ML Aggarwal Class 11 Maths Solutions Chapter 06 Quadratic Equations

Linear Inequalities

Introduction

You know about equations where two parts of a statement (called left hand side, L.H.S. and right hand side, R.H.S.) are joined by an equals sign (=). You have worked through many equations with one or two variables and solved word problems by turning them into equations. A natural question comes up: Can every word problem be written as an equation? The answer is no. Instead, you may face statements that use inequality signs (or symbols). These signs are

<, >, ≤, ≥

In this chapter, you will learn how to solve (using algebra or graphs) linear inequalities in one or two variables.

6.1 Linear Inequalities in One Variable

When two real numbers or two algebraic expressions are joined by the sign <, >, ≤ or ≥, this is called an inequality.

Examples of inequalities include: 7 < 11, 5½ > - 3, x < 3, x + 5 ≤ 7, 2x - 3 > 8, 3y + 5 ≥ 11, \( \frac{y-3}{2} < 2y+1 \).

The statements 7 < 11 and 5½ > - 3 are examples of numerical inequalities, while x < 3, x + 5 ≤ 7, 2x - 3 > 8, 3y + 5 ≥ 11, \( \frac{y-3}{2} < 2y+1 \) are examples of literal inequalities.

Any linear inequality in one variable can be written in one of these forms: \( ax + b < 0 \), \( ax + b ≤ 0 \), \( ax + b > 0 \), \( ax + b ≥ 0 \), where a and b are real numbers and \( a ≠ 0 \).

The inequalities x < 3, 2x - 3 > 8, and \( \frac{y-3}{2} < 2y+1 \) are called strict inequalities. The inequalities x + 5 ≤ 7 and 3y + 5 ≥ 11 are called slack inequalities.

Replacement set: The set from which we pick values for the variable in the inequality is called the replacement set.

Solution set: A solution to an inequality is a value (picked from the replacement set) that makes the inequality true when substituted for the variable. The set of all such solutions is called the solution set of the inequality.

For example, take the inequality x < 4.

Notice that the solution set depends on the replacement set.

Remark: If no replacement set is given, we take it to be R (the set of all real numbers).

6.1.1 Solving linear inequalities in one variable

The rules for solving inequalities are the same as those for solving equations, with one important exception: when you multiply or divide by a negative number, the inequality sign flips.

You can use any of the following methods on an inequality:

(i) Add or subtract the same number or expression from both sides.

(ii) Move any number or expression from one side to the other by flipping its sign.

(iii) Multiply or divide both sides by the same positive number.

However, when you multiply or divide by the same negative number, you must flip the inequality sign.

Examples:

(i) If x < 2, then - x > - 2 (Multiply by - 1)

(ii) If 3x - 1 ≥ 5, then - 4 (3x - 1) ≤ - 20 (Multiply by - 4)

(iii) If - 6x ≤ 12, then x ≥ - 2 (Divide by - 6)

Always flip the inequality sign when you multiply or divide by a negative number.

6.1.2 Steps to solve a linear inequality in one variable

(i) Simplify both sides by removing brackets and collecting like terms.

(ii) Clear fractions (or decimals) by multiplying both sides by an appropriate number (L.C.M. of the fractions or a power of 10 for decimals).

(iii) Move all variable terms to one side and all constant terms to the other. Combine like terms.

(iv) Make the coefficient of the variable equal to 1.

(v) Pick the solution set from the replacement set.

Illustrative Examples

Example 1. Given x ∈ {-3, -4, -5, -6} and 9 ≤ 1 - 2x, find the possible values of x. Also show its solution set on the number line.
Answer: From 9 ≤ 1 - 2x, we add 2x to both sides: 2x + 9 ≤ 1. Next, we add -9 to both sides: 2x ≤ -8. Dividing by 2 gives: x ≤ -4. Since x must be from {-3, -4, -5, -6}, the solution set is {-4, -5, -6}. The solution set is shown by solid dots on the number line at -6, -5, and -4.

Exam Tip: Always check each value from the replacement set in the original inequality to verify your solution.

Example 2. Solve the inequality 3 - 2x ≥ x - 32, given that (i) x ∈ N (ii) x ∈ I.
Answer: We start with 3 - 2x ≥ x - 32. Adding -3 to both sides gives: -2x ≥ x - 35. Adding -x to both sides gives: -3x ≥ -35. Multiplying both sides by -1/3 (and flipping the inequality) gives: x ≤ 35/3.
(i) When x ∈ N, the solution set is {1, 2, 3, ..., 11}.
(ii) When x ∈ I, the solution set is {..., -3, -2, -1, 0, 1, 2, ..., 11}.
In simple words: First, gather all x terms on one side and numbers on the other. Then divide to find x. Remember: if you divide by a negative number, flip the sign.

Exam Tip: The solution depends on what set (N or I) you are working with - always note the replacement set given in the question.

Example 3. Solve 5x - 3 < 3x + 1 when (i) x is an integer (ii) x is a real number.
Answer: From 5x - 3 < 3x + 1, we move 3 to the right side: 5x < 3x + 4. Moving 3x to the left side: 2x < 4. Dividing by 2: x < 2.
(i) When x is an integer, the solution set is {..., -3, -2, -1, 0, 1}.
(ii) When x is a real number, the solution includes all real numbers less than 2, written as {x : x ∈ R, x < 2} or (- ∞, 2).
In simple words: Move numbers and x terms to get x by itself on one side. When x is real, the answer is all numbers smaller than 2.

Exam Tip: Notice how the solution changes based on the replacement set - integers give a list, while real numbers give an interval.

Example 4. Solve the following inequalities for real x: (i) 3(2 - x) ≥ 2(1 - x) (ii) \( \frac{2x-1}{3} ≥ \frac{3x-2}{4} - \frac{2-x}{5} \)
Answer:
(i) Expanding 3(2 - x) ≥ 2(1 - x) gives 6 - 3x ≥ 2 - 2x. Adding 2x to both sides: 6 - x ≥ 2. Subtracting 6: -x ≥ -4. Dividing by -1 (and flipping): x ≤ 4. The solution set is {x : x ∈ R, x ≤ 4} or (- ∞, 4].
(ii) To remove fractions, multiply both sides by 60 (the L.C.M. of 3, 4, 5): 20(2x - 1) ≥ 15(3x - 2) - 12(2 - x). Expanding: 40x - 20 ≥ 45x - 30 - 24 + 12x. Simplifying: 40x - 20 ≥ 57x - 54. Rearranging: 40x - 57x ≥ -54 + 20, so -17x ≥ -34. Dividing by -17 (and flipping): x ≤ 2. The solution set is {x : x ∈ R, x ≤ 2} or (- ∞, 2].
In simple words: Clear brackets first, then clear fractions. Move all x terms to one side and numbers to the other. When dividing by a negative, flip the sign.

Exam Tip: Multiply by the L.C.M. to clear all fractions at once - this avoids errors from working with multiple fractions separately.

Example 5. Solve \( \frac{2x+1}{3} ≥ \frac{3x-2}{5} \), x ∈ R. Show the solution set on the number line.
Answer: Multiply both sides by 15 (L.C.M. of 3 and 5): 5(2x + 1) ≥ 3(3x - 2), which gives 10x + 5 ≥ 9x - 6. Moving 9x to the left and 5 to the right: 10x - 9x ≥ -6 - 5, so x ≥ -11. The solution set is {x : x ∈ R, x ≥ -11} or [-11, ∞). On the number line, show a solid dot at -11 and a thick line extending right to show all numbers greater than or equal to -11.
In simple words: Multiply to remove fractions. Then solve like a normal inequality. A solid dot means that number is included.

Exam Tip: Use a solid dot when the inequality has ≤ or ≥, and an open circle when it has < or >.

Example 6. Solve \( \frac{x}{4} > \frac{5x-2}{3} - \frac{7x-3}{5} \) and show the solution set on the number line.
Answer: Multiply both sides by 60 (L.C.M. of 4, 3, 5): 15x > 20(5x - 2) - 12(7x - 3). Expanding: 15x > 100x - 40 - 84x + 36. Simplifying: 15x > 16x - 4. Rearranging: 15x - 16x > -4, so -x > -4. Dividing by -1 (and flipping): x < 4. The solution set is {x : x ∈ R, x < 4} or (- ∞, 4). Show an open circle at 4 (since 4 is not included) and a thick line extending left.
In simple words: Clear fractions by multiplying. Simplify and solve. Use an open circle when the number itself is not included in the answer.

Exam Tip: An open circle at a boundary means that boundary value is NOT part of the solution.

Example 7. To receive grade 'A' in a maths course, one must get an average of at least 90 marks in five tests (each worth 100 marks). Ragini's marks in the first four tests are 87, 92, 94, and 95. What is the minimum mark she must get in the fifth test to earn grade 'A'?
Answer: Let x be the marks Ragini gets in the fifth test. The average of five tests is: (87 + 92 + 94 + 95 + x)/5 ≥ 90. This simplifies to: (368 + x)/5 ≥ 90. Multiplying both sides by 5: 368 + x ≥ 450. Subtracting 368: x ≥ 82. Ragini must get a minimum of 82 marks in the fifth test to achieve grade 'A'.
In simple words: Add up the marks she has. Write an average using these plus the unknown fifth mark. Solve to find what that unknown must be.

Exam Tip: Set up the inequality using the word "average" carefully - divide the sum by the number of tests and compare to the required average.

