4, 3] Solution (A) is the correct choice Common solution of the inequalities is from – ∞ to – 4 and 3 to ∞.

If x + 3 ≥ 10 , then

• (A)x ∈ (– 13, 7]
• (B)x ∈ (– 13, 7]
• (C)x ∈ (– ∞, – 13] ∪ [7, ∞)
• (D)x ∈ [– ∞, – 13] ∪ [7, ∞) Solution (D) is the correct choice, since x + 3 ≥ 10 , ⇒ x + 3 ≤ – 10 or x + 3 ≥ 10 ⇒ x ≤ – 13 or x ≥ 7 ⇒ x ∈ (– ∞, – 13] ∪ [7, ∞)

State whether the following statements are True or False.

• (i)If x > y and b < 0, then bx < by
• (ii)If xy > 0, then x > 0, and y < 0
• (iii)If xy < 0, then x > 0, and y > 0
• (iv)If x > 5 and x > 2, then x ∈ (5, ∞) (v) If | x | < 5, then x ∈ (– 5, 5) (vi) Graph of x > – 2 is (vii) Solution set of x – y ≤ 0 is Fig 6.3 Solution (i) True, because the sign of inequality is reversed when we multiply both sides of an inequality by a negative quantity. (ii) False, product of two numbers is positive if they have the same sign. (iii) False, product of two numbers is negative if they have opposite signs. (iv) True Fig 6.4 (v) True if | x | < 5 ⇒ – 5 < x < 5 ⇒ x ∈ (– 5, 5). (vi) False, because for x > – 2, the line x = –2 has to be dotted, i.e., the region does not include the points on the line x = –2 (vii) False, because (1, 0) does not satisfy the given inequality and it is a point in shaded portion.

Fill in the blanks in the following:

• (i)If x ≥ – 3, then x + 5 ................... 2
• (ii)If – x ≤ – 4, then 2x ................... 8
• (iii)If < 0, then x ................... 2 x−2 a b
• (iv)If a < b and c < 0, then ............ c c (v) If x −1 ≤ 2 , then – 1..... x .... 3 (vi) If |3x – 7| > 2, then x .... or x ....3 (vii) If p > 0 and q < 0, then p + q ... p Solution (i) (≥), because same number can be added to both sides of inequality without changing the sign of inequality. (ii) (≥), after multiplying both sides by – 2, the sign of inequality is reversed. (iii) (<), because if < 0 and a > 0, then b < 0. (iv) (>), if both sides are divided by the same negative quantity, then the sign of inequality is reversed. LINEAR INEQUALITIES 107 (v) (≤, ≤ ), x −1 ≤ 2 ⇒ – 2 ≤ x – 1 ≤ 2 ⇒ –1 ≤ x ≤ 3. (vi) (<, > ), |3x – 7| > 2 ⇒ 3x – 7 < – 2 or 3x – 7 > 2 ⇒ x< or x > 3 (vii) (<), as p is positive and q is negative, therefore, p + q is always smaller than p.