Class 11 : Maths (In English) – Lesson 5. Linear Inequalities
EXPLANATION & SUMMARY
🔷 Explanation (~1700 Words)
🔹 1️⃣ Introduction
🔵 In mathematics, inequality is a statement that compares two quantities using symbols like
➡️ < (less than), > (greater than), ≤ (less than or equal to), ≥ (greater than or equal to).
🧠 Unlike equations (which show equality), inequalities show that one side is larger or smaller than the other.
✏️ Example:
x + 3 > 7 means the value of x is greater than 4.
🔹 2️⃣ Types of Inequalities
🟢 (a) Numerical inequalities: involve only numbers.
✔️ Example: 3 < 5 is always true.
🟡 (b) Literal inequalities: involve variables.
✔️ Example: x + 2 < 5 depends on the value of x.
🔴 (c) Linear inequalities: where the power of the variable is 1.
✔️ Example: 2x + 3 ≤ 7.
🔹 3️⃣ Symbols Used
🔵 < : less than
🟢 > : greater than
🟠 ≤ : less than or equal to
🔴 ≥ : greater than or equal to
🔹 4️⃣ Solution of an Inequality
🧭 The solution is the set of all values of the variable which satisfy the inequality.
✏️ Example:
Solve x + 2 < 5
➡️ Subtract 2 from both sides: x < 3
✔️ So, solution set: all real numbers less than 3.
🔹 5️⃣ Basic Rules for Solving Inequalities
💡 Rule 1: You can add or subtract the same number on both sides without changing the inequality.
✏️ Example:
x + 3 < 7
Subtract 3 ⇒ x < 4 ✅
💡 Rule 2: You can multiply or divide both sides by the same positive number without changing the sign.
✏️ Example:
2x ≤ 8
Divide by 2 ⇒ x ≤ 4 ✅
💡 Rule 3 (⚠️ Important):
When multiplying or dividing both sides by a negative number, the inequality sign reverses.
✏️ Example:
–2x > 6
Divide by –2 ⇒ x < –3 ✅
🔹 6️⃣ Representation of Solution on Number Line
📈 Represent by a line:
Open circle (○) for < or >
Filled circle (●) for ≤ or ≥
✏️ Example:
x < 3 → draw an open circle at 3 and shade all points to the left.
x ≥ 2 → draw a filled circle at 2 and shade all points to the right.
🔹 7️⃣ Linear Inequalities in One Variable
🧠 Form: ax + b < c
✔️ Solving gives a range for x.
✏️ Example:
3x – 5 ≤ 4
➡️ 3x ≤ 9
➡️ x ≤ 3
✔️ Solution: all x ≤ 3.
🔹 8️⃣ Linear Inequalities in Two Variables
🧭 Form: ax + by < c
Solution is a region in the xy-plane.
✏️ Example:
x + y ≤ 4
✔️ Boundary line: x + y = 4
✔️ Region: all points (x, y) satisfying x + y < 4, shaded region below the line.
💡 Use test point (0, 0):
If it satisfies inequality → region includes origin.
🔹 9️⃣ Graphical Representation (Steps)
➤ Step 1: Write the inequality.
➤ Step 2: Replace sign with “=” to find boundary line.
➤ Step 3: Draw boundary line (solid for ≤ or ≥, dotted for < or >).
➤ Step 4: Choose a test point (usually 0,0).
➤ Step 5: If it satisfies, shade that region.
✏️ Example:
2x + 3y ≤ 6
Boundary: 2x + 3y = 6
Test (0,0): 0 ≤ 6 → ✅ so shade region containing origin.
🔹 10️⃣ System of Linear Inequalities
📘 A set of two or more inequalities in x and y.
Solution = common region satisfying all inequalities.
✏️ Example:
x + y ≤ 5,
x ≥ 0, y ≥ 0
✅ Solution: region bounded by line x + y = 5 and coordinate axes.
🔹 11️⃣ Word Problems
🧠 Convert real-life situations into inequalities and solve.