Example 8. A company makes cassettes. Its cost and revenue functions are C(x) = 26000 + 30x and R(x) = 43x, where x is the number of cassettes made and sold each week. How many cassettes must the company sell to make a profit?
Answer: Profit occurs when revenue is greater than cost. So we need R(x) > C(x), which means 43x > 26000 + 30x. Subtracting 30x from both sides: 13x > 26000. Dividing by 13: x > 2000. The company must sell more than 2000 cassettes to earn a profit.
In simple words: Profit means money in minus money out. Set revenue bigger than cost and solve.

Exam Tip: Remember that profit = revenue - cost, so profit is positive when revenue > cost.

Example 9. The longest side of a triangle is 3 times the shortest side. The third side is 2 cm shorter than the longest side. If the perimeter is at least 61 cm, what is the minimum length of the shortest side?
Answer: Let x cm be the length of the shortest side. Then the longest side is 3x cm and the third side is (3x - 2) cm. The perimeter is x + 3x + (3x - 2) = 7x - 2 cm. For the perimeter to be at least 61 cm: 7x - 2 ≥ 61. Adding 2: 7x ≥ 63. Dividing by 7: x ≥ 9. The minimum length of the shortest side is 9 cm.
In simple words: Write each side in terms of the shortest side. Add them to get the perimeter. Then use the condition about the perimeter to find the shortest side.

Exam Tip: Always express all sides in terms of one variable, then use the perimeter (or other) condition to solve.

Exercise 6.1

Very short answer type questions (1 to 13):

Question 1. Solve the inequality 3x - 5 ≤ 13 - x, where x ∈ {1, 2, 3, ..., 10}.
Answer: From 3x - 5 ≤ 13 - x, add x to both sides: 4x - 5 ≤ 13. Add 5: 4x ≤ 18. Divide by 4: x ≤ 4.5. Since x must be a natural number from {1, 2, 3, ..., 10}, the values are {1, 2, 3, 4}.
In simple words: Rearrange so x is on one side. Get x by itself. Then pick numbers from your set that work.

Exam Tip: Always check your boundary value in the original inequality to confirm it works.

Question 2. Solve 11 - 5x > 3 - 2x, x ∈ W.
Answer: From 11 - 5x > 3 - 2x, add 5x to both sides: 11 > 3 + 3x. Subtract 3: 8 > 3x. Divide by 3: 8/3 > x, or x < 8/3 ≈ 2.67. Since x ∈ W (whole numbers: 0, 1, 2, 3, ...), the solution is {0, 1, 2}.
In simple words: Move all x terms to one side and numbers to the other. Then find which whole numbers satisfy the result.

Exam Tip: Whole numbers include 0 - do not forget to check if 0 is a valid solution.

Question 3. If x is a negative integer, find the solution set of \( \frac{2}{5} + \frac{1}{5}(2x+5) ≥ 0 \).
Answer: From \( \frac{2}{5} + \frac{1}{5}(2x+5) ≥ 0 \), multiply by 5: 2 + (2x + 5) ≥ 0. Simplify: 2x + 7 ≥ 0. Subtract 7: 2x ≥ -7. Divide by 2: x ≥ -3.5. Since x is a negative integer, the values are {-3, -2, -1}.
In simple words: Clear the fractions first. Then solve for x. Pick negative integers that satisfy the answer.

Exam Tip: "Negative integer" means x < 0, so -1, -2, -3, ... are allowed, but 0 and positive integers are not.

Question 4. Solve 30x < 200, when (i) x is a natural number (ii) x is an integer.
Answer: From 30x < 200, divide by 30: x < 200/30 = 20/3 ≈ 6.67.
(i) When x is a natural number: {1, 2, 3, 4, 5, 6}.
(ii) When x is an integer: {..., -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}.
In simple words: Divide to get x. Then list which numbers of that type are smaller than 6.67.

Exam Tip: Natural numbers start at 1, but integers include 0 and all negative numbers.

Question 5. Solve -12x > 30, when (i) x is a natural number (ii) x is an integer.
Answer: From -12x > 30, divide by -12 (and flip the sign): x < -30/12 = -2.5.
(i) When x is a natural number: The solution set is empty (∅), because no natural number is less than -2.5.
(ii) When x is an integer: {..., -5, -4, -3}.
In simple words: When you divide by a negative, flip the sign. Natural numbers are never negative, so there is no solution for part (i).

Exam Tip: Remember to flip the inequality when dividing by a negative number, and remember what each set type includes.

Question 6. Solve 2 + 23x < 99 - x, when (i) x is a natural number (ii) x is an integer.
Answer: From 2 + 23x < 99 - x, add x to both sides: 2 + 24x < 99. Subtract 2: 24x < 97. Divide by 24: x < 97/24 ≈ 4.04.
(i) When x is a natural number: {1, 2, 3, 4}.
(ii) When x is an integer: {..., -3, -2, -1, 0, 1, 2, 3, 4}.
In simple words: Combine the x terms on one side. Divide to isolate x. Then list valid numbers.

Exam Tip: After dividing, round to see the approximate value, then list all integers or natural numbers less than that.

Question 7. Solve 7 - 7x ≥ 37, when (i) x is a natural number (ii) x is an integer.
Answer: From 7 - 7x ≥ 37, subtract 7 from both sides: -7x ≥ 30. Divide by -7 (and flip): x ≤ -30/7 ≈ -4.29.
(i) When x is a natural number: The solution set is empty (∅), because no natural number is ≤ -4.29.
(ii) When x is an integer: {..., -7, -6, -5}.
In simple words: Isolate the x term. Divide by its coefficient, flipping if it is negative.

Exam Tip: After flipping, be careful about whether the boundary itself is included (≤ or ≥ means it is).

Question 8. Solve 2 < 3x + 7, when (i) x is an integer (ii) x is a real number.
Answer: From 2 < 3x + 7, subtract 7: -5 < 3x. Divide by 3: -5/3 < x, or x > -5/3 ≈ -1.67.
(i) When x is an integer: {-1, 0, 1, 2, 3, ...}.
(ii) When x is a real number: {x : x ∈ R, x > -5/3} or (-5/3, ∞).
In simple words: Get x by itself. Then for integers, list those that work. For reals, describe the interval.

Exam Tip: The boundary (-5/3) is NOT included because the inequality is < (strict), not ≤.

Question 9. Solve 3x + 8 > 2 when (i) x is an integer (ii) x is a real number.
Answer: From 3x + 8 > 2, subtract 8: 3x > -6. Divide by 3: x > -2.
(i) When x is an integer: {-1, 0, 1, 2, 3, ...}.
(ii) When x is a real number: {x : x ∈ R, x > -2} or (-2, ∞).
In simple words: Subtract constants and divide to get x. Check: does -2 work? No, because 3(-2) + 8 = 2, not > 2.

Exam Tip: Always verify whether the boundary is included by testing it in the original inequality.

Question 10. Solve 7x - 1 < 5x + 3 when (i) x is an integer (ii) x is a real number.
Answer: From 7x - 1 < 5x + 3, subtract 5x: 2x - 1 < 3. Add 1: 2x < 4. Divide by 2: x < 2.
(i) When x is an integer: {..., -1, 0, 1}.
(ii) When x is a real number: {x : x ∈ R, x < 2} or (-∞, 2).
In simple words: Gather x terms on one side. Simplify and divide to find the boundary. List values that satisfy it.

Exam Tip: When the inequality is strict (<), the boundary is NOT part of the solution, so use (-∞, 2), not (-∞, 2].

Solve the following (11 to 20) inequalities for real x:

Question 11. (i) 4x + 3 ≤ 6x + 7 (ii) 3x - 7 > 5x - 1.
Answer:
(i) From 4x + 3 ≤ 6x + 7, subtract 4x: 3 ≤ 2x + 7. Subtract 7: -4 ≤ 2x. Divide by 2: -2 ≤ x, or x ≥ -2. Solution: {x : x ∈ R, x ≥ -2} or [-2, ∞).
(ii) From 3x - 7 > 5x - 1, subtract 3x: -7 > 2x - 1. Add 1: -6 > 2x. Divide by 2: -3 > x, or x < -3. Solution: {x : x ∈ R, x < -3} or (-∞, -3).
In simple words: Move all x terms to one side and numbers to the other. Simplify and divide.

Exam Tip: Use a solid bracket [·] when the boundary is included (≤ or ≥) and a round bracket (·) when it is not (< or >).

Question 12. (i) -(x - 3) + 4 > -2x + 5 (ii) 3(x - 1) ≤ 2(x - 3).
Answer:
(i) Expand -(x - 3) + 4 > -2x + 5 to get -x + 3 + 4 > -2x + 5. Simplify: -x + 7 > -2x + 5. Add 2x: x + 7 > 5. Subtract 7: x > -2. Solution: {x : x ∈ R, x > -2} or (-2, ∞).
(ii) Expand 3(x - 1) ≤ 2(x - 3) to get 3x - 3 ≤ 2x - 6. Subtract 2x: x - 3 ≤ -6. Add 3: x ≤ -3. Solution: {x : x ∈ R, x ≤ -3} or (-∞, -3].
In simple words: Expand brackets first. Then move x and numbers to opposite sides. Simplify to find x.

Exam Tip: When expanding a negative sign in front of brackets, reverse all signs inside.

Question 13. (i) 2(2x + 3) - 10 ≤ 6(x - 2) (ii) 37 - (3x + 5) ≥ 9x - 8(x - 3).
Answer:
(i) Expand: 4x + 6 - 10 ≤ 6x - 12. Simplify: 4x - 4 ≤ 6x - 12. Subtract 4x: -4 ≤ 2x - 12. Add 12: 8 ≤ 2x. Divide by 2: 4 ≤ x, or x ≥ 4. Solution: {x : x ∈ R, x ≥ 4} or [4, ∞).
(ii) Expand: 37 - 3x - 5 ≥ 9x - 8x + 24. Simplify: 32 - 3x ≥ x + 24. Add 3x: 32 ≥ 4x + 24. Subtract 24: 8 ≥ 4x. Divide by 4: 2 ≥ x, or x ≤ 2. Solution: {x : x ∈ R, x ≤ 2} or (-∞, 2].
In simple words: Clear all brackets by expanding. Collect x on one side and numbers on the other. Divide to find x.

Exam Tip: Double-check your expansion - a single sign error early on will make the whole answer wrong.

Question 14. (i) \( x + \frac{x}{2} + \frac{x}{3} < 11 \) (ii) \( \frac{x}{3} > \frac{x}{2} + 1 \).
Answer:
(i) Multiply by 6 (L.C.M.): 6x + 3x + 2x < 66. Simplify: 11x < 66. Divide by 11: x < 6. Solution: {x : x ∈ R, x < 6} or (-∞, 6).
(ii) Multiply by 6: 2x > 3x + 6. Subtract 2x: 0 > x + 6. Subtract 6: -6 > x, or x < -6. Solution: {x : x ∈ R, x < -6} or (-∞, -6).
In simple words: Multiply by the L.C.M. to clear all fractions. Then solve as a regular inequality.

Exam Tip: Always find the L.C.M. of all denominators to clear fractions in one step.

Question 15. (i) \( \frac{4+2x}{3} ≥ \frac{x}{2} - 3 \) (ii) \( \frac{5-2x}{3} ≤ \frac{x}{6} - 5 \).
Answer:
(i) Multiply by 6: 2(4 + 2x) ≥ 3x - 18. Expand: 8 + 4x ≥ 3x - 18. Subtract 3x: 8 + x ≥ -18. Subtract 8: x ≥ -26. Solution: {x : x ∈ R, x ≥ -26} or [-26, ∞).
(ii) Multiply by 6: 2(5 - 2x) ≤ x - 30. Expand: 10 - 4x ≤ x - 30. Add 4x: 10 ≤ 5x - 30. Add 30: 40 ≤ 5x. Divide by 5: 8 ≤ x, or x ≥ 8. Solution: {x : x ∈ R, x ≥ 8} or [8, ∞).
In simple words: Clear fractions by multiplying. Expand and rearrange to get x by itself.

Exam Tip: After clearing fractions, treat it like any other inequality.

Question 16. (i) \( \frac{2x-3}{4} + 8 ≥ 2 + \frac{4x}{3} \) (ii) \( \frac{1}{2}\left(\frac{3x}{5}+4\right) ≥ \frac{1}{3}(x-6) \).
Answer:
(i) Multiply by 12: 3(2x - 3) + 96 ≥ 24 + 16x. Expand: 6x - 9 + 96 ≥ 24 + 16x. Simplify: 6x + 87 ≥ 24 + 16x. Subtract 6x: 87 ≥ 24 + 10x. Subtract 24: 63 ≥ 10x. Divide by 10: 6.3 ≥ x, or x ≤ 6.3. Solution: {x : x ∈ R, x ≤ 6.3} or (-∞, 6.3].
(ii) Expand: \( \frac{3x}{10} + 2 ≥ \frac{x}{3} - 2 \). Multiply by 30: 9x + 60 ≥ 10x - 60. Subtract 9x: 60 ≥ x - 60. Add 60: 120 ≥ x, or x ≤ 120. Solution: {x : x ∈ R, x ≤ 120} or (-∞, 120].
In simple words: Multiply to remove fractions. Expand brackets carefully. Rearrange to get x.

Exam Tip: When you have nested fractions (fractions inside brackets), expand those first, then clear the remaining fractions.

Question 17. Solve the inequality 3x - 11 < 3 where x ∈ {1, 2, 3, ..., 10}. Also show its solution on a number line.
Answer: From 3x - 11 < 3, add 11: 3x < 14. Divide by 3: x < 14/3 ≈ 4.67. Since x ∈ {1, 2, 3, ..., 10}, the solution is {1, 2, 3, 4}. On the number line, mark solid dots above 1, 2, 3, and 4.
In simple words: Solve the inequality normally. Then pick from your set all numbers that satisfy it.

Exam Tip: Always show the replacement set on the number line using dots or marks to indicate which values are in the solution.

Question 18. Solve 5 - 4x > 2 - 3x, x ∈ W. Also show its solution on the number line.
Answer: From 5 - 4x > 2 - 3x, add 4x: 5 > 2 + x. Subtract 2: 3 > x, or x < 3. Since x ∈ W (whole numbers), the solution is {0, 1, 2}. On the number line, mark solid dots above 0, 1, and 2.
In simple words: Simplify to find the boundary. List all whole numbers less than that boundary.

Exam Tip: Whole numbers start at 0 (not 1), so check whether 0 satisfies the inequality.

Question 19. List the solution set of \( \frac{11-2x}{5} ≥ \frac{9-3x}{8} + \frac{3}{4} \), x ∈ N.
Answer: Multiply by 40 (L.C.M.): 8(11 - 2x) ≥ 5(9 - 3x) + 30. Expand: 88 - 16x ≥ 45 - 15x + 30. Simplify: 88 - 16x ≥ 75 - 15x. Add 16x: 88 ≥ 75 + x. Subtract 75: 13 ≥ x, or x ≤ 13. Since x ∈ N, the solution is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.
In simple words: Clear all fractions by multiplying. Rearrange to isolate x. List natural numbers that satisfy it.

Exam Tip: For compound fractions, always multiply by the L.C.M. of ALL denominators, not just the largest one.

Question 20. If x ∈ W, find the solution set of \( \frac{3}{5}x - \frac{2x-1}{3} > 1 \).
Answer: Multiply by 15 (L.C.M.): 9x - 5(2x - 1) > 15. Expand: 9x - 10x + 5 > 15. Simplify: -x + 5 > 15. Subtract 5: -x > 10. Divide by -1 (and flip): x < -10. Since x ∈ W (non-negative), the solution set is empty (∅).
In simple words: Clear fractions and simplify. The answer says x must be less than -10, but whole numbers are never negative, so no whole numbers work.

Exam Tip: When the solution doesn't fit the replacement set, the answer is an empty set (∅).

Question 21. Solve the following inequalities and show the graph of the solution in each case on the number line:

(i) 7x + 3 < 5x + 9
Answer: From 7x + 3 < 5x + 9, subtract 5x: 2x + 3 < 9. Subtract 3: 2x < 6. Divide by 2: x < 3. On the number line, draw an open circle at 3 and shade the region to the left.
In simple words: Move x terms to one side and numbers to the other. Divide to get x. Mark the boundary with an open circle (since < is strict).

Exam Tip: Use an open circle for < or >, and a solid circle for ≤ or ≥.

(ii) 3(1 - x) < 2(x + 4)
Answer: Expand: 3 - 3x < 2x + 8. Add 3x: 3 < 5x + 8. Subtract 8: -5 < 5x. Divide by 5: -1 < x, or x > -1. On the number line, draw an open circle at -1 and shade the region to the right.
In simple words: Expand both sides. Rearrange so x is on one side. Mark and shade appropriately.

Exam Tip: After rearranging, rewrite the inequality in standard form (x on the left) to make graphing easier.

(iii) x + 5 ≤ 2x + 3
Answer: Subtract x: 5 ≤ x + 3. Subtract 3: 2 ≤ x, or x ≥ 2. On the number line, draw a solid circle at 2 and shade the region to the right.
In simple words: Move x and constants to opposite sides. Mark with a solid circle since ≤ includes the boundary.

Exam Tip: When rewriting 5 ≤ x + 3 as x ≥ 2, remember that the inequality still reads the same way (smaller value on the left).

(iv) \( \frac{4x-10}{3} ≤ \frac{5x-7}{2} \)
Answer: Multiply by 6: 2(4x - 10) ≤ 3(5x - 7). Expand: 8x - 20 ≤ 15x - 21. Subtract 8x: -20 ≤ 7x - 21. Add 21: 1 ≤ 7x. Divide by 7: 1/7 ≤ x, or x ≥ 1/7. On the number line, draw a solid circle at 1/7 and shade to the right.
In simple words: Clear fractions by multiplying. Expand and rearrange. Mark the boundary with a solid circle and shade.

Exam Tip: When the boundary is a fraction, mark it as accurately as you can on the number line.

(v) \( \frac{3x}{5} - \frac{2x-1}{3} > 1 \)
Answer: Multiply by 15: 9x - 5(2x - 1) > 15. Expand: 9x - 10x + 5 > 15. Simplify: -x + 5 > 15. Subtract 5: -x > 10. Divide by -1 (and flip): x < -10. On the number line, draw an open circle at -10 and shade to the left.
In simple words: Clear fractions and simplify. When you divide by a negative, flip the inequality. Mark with an open circle and shade.