✏️ Example:
A number is greater than 3 and less than or equal to 7.
⇒ 3 < x ≤ 7 ✅
Another Example:
A person can spend at most ₹100 on two items costing ₹x and ₹y.
⇒ x + y ≤ 100 ✅
🔹 12️⃣ Solution Set
✔️ Set builder form: {x ∈ R | condition}
✔️ Interval form:
x < a → (–∞, a)
x ≤ a → (–∞, a]
a < x < b → (a, b)
a ≤ x ≤ b → [a, b]
✏️ Example:
x < 3 → (–∞, 3)
2 ≤ x ≤ 5 → [2, 5]
🔹 13️⃣ Operations on Intervals
🧠 Union (∪): combines sets
🧠 Intersection (∩): common part
✏️ Example:
A = (–∞, 4), B = (2, ∞)
A ∩ B = (2, 4)
🔹 14️⃣ Double Inequalities
Form: a < x < b
Solve like normal equations.
✏️ Example:
2 < 3x + 2 < 8
➡️ Subtract 2: 0 < 3x < 6
➡️ Divide by 3: 0 < x < 2
✔️ Solution: (0, 2)
🔹 15️⃣ Properties
✔️ Transitive: If a < b and b < c ⇒ a < c
✔️ Addition: a < b ⇒ a + c < b + c
✔️ Multiplication: If c > 0, a < b ⇒ ac < bc
If c < 0, inequality reverses.
🔹 16️⃣ Application in Real Life
📚 Used in:
Budget planning
Production limits
Constraints in optimization (Linear Programming)
Engineering and economics
🔹 17️⃣ Common Mistakes
⚠️ Don’t forget to reverse sign when multiplying/dividing by negative.
⚠️ Don’t confuse open and closed intervals.
⚠️ Always check test points when graphing.
🔹 18️⃣ Summary Table
A linear inequality is a mathematical statement showing the relationship between two expressions using the symbols <, >, ≤, or ≥. Unlike equations, inequalities express that one quantity is smaller or larger than another, not necessarily equal.
A linear inequality in one variable has the form ax + b < c or ax + b ≤ c, where the variable has power 1. Its solution is a set of real numbers satisfying the condition.
For two variables, the inequality is ax + by < c or ax + by ≤ c, and the solution is a region in the coordinate plane.
Rules for solving inequalities:
The same number can be added or subtracted on both sides without changing the sign.
Multiplying or dividing by a positive number keeps the sign unchanged.
Multiplying or dividing by a negative number reverses the inequality sign.
Representation on a number line:
Use open circle (○) for < or >.
Use filled circle (●) for ≤ or ≥.
Shade left or right according to the inequality direction.
Solution set forms:
Set-builder form: {x ∈ R | condition}
Interval form:
(a, b) → open interval, [a, b] → closed interval.
Example: x ≤ 3 → (–∞, 3]
Graphical representation in two variables:
Replace inequality with “=” to draw boundary line.
Use solid line for ≤, ≥; dotted line for <, >.
Choose a test point to find the solution region.
Operations on intervals:
Union (∪) combines sets; Intersection (∩) gives common part.
Applications: Used in budgeting, profit limits, and constraints in optimization.
Key points:
Reverse sign when multiplying/dividing by negative.
Represent solution properly on number line or graph.
Goal: Identify the complete set of values satisfying the inequality.
🔹 19️⃣ Conclusion
🎯 Linear inequalities describe ranges of possible values, not exact values.
🎯 Mastery of solving and representing them is key to advanced topics like Linear Programming.
🧭 Summary (~300 Words)
🔹 Inequality: comparison of two expressions using <, >, ≤, ≥.
🔹 Linear inequality: variable’s highest power is 1.
🔹 Solution: all values satisfying the condition.
🔹 Rules:
✔ Add/Subtract same value → no change
✔ Multiply/Divide by positive → no change
✔ Multiply/Divide by negative → reverse sign
🔹 Representation:
– On number line using open/closed circles.