Exam Tip: Always flip the inequality sign when multiplying or dividing by a negative.

(vi) \( \frac{3x-4}{2} ≥ \frac{x+1}{4} - 1 \)
Answer: Multiply by 4: 2(3x - 4) ≥ (x + 1) - 4. Expand: 6x - 8 ≥ x + 1 - 4. Simplify: 6x - 8 ≥ x - 3. Subtract x: 5x - 8 ≥ -3. Add 8: 5x ≥ 5. Divide by 5: x ≥ 1. On the number line, draw a solid circle at 1 and shade to the right.
In simple words: Clear fractions. Expand both sides. Gather x on one side and constants on the other. Mark and shade.

Exam Tip: A solid circle means the boundary value is included in the solution.

Question 22. Solve 3(2 - x) ≥ 2(1 - x), given that x ∈ R+ where R+ = {x : x ∈ R and x > 0}. Also show the solution on the number line.
Answer: Expand: 6 - 3x ≥ 2 - 2x. Add 2x: 6 - x ≥ 2. Subtract 6: -x ≥ -4. Divide by -1 (and flip): x ≤ 4. Since x ∈ R+ (x > 0), the solution is 0 < x ≤ 4. On the number line, draw an open circle at 0 and a solid circle at 4, shading between them.
In simple words: Solve normally. Then intersect with the replacement set: x must be positive and ≤ 4.

Exam Tip: The replacement set R+ excludes 0, so use an open circle at 0 even though the solution is ≤ 4.

Question 23. A student scored 62 and 48 in two tests. What is the minimum mark needed in the third test to have an average of at least 60 marks?
Answer: Let x be the mark in the third test. The average is (62 + 48 + x)/3 ≥ 60. Simplify: (110 + x)/3 ≥ 60. Multiply by 3: 110 + x ≥ 180. Subtract 110: x ≥ 70. The student must score a minimum of 70 marks in the third test.
In simple words: Add the marks from the first two tests. Write the average formula. Set it ≥ 60 and solve for the third test mark.

Exam Tip: Always write the average as (sum of all marks) / (number of tests), then set up and solve the inequality.

Question 24. Sukriti scored 75 and 70 in first two unit tests. What is the minimum mark she needs in the third test to have an average of at least 60 marks?
Answer: Let x be the mark in the third test. The average is (75 + 70 + x)/3 ≥ 60. Simplify: (145 + x)/3 ≥ 60. Multiply by 3: 145 + x ≥ 180. Subtract 145: x ≥ 35. Sukriti needs a minimum of 35 marks in the third test.
In simple words: Use the average formula with the given marks and the requirement of ≥ 60.

Exam Tip: Note that here she already has good marks, so the minimum needed in the third test is lower.

Question 25. The cost and revenue functions of a product are C(x) = 20x + 4000 and R(x) = 60x + 2000, where x is the number of items produced and sold. How many items must be sold to make a profit?
Answer: Profit occurs when R(x) > C(x), so 60x + 2000 > 20x + 4000. Subtract 20x: 40x + 2000 > 4000. Subtract 2000: 40x > 2000. Divide by 40: x > 50. The company must sell more than 50 items to earn a profit.
In simple words: Profit means money in > money out. Set revenue > cost and solve.

Exam Tip: A profit occurs when the revenue function value is larger than the cost function value.

6.2 Solution of a System of Linear Inequalities in One Variable

To solve a system of linear inequalities in one variable, follow these steps:

(i) Work through each linear inequality one at a time.

(ii) Find which values of the variable satisfy every single one of the given inequalities.

(iii) The set of values that satisfy all of them at the same time makes up the required solution of the system.

Illustrative Examples

Example 1. Solve the following system of linear inequalities: 3x - 1 ≥ 5, 2x - 3 > 7.
Answer: Solve each separately.
From 3x - 1 ≥ 5: add 1 to get 3x ≥ 6, divide by 3 to get x ≥ 2 ... (i)
From 2x - 3 > 7: add 3 to get 2x > 10, divide by 2 to get x > 5 ... (ii)
Looking at (i) and (ii), the values of x that work for both are those with x > 5. So the solution set is {x : x ∈ R, x > 5} or (5, ∞).
In simple words: Solve each inequality. Then pick the values that satisfy both. Think: which numbers are ≥ 2 AND > 5 at the same time? Answer: numbers > 5.

Exam Tip: When combining two inequalities, you need to find the overlap - values that satisfy BOTH.

Example 2. Solve the following system of linear inequalities: 4x - 5 < 11, -3x - 4 ≥ 8.
Answer: Solve each separately.
From 4x - 5 < 11: add 5 to get 4x < 16, divide by 4 to get x < 4 ... (i)
From -3x - 4 ≥ 8: add 4 to get -3x ≥ 12, divide by -3 (flip) to get x ≤ -4 ... (ii)
Looking at (i) and (ii), we need x < 4 AND x ≤ -4. The overlap is x ≤ -4. So the solution set is {x : x ∈ R, x ≤ -4} or (-∞, -4].
In simple words: From (i), x must be less than 4. From (ii), x must be ≤ -4. Numbers that satisfy both must be ≤ -4.

Exam Tip: Sketch both regions on a number line - the solution is where they overlap.

Example 3. Solve the following system of linear inequalities: 2(x + 1) ≤ x + 5, 3(x + 2) > 2 - x. Also show the solution graphically on the number line.
Answer: Solve each separately.
From 2(x + 1) ≤ x + 5: expand to 2x + 2 ≤ x + 5, subtract x to get x + 2 ≤ 5, subtract 2 to get x ≤ 3 ... (i)
From 3(x + 2) > 2 - x: expand to 3x + 6 > 2 - x, add x to get 4x + 6 > 2, subtract 6 to get 4x > -4, divide by 4 to get x > -1 ... (ii)
Looking at (i) and (ii), the overlap is: -1 < x ≤ 3. So the solution set is {x : x ∈ R, -1 < x ≤ 3} or (-1, 3]. On the number line, draw an open circle at -1 and a solid circle at 3, with shading between them.
In simple words: First inequality says x ≤ 3. Second says x > -1. Together: -1 < x ≤ 3.

Exam Tip: Use an open circle where the boundary is not included and a solid circle where it is.

Example 4. Solve 2 ≤ 3x - 4 ≤ 5.
Answer: This is two inequalities written together: 2 ≤ 3x - 4 AND 3x - 4 ≤ 5. Add 4 throughout: 2 + 4 ≤ 3x ≤ 5 + 4, so 6 ≤ 3x ≤ 9. Divide by 3: 2 ≤ x ≤ 3. The solution set is [2, 3].
In simple words: When a number appears between two inequalities like this, treat both sides at once. Add, subtract, multiply, or divide the same thing to all three parts.

Exam Tip: Work on all three parts together - if you add 4 to the middle, add it to both the left and right parts too.

Example 5. Solve 6 ≤ -3(2x - 4) < 12.
Answer: Expand the middle: 6 ≤ -6x + 12 < 12. Subtract 12 from all parts: 6 - 12 ≤ -6x < 12 - 12, so -6 ≤ -6x < 0. Divide by -6 (and flip both inequalities): 1 ≥ x > 0, or 0 < x ≤ 1. The solution set is (0, 1].
In simple words: Expand the middle expression. Simplify all three parts. When dividing by a negative, flip all the signs.

Exam Tip: Be very careful with the inequality direction when dividing by negative numbers - flip all of them.

Example 6. Solve the following inequalities: (i) \( -12 < 4 - \frac{3x}{5} ≤ 2 \) (ii) \( -3 ≤ 4 - \frac{7x}{2} ≤ 18 \)
Answer:
(i) Subtract 4 from all parts: -12 - 4 < \( -\frac{3x}{5} \) ≤ 2 - 4, so -16 < \( -\frac{3x}{5} \) ≤ -2. Multiply by -5/3 (flip all signs): -16 × (-5/3) > x ≥ -2 × (-5/3), which is 80/3 > x ≥ 10/3, or 10/3 ≤ x < 80/3. Solution: [10/3, 80/3).
(ii) Subtract 4 from all parts: -3 - 4 ≤ \( -\frac{7x}{2} \) ≤ 18 - 4, so -7 ≤ \( -\frac{7x}{2} \) ≤ 14. Divide by -7/2 (flip all signs): -7 ÷ (-7/2) ≥ x ≤ 14 ÷ (-7/2), which is 2 ≥ x ≥ -4, or -4 ≤ x ≤ 2. Solution: [-4, 2].
In simple words: Isolate the fraction in the middle. Then multiply/divide to clear it, flipping inequalities if needed.

Exam Tip: When multiplying or dividing all parts by a negative number, reverse all inequality signs simultaneously.

Example 7. Find the values of x satisfying \( -\frac{1}{5} ≤ \frac{3x}{10} + 1 < \frac{2}{5} \), x ∈ R. Show the solution set on the number line.
Answer: Subtract 1 from all parts: \( -\frac{1}{5} - 1 ≤ \frac{3x}{10} < \frac{2}{5} - 1 \), so \( -\frac{6}{5} ≤ \frac{3x}{10} < -\frac{3}{5} \). Multiply by 10/3 (positive, so no flip): -6/5 × 10/3 ≤ x < -3/5 × 10/3, which is -4 ≤ x < -2. Solution: {x : x ∈ R, -4 ≤ x < -2} or [-4, -2). On the number line, draw a solid circle at -4 and an open circle at -2, shading between them.
In simple words: Remove the "1" by subtracting. Then remove the fraction by multiplying by its reciprocal (which is positive, so no flip).

Exam Tip: Check the sign of what you are multiplying or dividing by - only flip if it is negative.

Example 8. Solve the following inequality and show the solution set on the number line: 2y - 3 < y + 2 ≤ 3y + 5.
Answer: This splits into two: 2y - 3 < y + 2 AND y + 2 ≤ 3y + 5. From the first: subtract y to get y - 3 < 2, add 3 to get y < 5. From the second: subtract y to get 2 ≤ 2y + 5, subtract 5 to get -3 ≤ 2y, divide by 2 to get -3/2 ≤ y. So the overlap is: -3/2 ≤ y < 5. Solution: {y : y ∈ R, -3/2 ≤ y < 5} or [-3/2, 5). On the number line, draw a solid circle at -3/2 and an open circle at 5, shading between them.
In simple words: Split into two separate inequalities. Solve each. Find the overlap - what satisfies both.

Exam Tip: When y appears in the middle of a chain, split it into two inequalities sharing that middle expression.

Example 9. Solve the following system of linear inequalities: 2(2x + 3) - 10 < 6(x - 2), \( \frac{2x-3}{4} + 6 ≥ 4 + \frac{4x}{3} \).
Answer: Solve each separately.
From 2(2x + 3) - 10 < 6(x - 2): expand to 4x + 6 - 10 < 6x - 12, simplify to 4x - 4 < 6x - 12, rearrange to -2x < -8, divide by -2 (flip) to get x > 4 ... (i)
From \( \frac{2x-3}{4} + 6 ≥ 4 + \frac{4x}{3} \): multiply by 12 to get 3(2x - 3) + 72 ≥ 48 + 16x, expand to 6x - 9 + 72 ≥ 48 + 16x, simplify to 6x + 63 ≥ 48 + 16x, rearrange to -10x ≥ -15, divide by -10 (flip) to get x ≤ 1.5 ... (ii)
Looking at (i) and (ii), we need x > 4 AND x ≤ 1.5. There is no overlap, so the solution set is empty: φ.
In simple words: Solve each one. The first says x > 4. The second says x ≤ 1.5. No number can be both, so there is no solution.

Exam Tip: When two regions do not overlap, the solution is the empty set. Draw the number line showing both regions separately, and you will see there is no common area.

Example 10. An electrician can be hired under two schemes: Scheme I: Rs 500 and Rs 70 per hour. Scheme II: Rs 120 per hour. For what number of hours does Scheme I pay better wages?
Answer: Let x be the number of hours. Under Scheme I, total wage = Rs (500 + 70x). Under Scheme II, total wage = Rs 120x. Scheme I gives better wages if: 500 + 70x > 120x. Rearrange: 500 > 50x. Divide by 50: 10 > x, or x < 10. Scheme I gives better wages for fewer than 10 hours of work.
In simple words: Write the total wage for each scheme. Set Scheme I's wage > Scheme II's wage. Solve for x.

Exam Tip: Always set up inequalities carefully - "better" means greater than, so use >.

Example 11. Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.
Answer: Let x be the smaller of the two consecutive even integers. Then the other is x + 2. We need: x > 5 AND x + (x + 2) < 23. From the second: 2x + 2 < 23, so 2x < 21, so x < 10.5. Combining: 5 < x < 10.5. Since x is even and must be in this range, possible values are x ∈ {6, 8, 10}. The corresponding second integers are x + 2 ∈ {8, 10, 12}. So the pairs are: (6, 8), (8, 10), (10, 12).
In simple words: Let the first number be x. The second is x + 2. Write both conditions as inequalities. Solve to find x, then list the valid pairs.

Exam Tip: When finding integer pairs, always check which values from your range actually satisfy both conditions.

Example 12. Find all pairs of consecutive odd positive integers, both of which are larger than 10, such that their sum is less than 40.
Answer: Let x be the smaller of the two consecutive odd integers. Then the other is x + 2. We need: x > 10 AND x + (x + 2) < 40. From the second: 2x + 2 < 40, so 2x < 38, so x < 19. Combining: 10 < x < 19. Since x is odd and must be in this range, possible values are x ∈ {11, 13, 15, 17}. The corresponding second integers are x + 2 ∈ {13, 15, 17, 19}. So the pairs are: (11, 13), (13, 15), (15, 17), (17, 19).
In simple words: Same approach as Example 11, but with odd numbers. Odd numbers are 1, 3, 5, 7, 9, 11, ...

Exam Tip: Consecutive odd integers differ by 2, just like consecutive even integers.

Example 13. An acid solution is to be kept between 30°C and 35°C. What is the range of temperature in Fahrenheit if the conversion formula is \( C = \frac{5}{9}(F - 32) \)?
Answer: Given: 30 < C < 35. Substitute \( C = \frac{5}{9}(F - 32) \): 30 < \frac{5}{9}(F - 32) < 35. Divide by 5: 6 < \frac{1}{9}(F - 32) < 7. Multiply by 9: 54 < F - 32 < 63. Add 32: 86 < F < 95. The temperature range is between 86°F and 95°F.
In simple words: Substitute the conversion formula into the inequality. Then solve by using inverse operations on all three parts.

Exam Tip: When using a conversion formula inside an inequality, substitute it directly and work with all parts simultaneously.

Example 14. IQ is calculated by \( \text{IQ} = \frac{\text{MA}}{\text{CA}} \times 100 \), where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12-year-old children, what is the range of their mental age?
Answer: Given: CA = 12 and 80 ≤ IQ ≤ 140. Substitute: \( 80 ≤ \frac{\text{MA}}{12} \times 100 ≤ 140 \), which simplifies to \( 80 ≤ \frac{25 \times \text{MA}}{3} ≤ 140 \). Multiply by 3/25: \( \frac{3}{25} \times 80 ≤ \text{MA} ≤ \frac{3}{25} \times 140 \), so 9.6 ≤ MA ≤ 16.8. The mental age ranges from 9.6 to 16.8 years.
In simple words: Substitute the known value (CA = 12) and the inequality bounds into the formula. Solve for MA by working through all three parts.

Exam Tip: Simplify the coefficients early (like 100/12 = 25/3) to make the algebra cleaner.

Example 15. The water in a pool is considered normal when the average pH reading of three daily measurements is between 8.2 and 8.5. If the first two readings are 8.48 and 8.35, what pH range in the third reading keeps the average normal?
Answer: Let x be the third pH reading. Average = (8.48 + 8.35 + x)/3 = (16.83 + x)/3. For normal pH: 8.2 < (16.83 + x)/3 < 8.5. Multiply by 3: 24.6 < 16.83 + x < 25.5. Subtract 16.83: 7.77 < x < 8.67. The third reading must be between 7.77 and 8.67.
In simple words: Write the average. Set it between 8.2 and 8.5. Solve for the third reading.

Exam Tip: Always write out the average formula carefully before substituting into the inequality.

Example 16. In drilling the world's deepest hole, the temperature T (in °C) at x km below Earth's surface was given by \( T = 30 + 25(x - 3) \), where 3 ≤ x ≤ 15. At what depth is the temperature between 155°C and 205°C?
Answer: Given: 155 < T < 205. Substitute \( T = 30 + 25(x - 3) \): 155 < 30 + 25(x - 3) < 205. Subtract 30: 125 < 25(x - 3) < 175. Divide by 25: 5 < x - 3 < 7. Add 3: 8 < x < 10. The depth is between 8 km and 10 km.
In simple words: Substitute the temperature formula into the inequality. Then isolate x using inverse operations.

Exam Tip: When you have a linear expression in the middle (like 25(x - 3)), divide by its coefficient first, then add/subtract the constant.

Example 17. A man cuts three pieces from a 91 cm cardboard. The second piece is 3 cm longer than the shortest. The third piece is twice as long as the shortest. If the third piece is at least 5 cm longer than the second, what are the possible lengths of the shortest piece?
Answer: Let x cm be the length of the shortest piece. Then: second piece = (x + 3) cm, third piece = 2x cm. Conditions: (i) total ≤ 91: x + (x + 3) + 2x ≤ 91, so 4x + 3 ≤ 91, so 4x ≤ 88, so x ≤ 22. (ii) third ≥ second + 5: 2x ≥ (x + 3) + 5, so 2x ≥ x + 8, so x ≥ 8. Combining: 8 ≤ x ≤ 22. The shortest piece must be between 8 cm and 22 cm long.
In simple words: Express each piece in terms of the shortest. Use the given conditions to write inequalities. Solve and find the overlap.

Exam Tip: Always translate word problems into inequalities carefully - "at least" means ≥, "not more than" means ≤.

Example 18. An 8% boric acid solution is diluted by mixing with a 2% solution. The mixture should be more than 4% but less than 6% boric acid. If we have 640 litres of 8% solution, how many litres of 2% solution must be added?
Answer: Let x litres of 2% solution be added. Total mixture = (x + 640) litres. Acid in mixture = 2% of x + 8% of 640. For 4% < acid % < 6%: 4% of (x + 640) < 2% of x + 8% of 640 < 6% of (x + 640). Simplifying: 0.04(x + 640) < 0.02x + 0.08(640) < 0.06(x + 640). This gives: 0.04x + 25.6 < 0.02x + 51.2 < 0.06x + 38.4. From the left: 0.02x < 25.6, so x < 1280. From the right: 0.02x + 51.2 < 0.06x + 38.4, so 12.8 < 0.04x, so x > 320. Therefore: 320 < x < 1280. Between 320 and 1280 litres of 2% solution must be added.
In simple words: Set up the inequality using percentages: % acid in mixture is between 4% and 6%. Solve for x.

Exam Tip: With mixture problems, track the acid separately: acid total = (% in solution 1 × volume 1) + (% in solution 2 × volume 2).

Exercise 6.2

Question 1. (i) 3x - 1 ≥ 5, x + 2 > -1 (ii) 2x - 7 < 11, 3x + 4 < -5.
Answer:
(i) From 3x - 1 ≥ 5: 3x ≥ 6, so x ≥ 2. From x + 2 > -1: x > -3. The overlap is x ≥ 2.
(ii) From 2x - 7 < 11: 2x < 18, so x < 9. From 3x + 4 < -5: 3x < -9, so x < -3. The overlap is x < -3.
In simple words: Solve each. Find where both are true at once.

Exam Tip: Overlapping regions means AND - you need both conditions satisfied.

Question 2. (i) 4 - 5x > -11, 4x + 11 ≤ -13 (ii) 5x + 1 > -24, 5x - 1 < 24.
Answer:
(i) From 4 - 5x > -11: -5x > -15, x < 3. From 4x + 11 ≤ -13: 4x ≤ -24, x ≤ -6. The overlap is x ≤ -6.
(ii) From 5x + 1 > -24: 5x > -25, x > -5. From 5x - 1 < 24: 5x < 25, x < 5. The overlap is -5 < x < 5.
In simple words: Rearrange both. Find the region where both hold.

Exam Tip: Remember to flip when multiplying/dividing by negative numbers.

Question 3. (i) 7x - 8 < 4x + 7, \( -\frac{x}{2} > 4 \) (ii) x + 2 ≤ 5, 3x - 4 > -2 + x.
Answer:
(i) From 7x - 8 < 4x + 7: 3x < 15, x < 5. From \( -\frac{x}{2} > 4 \): x < -8. The overlap is x < -8.
(ii) From x + 2 ≤ 5: x ≤ 3. From 3x - 4 > -2 + x: 2x > 2, x > 1. The overlap is 1 < x ≤ 3.
In simple words: First solve each inequality. Then find the overlapping region.

Exam Tip: When one answer is "all numbers less than -8" and the other is "all numbers less than 5", the overlap is "all less than -8".

Question 4. (i) 3x - 7 < 5 + x, 11 - 5x ≤ 1 (ii) 4x + 5 > 3x, -(x + 3) + 4 ≤ -2x + 5.
Answer:
(i) From 3x - 7 < 5 + x: 2x < 12, x < 6. From 11 - 5x ≤ 1: -5x ≤ -10, x ≥ 2. The overlap is 2 ≤ x < 6.
(ii) From 4x + 5 > 3x: x > -5. From -(x + 3) + 4 ≤ -2x + 5: -x - 3 + 4 ≤ -2x + 5, so -x + 1 ≤ -2x + 5, x ≤ 4. The overlap is -5 < x ≤ 4.
In simple words: Rearrange each separately. Then find what satisfies both.

Exam Tip: When one inequality gives x ≥ 2 and the other x < 6, the overlap is 2 ≤ x < 6 (solid dot at 2, open at 6).

Question 5. (i) \( -2 - \frac{x}{4} ≤ \frac{1+x}{3} \), 3 - x < 4(x - 3) (ii) 5x - 7 < 3(x + 3), \( 1 - \frac{3x}{2} ≥ x - 4 \).
Answer:
(i) From \( -2 - \frac{x}{4} ≤ \frac{1+x}{3} \): multiply by 12 to get -24 - 3x ≤ 4(1 + x), so -24 - 3x ≤ 4 + 4x, so -28 ≤ 7x, x ≥ -4. From 3 - x < 4x - 12: 15 < 5x, x > 3. The overlap is x > 3.
(ii) From 5x - 7 < 3x + 9: 2x < 16, x < 8. From \( 1 - \frac{3x}{2} ≥ x - 4 \): multiply by 2 to get 2 - 3x ≥ 2x - 8, so 10 ≥ 5x, x ≤ 2. The overlap is x ≤ 2.
In simple words: Clear fractions first if needed. Then rearrange and find the overlap.

Exam Tip: Always multiply by the L.C.M. of all denominators to clear fractions efficiently.

Question 6. (i) -4x + 1 ≥ 0, 3 - 4x < 0 (ii) 4x + 3 ≥ 2x + 17, 3x - 5 < -2.
Answer:
(i) From -4x + 1 ≥ 0: -4x ≥ -1, x ≤ 1/4. From 3 - 4x < 0: -4x < -3, x > 3/4. There is no overlap (no x can be ≤ 1/4 AND > 3/4), so the solution is ∅.
(ii) From 4x + 3 ≥ 2x + 17: 2x ≥ 14, x ≥ 7. From 3x - 5 < -2: 3x < 3, x < 1. There is no overlap, so the solution is ∅.
In simple words: When the regions do not overlap (like x ≤ 1/4 and x > 3/4), there is no solution.

Exam Tip: An empty solution set (∅) is valid - it means the two inequalities contradict each other.

Question 7. Solve the following system of inequalities: (i) 3x - 7 > 2(x - 6), 6 - x > 11 - 2x (ii) 5(2x - 7) - 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47. Also show their solutions on the number line.
Answer:
(i) From 3x - 7 > 2x - 12: x > -5. From 6 - x > 11 - 2x: x > 5. The overlap is x > 5. On the number line, draw an open circle at 5 and shade right.
(ii) From 5(2x - 7) - 3(2x + 3) ≤ 0: 10x - 35 - 6x - 9 ≤ 0, so 4x - 44 ≤ 0, x ≤ 11. From 2x + 19 ≤ 6x + 47: -4x ≤ 28, x ≥ -7. The overlap is -7 ≤ x ≤ 11. On the number line, draw a solid circle at -7 and a solid circle at 11, shading between them.
In simple words: Solve each. Find the overlap. Draw the region on the number line.

Exam Tip: Use open circles for < or >, and solid circles for ≤ or ≥.

Question 8. (i) -8 ≤ 5x - 3 < 7 (ii) -2 ≤ 6x - 1 < 2.
Answer:
(i) Add 3: -5 ≤ 5x < 10. Divide by 5: -1 ≤ x < 2. Solution: [-1, 2).
(ii) Add 1: -1 ≤ 6x < 3. Divide by 6: -1/6 ≤ x < 1/2. Solution: [-1/6, 1/2).
In simple words: Apply the same operation to all three parts.

Exam Tip: Treat chained inequalities as one unit - do the same thing to all parts.

Question 9. (i) -2 < 1 - 3x < 7 (ii) -12 < 3x - 5 ≤ 4.
Answer:
(i) Subtract 1: -3 < -3x < 6. Divide by -3 (flip): 1 > x > -2, or -2 < x < 1. Solution: (-2, 1).
(ii) Add 5: -7 < 3x ≤ 9. Divide by 3: -7/3 < x ≤ 3. Solution: (-7/3, 3].
In simple words: Isolate x in the middle. When dividing by negative, flip both inequality signs.

Exam Tip: Always check after rewriting - if you divided by a negative, both inequalities should flip.

Question 10. (i) \( 7 ≤ \frac{3x+11}{2} ≤ 11 \) (ii) \( -5 ≤ \frac{5-3x}{2} ≤ 8 \).
Answer:
(i) Multiply by 2: 14 ≤ 3x + 11 ≤ 22. Subtract 11: 3 ≤ 3x ≤ 11. Divide by 3: 1 ≤ x ≤ 11/3. Solution: [1, 11/3].
(ii) Multiply by 2: -10 ≤ 5 - 3x ≤ 16. Subtract 5: -15 ≤ -3x ≤ 11. Divide by -3 (flip): 5 ≥ x ≥ -11/3, or -11/3 ≤ x ≤ 5. Solution: [-11/3, 5].
In simple words: Multiply to clear fractions. Then isolate x. If dividing by negative, flip.

Exam Tip: After dividing by -3, rewrite as -11/3 ≤ x ≤ 5 to show the interval in standard form.

Question 11. (i) \( -15 < \frac{3(x-2)}{5} ≤ 0 \) (ii) \( -5 ≤ \frac{2-3x}{4} ≤ 9 \).
Answer:
(i) Multiply by 5: -75 < 3(x - 2) ≤ 0. Divide by 3: -25 < x - 2 ≤ 0. Add 2: -23 < x ≤ 2. Solution: (-23, 2].
(ii) Multiply by 4: -20 ≤ 2 - 3x ≤ 36. Subtract 2: -22 ≤ -3x ≤ 34. Divide by -3 (flip): 22/3 ≥ x ≥ -34/3, or -34/3 ≤ x ≤ 22/3. Solution: [-34/3, 22/3].
In simple words: Clear fractions first. Then isolate the variable. Remember to flip when dividing by negative.

Exam Tip: Simplify fractions like 22/3 ≈ 7.33 to help you visualize where they lie on a number line.

Question 12. (i) \( -12 ≤ \frac{4-3x}{5} < 2 \) (ii) 3(2 - x) ≤ 2x + 1 < 15.
Answer:
(i) Multiply by 5: -60 ≤ 4 - 3x < 10. Subtract 4: -64 ≤ -3x < 6. Divide by -3 (flip): 64/3 ≥ x > -2, or -2 < x ≤ 64/3. Solution: (-2, 64/3].
(ii) Expand 3(2 - x): 6 - 3x ≤ 2x + 1 < 15. From left: 5 ≤ 5x, so x ≥ 1. From right: 2x < 14, so x < 7. The overlap is 1 ≤ x < 7. Solution: [1, 7).
In simple words: For (i), multiply and isolate x. For (ii), split into two and find the overlap.

Exam Tip: Chained inequalities can be solved as one unit (multiply all parts) or split into two (if there's a variable term in the middle).

Question 13. Solve the inequality and show the solution on the number line: \( -\frac{2}{3} < \frac{x}{3} + 1 ≤ \frac{2}{3} \), x ∈ R.
Answer: Subtract 1 from all parts: \( -\frac{2}{3} - 1 < \frac{x}{3} ≤ \frac{2}{3} - 1 \), so \( -\frac{5}{3} < \frac{x}{3} ≤ -\frac{1}{3} \). Multiply by 3: -5 < x ≤ -1. Solution: {x : x ∈ R, -5 < x ≤ -1} or (-5, -1]. On the number line, draw an open circle at -5 and a solid circle at -1, shading between them.
In simple words: Remove the constant from the middle expression. Then multiply to clear the fraction.

Exam Tip: When the inequality has a + or - constant with the fraction, subtract/add that first.

Question 14. Find the range of values of x satisfying \( -\frac{1}{3} ≤ \frac{2x-1}{3} - \frac{1}{6} < \frac{1}{6} \), x ∈ R. Show these values on the real number line.
Answer: Add 1/6 to all parts: \( -\frac{1}{3} + \frac{1}{6} ≤ \frac{2x-1}{3} < \frac{1}{6} + \frac{1}{6} \), so \( -\frac{1}{6} ≤ \frac{2x-1}{3} < \frac{1}{3} \). Multiply by 3: \( -\frac{1}{2} ≤ 2x - 1 < 1 \). Add 1: \( \frac{1}{2} ≤ 2x < 2 \). Divide by 2: \( \frac{1}{4} ≤ x < 1 \). Solution: [1/4, 1). On the number line, draw a solid circle at 1/4 and an open circle at 1, shading between them.
In simple words: Isolate the fraction step by step. Add, multiply, and divide as needed.

Exam Tip: Work carefully through each step - one arithmetic error will shift the entire answer.

Question 15. Find the range of values of x satisfying \( -2\frac{2}{3} ≤ x + \frac{1}{3} < 3\frac{1}{3} \), x ∈ R. Show the values on the number line.
Answer: Convert mixed numbers: \( -\frac{8}{3} ≤ x + \frac{1}{3} < \frac{10}{3} \). Subtract 1/3: \( -\frac{8}{3} - \frac{1}{3} ≤ x < \frac{10}{3} - \frac{1}{3} \), so \( -3 ≤ x < 3 \). Solution: [-3, 3). On the number line, draw a solid circle at -3 and an open circle at 3, shading between them.
In simple words: Convert mixed numbers to improper fractions. Then subtract the constant from the middle expression.

Exam Tip: Always convert mixed numbers early to avoid arithmetic errors.

Question 16. Find all pairs of consecutive odd positive integers, both of which are smaller than 10, such that their sum is more than 11.
Answer: Let x be the smaller odd integer. Then the other is x + 2. We need: x < 10, x + 2 < 10, and x + (x + 2) > 11. From the last: 2x + 2 > 11, so x > 4.5. Since x is an odd positive integer and 4.5 < x < 8 (from x + 2 < 10), the possible values are x ∈ {5, 7}. The pairs are: (5, 7) and (7, 9).
In simple words: Set up both conditions. Find x from the sum condition and combine with the "smaller than 10" condition.

Exam Tip: "Smaller than 10" for both means x + 2 < 10, so x < 8.

Question 17. A solution is to be kept between 68°F and 77°F. What is the range in temperature in Celsius if the conversion formula is \( F = \frac{9}{5}C + 32 \)?
Answer: Given: 68 < F < 77. Substitute \( F = \frac{9}{5}C + 32 \): 68 < \frac{9}{5}C + 32 < 77. Subtract 32: 36 < \frac{9}{5}C < 45. Multiply by 5/9: 20 < C < 25. The temperature range is between 20°C and 25°C.
In simple words: Substitute the Fahrenheit formula into the inequality. Solve for C.

Exam Tip: When multiplying by 5/9 (a positive number), the inequality signs do not flip.

Question 18. A solution is to be kept between 40°C and 45°C. What is the range of temperature in Fahrenheit if the conversion formula is \( F = \frac{9}{5}C + 32 \)?
Answer: Given: 40 < C < 45. Substitute into \( F = \frac{9}{5}C + 32 \): multiply the inequality by 9/5: 72 < \frac{9}{5}C < 81. Add 32: 104 < F < 113. The temperature range is between 104°F and 113°F.
In simple words: Apply the conversion formula to all parts of the Celsius range.

Exam Tip: To convert a range, apply the formula to the endpoints: C = 40 → F = 104, and C = 45 → F = 113.

Question 19. A manufacturer has 600 litres of 12% acid solution. How many litres of 30% acid solution must be added so that the resulting mixture has more than 15% but less than 18% acid?
Answer: Let x litres of 30% solution be added. Total volume = (x + 600) litres. Acid in mixture = 30% of x + 12% of 600. For the mixture to be between 15% and 18%: 15% of (x + 600) < 30% of x + 12% of 600 < 18% of (x + 600). Simplify: 0.15(x + 600) < 0.30x + 0.12(600) < 0.18(x + 600). From left: 0.15x + 90 < 0.30x + 72, so 18 < 0.15x, so x > 120. From right: 0.30x + 72 < 0.18x + 108, so 0.12x < 36, so x < 300. Therefore: 120 < x < 300. Between 120 and 300 litres of 30% solution must be added.
In simple words: Acid in mixture = (% 1 × volume 1) + (% 2 × volume 2). Set this between 15% and 18% of total volume.

Exam Tip: Always track: acid in = acid from solution 1 + acid from solution 2, divided by total volume, gives % acid in mixture.

Question 20. How many litres of water will have to be added to 1125 litres of 45% acid solution so that the mixture contains more than 25% but less than 30% acid?
Answer: Let x litres of water be added. Total volume = (x + 1125) litres. Acid in mixture = 45% of 1125 (water adds 0% acid). For 25% < acid% < 30%: 0.25(x + 1125) < 0.45(1125) < 0.30(x + 1125). From left: 0.25x + 281.25 < 506.25, so 0.25x < 225, so x < 900. From right: 506.25 < 0.30x + 337.5, so 168.75 < 0.30x, so x > 562.5. Therefore: 562.5 < x < 900. Between 562.5 and 900 litres of water must be added.
In simple words: Water has 0% acid. The acid amount stays constant at 45% of 1125. Diluting makes the % smaller. Find when it is between 25% and 30%.

Exam Tip: When diluting with water, the acid amount does not change - only the total volume increases, making the percentage smaller.

6.3 Graphical Solution of Linear Inequalities

A statement of any one of these types:

(i) x < a (ii) x ≤ a (iii) x > a (iv) x ≥ a or

(i) y < b (ii) y ≤ b (iii) y > b (iv) y ≥ b

where a and b are real numbers, is called a linear inequality in one variable (x or y).

A statement of any one of these types:

(i) ax + by < c (ii) ax + by ≤ c (iii) ax + by > c (iv) ax + by ≥ c

where a, b, c are real numbers and a and b are non-zero, is called a linear inequality in two variables (x and y).

A straight line l divides the cartesian plane into two regions. Each region is called a half-plane. A vertical line divides the plane into left half and right half planes. A non-vertical line divides the plane into lower half and upper half planes (see Fig. 6.10 (i) and (ii)).

Graphical Solution of a Linear Inequality

An ordered pair (α, β) of real numbers may or may not satisfy a given inequality.

The set of all ordered pairs that satisfy a given inequality is called the solution set of that inequality.

Since there is a one-to-one correspondence between ordered pairs and points on a coordinate plane, we can represent the solution set by points on the plane.

The region of the plane containing all the points whose coordinates satisfy a given inequality is called the solution region (or graph) of the inequality.

To find the graphical solution of an inequality in one variable:

(i) Draw the straight line x = a (or y = b) as needed.

Question. To find the graphical solution of an inequality in one variable
(ii) The straight line x = a divides the co-ordinate plane into two halves.
(iii) One half is the graph of x < a and the other half is the graph of x > a. Shade the solution region of the given inequality.
(iv) If an inequality is of the form x ≤ a or x ≥ a, then the points on the line x = a are also included in the solution region and draw a dark line in the solution region.
(v) If an inequality is of the form x < a or x > a, then the points on the line x = a are not included in the solution region and draw a broken line in the solution region.
Answer: These are the key steps for graphing a single-variable inequality on a number line or coordinate plane. Start by drawing the boundary line x = a, which separates the plane into two regions. For inequalities using ≤ or ≥, include the boundary line itself by drawing it as a solid line. For inequalities using < or >, exclude the boundary by drawing it as a dashed line. Then shade the appropriate half-plane based on which values satisfy the inequality.
In simple words: Draw a line at x = a. If the inequality has "equal to" (≤ or ≥), make the line solid. If it does not have "equal to" (< or >), make the line dashed. Then shade the side where the inequality is true.

Exam Tip: Always check which side to shade by picking a test point (like the origin) and seeing if it makes the inequality true. Solid vs. dashed lines are critical for marks - examiners check this carefully.

Question. Example 1: Let us consider the inequality x ≥ 1 in one variable.
Answer: Draw the vertical line x = 1, which splits the plane into two regions. The values satisfying x ≥ 1 are all points that lie to the right of the line or on the line itself. Because the inequality includes "equal to," use a solid line to show that points on x = 1 are part of the solution. The shaded region representing the solution covers all points from x = 1 extending to the right (Fig. 6.11).
In simple words: Draw a solid vertical line at x = 1. Shade everything to the right of it, including the line. This shows all numbers bigger than or equal to 1.

Exam Tip: Verify your shading by testing a point like x = 2: since 2 ≥ 1 is true, x = 2 must be in the shaded region. If it is not, your shading is wrong.

Question. Example 2: Let us consider the inequality 2y - 3 < 0 in one variable.
Answer: Draw the horizontal line 2y - 3 = 0, which gives y = 3/2. This line divides the plane into two halves. Points that lie below this horizontal line satisfy the inequality 2y - 3 < 0. Because the inequality uses only < (no "equal to"), draw the line as dashed to show that points on y = 3/2 are not part of the solution. The shaded region below the line represents all solutions (Fig. 6.12).
In simple words: Draw a dashed horizontal line at y = 3/2. Shade everything below it. Do not shade the line itself because the inequality does not include "equal to".

Exam Tip: Use a test point like (0, 0): since 2(0) - 3 = -3 < 0 is true, the origin should be in your shaded region. This confirms you shaded the correct half.

Question. To find the graphical solution of an inequality in two variables
(i) Draw the straight line ax + by = c.
(ii) The straight line ax + by = c divides the co-ordinate plane into two halves.
(iii) One half is the graph of ax + by < c and the other half is the graph of ax + by > c.
(iv) In order to identify the half plane represented by an inequality, take any point (α, β) not lying on the line ax + by = c and check whether it satisfies the inequality or not. If it satisfies, then the inequality represents the half plane containing the point and shade this region; otherwise, the inequality represents that half plane which does not contain the point within it. For convenience, we take the point (0, 0). However, if (0, 0) lies on the line then take any other point of the plane not lying on the line.
(v) If an inequality is of the form ax + by ≤ c or ax + by ≥ c, then the points on the line ax + by = c are also included in the solution region and draw a dark line in the solution region.
(vi) If an inequality is of the form ax + by < c or ax + by > c, then the points on the line ax + by = c are not to be included in the solution region and draw a broken line in the solution region.
Answer: To graph an inequality in two variables, first graph the boundary line ax + by = c by finding two points on it. This line splits the plane into two half-planes. To find which half-plane contains solutions, select a test point not on the line - the origin (0, 0) works well unless it lies on the boundary. Substitute the test point into the inequality; if true, that half-plane contains solutions; if false, the other half-plane does. Draw the boundary line solid if the inequality includes ≤ or ≥, and dashed if it uses < or >. Then shade the correct region.
In simple words: Draw the line ax + by = c. Pick a point not on the line and test it. If it makes the inequality true, shade that side. If false, shade the other side. Use a solid line if "equal to" is included, dashed if not.

Exam Tip: Always pick (0, 0) as your test point first - it is easiest to substitute. Only use a different point if (0, 0) sits exactly on the boundary line.

Question. Example 1: Let us consider the inequality 3x + 4y ≥ 12 in two variables.
Answer: Begin by drawing the line 3x + 4y = 12. To find two points on this line, set y = 0 to get x = 4, giving point A (4, 0); then set x = 0 to get y = 3, giving point B (0, 3). Draw the line through these points. Next, test the origin (0, 0): substituting gives 3(0) + 4(0) = 0, and 0 < 12 is false, so (0, 0) does not satisfy the inequality. This means the solution region is the half-plane that does not contain the origin - the region above the line. Since the inequality includes ≥, use a solid line and include points on the boundary. The shaded region above the line (including the line itself) represents all solutions (Fig. 6.13).
In simple words: Draw a solid line through (4, 0) and (0, 3). Test (0, 0): it fails the inequality, so shade the side that does not have (0, 0). That is the upper side of the line.

Exam Tip: Always show your test point calculation step - examiners reward this. If your shaded region does not match the side of your test point, you have made an error.

Question. Example 2: Consider the inequality 3x - 4y < 12 in two variables.
Answer: First, draw the line 3x - 4y = 12. Setting y = 0 gives x = 4, so point A is (4, 0). Setting x = 0 gives y = -3, so point B is (0, -3). Draw the line through these two points. Now test (0, 0): substituting gives 3(0) - 4(0) = 0, and 0 < 12 is true, so (0, 0) does satisfy the inequality. The solution region is therefore the half-plane containing the origin - the region above the line. Since the inequality uses only < (no "equal to"), draw the line as dashed to show that boundary points are excluded. The shaded region above the line (not including the line itself) represents the solution set (Fig. 6.14).
In simple words: Draw a dashed line through (4, 0) and (0, -3). Test (0, 0): it works, so shade the side that has (0, 0). That is above the line. The dashed line means do not include points on it.

Exam Tip: Dashed vs. solid lines are worth checking twice - a dashed line is a common exam trap. Always mark whether your boundary line is included (solid) or excluded (dashed).

Question. Example 1. Solve the inequality 2x - 3y > 6 graphically.
Answer: To solve 2x - 3y > 6 graphically, first rewrite the boundary as 2x - 3y = 6, or y = (2x - 6) / 3. Build a table: when x = 0, y = -2; when x = 3, y = 0. Plot the points A (0, -2) and B (3, 0) and draw the line through them. This line splits the plane into two regions. Test the origin (0, 0) in the original inequality: 2(0) - 3(0) = 0, and 0 > 6 is false. Since the test point fails, the solution region is on the side of the plane that does not contain the origin. Draw the line as dashed because the inequality uses > (no "equal to"). Shade the region below the line, which forms the solution set (Fig. 6.15).
In simple words: Graph y = (2x - 6) / 3 as a dashed line through (0, -2) and (3, 0). Test (0, 0): it fails, so shade the side without it. That is the lower region.

Exam Tip: When solving these problems, always create a small table of x and y values first - it makes plotting the line quick and accurate. Write out your test point check so the examiner sees your reasoning.

Exercise 6.1

Exercise 6.2

Exercise 6.3

The solution set comprises all points in the shaded area of the coordinate plane. A dotted line indicates that points lying on the line are not part of the solution set.

Question 1. Solve y < 2 graphically.
Answer: Draw the horizontal line y = 2, which is a dashed line because the inequality does not include equality. The region below this line (all points with y-coordinate less than 2) forms the solution set. Shade this entire region to show all solutions.

Exam Tip: Use a test point like (0, 0). Since 0 < 2 is true, (0, 0) must be in your shaded region. If it is not, your shading is incorrect.

Question 2. Solve y + 3 ≥ 0 graphically.
Answer: Rewrite as y ≥ -3. Draw the horizontal line y = -3 as a solid line (since ≥ includes equality). All points on and above this line satisfy the inequality. Shade this entire region to display the solution set.

Exam Tip: Always rewrite the inequality in standard form first to avoid errors. For y + 3 ≥ 0, moving the 3 to the right gives y ≥ -3 immediately.

Question 3. Solve x < -3 graphically.
Answer: Draw the vertical line x = -3 as a dashed line (since < does not include equality). All points to the left of this line (where x is less than -3) form the solution. Shade the left half-plane to show all solutions.

Exam Tip: For vertical lines, always remember: x = a means a vertical line at x = a; y = b means a horizontal line at y = b. Confusing these costs easy marks.

Question 4. Solve \( x \geq -\frac{9}{5} \) graphically.
Answer: Draw the vertical line \( x = -\frac{9}{5} \) as a solid line (since ≥ includes equality). All points on and to the right of this line satisfy the inequality. Shade this region to represent the complete solution set.

Exam Tip: Convert fractions to decimals to plot accurately: \( -\frac{9}{5} = -1.8 \). This helps position the line correctly on the coordinate plane.

Question 5. Solve x + y ≤ 5 graphically.
Answer: Draw the line x + y = 5 (which passes through (5, 0) and (0, 5)) as a solid line because the inequality includes ≤. Test (0, 0): since 0 + 0 = 0 ≤ 5 is true, the origin lies in the solution region. Shade the area below and to the left of the line, including the line itself, to show all solutions.

Exam Tip: Always find two clear points on the boundary line by setting x = 0 and y = 0 separately. This makes drawing the line quick and accurate.

Question 6. Solve 3x + 2y > 6 graphically.
Answer: Draw the line 3x + 2y = 6 as a dashed line (since > does not include equality). To find two points: set x = 0 to get y = 3, and set y = 0 to get x = 2. The line passes through (2, 0) and (0, 3). Test (0, 0): 3(0) + 2(0) = 0, and 0 > 6 is false, so the origin is not in the solution. Shade the region above the line that does not contain the origin.

Exam Tip: Always perform the test-point check and show it in your answer. Examiners look for this step and give credit for it, even if shading is slightly imperfect.