– In 2D plane as a shaded region.
🔹 Intervals:
( ) → open interval, [ ] → closed interval.
Examples:
x < 2 → (–∞, 2)
2 ≤ x ≤ 5 → [2, 5]
🔹 System of inequalities: region common to all conditions.
🔹 Word problems: translate sentences to inequalities.
🔹 Operations: Union (∪), Intersection (∩).
🔹 Double inequalities: solve step by step.
🔹 Real-life uses: budgeting, limits, optimization.
💡 Key Point: Reverse sign if multiplied/divided by negative.
🧠 Goal: Understand solution set and represent it visually and algebraically.
📝 Quick Recap
✔️ Inequality shows comparison, not equality
✔️ Solution set: all x satisfying inequality
✔️ Reverse sign when multiplying/dividing by negative
✔️ Represent on number line or graph
✔️ Use intervals (open/closed) for sets
✔️ Common region = solution of multiple inequalities
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QUESTIONS FROM TEXTBOOK
📗 Exercise 5.1
🔵 Question 1
Solve 24x < 100, when
(i) x is a natural number.
(ii) x is an integer.
🟢 Answer
➡️ 24x < 100
➡️ x < 100/24 = 25/6 ≈ 4.166…
✔️ (i) Natural numbers satisfying: {1, 2, 3, 4}
✔️ (ii) Integers satisfying: all integers ≤ 4 (…, −3, −2, −1, 0, 1, 2, 3, 4)
🔵 Question 2
Solve −12x > 30, when
(i) x is a natural number.
(ii) x is an integer.
🟢 Answer
➡️ −12x > 30
➡️ Divide by −12 (reverse sign): x < −30/12 = −5/2
✔️ (i) Natural numbers < −5/2 → none (∅)
✔️ (ii) Integers: x ≤ −3 (…, −5, −4, −3)
🔵 Question 3
Solve 5x − 3 < 7, when
(i) x is an integer.
(ii) x is a real number.
🟢 Answer
➡️ 5x − 3 < 7
➡️ 5x < 10
➡️ x < 2
✔️ (i) Integers: x ≤ 1 (…, −1, 0, 1)
✔️ (ii) Reals: (−∞, 2)
🔵 Question 4
Solve 3x + 8 > 2, when
(i) x is an integer.
(ii) x is a real number.
🟢 Answer
➡️ 3x + 8 > 2
➡️ 3x > −6
➡️ x > −2
✔️ (i) Integers: x ≥ −1 (−1, 0, 1, 2, …)
✔️ (ii) Reals: (−2, ∞)
🔵 Question 5:
4x + 3 < 5x + 7
🟢 Answer:
➡️ Bring terms together: 4x – 5x < 7 – 3
➡️ -x < 4
➡️ Multiply by (-1) and reverse sign: x > -4
✔️ Solution: x ∈ (-4, ∞)
🔵 Question 6:
3x – 7 > 5x – 1
🟢 Answer:
➡️ 3x – 5x > -1 + 7
➡️ -2x > 6
➡️ Multiply by (-1): 2x < -6
➡️ x < -3
✔️ Solution: x ∈ (-∞, -3)
🔵 Question 7:
3(x – 1) ≤ 2(x – 3)
🟢 Answer:
➡️ 3x – 3 ≤ 2x – 6
➡️ 3x – 2x ≤ -6 + 3
➡️ x ≤ -3
✔️ Solution: x ∈ (-∞, -3]
🔵 Question 8:
3(2 – x) ≥ 2(1 – x)
🟢 Answer:
➡️ 6 – 3x ≥ 2 – 2x
➡️ -3x + 2x ≥ 2 – 6
➡️ -x ≥ -4
➡️ x ≤ 4
✔️ Solution: x ∈ (-∞, 4]
🔵 Question 9:
x + (x/2) + (x/3) < 11
🟢 Answer:
➡️ Take LCM 6 → (6x + 3x + 2x)/6 < 11
➡️ 11x/6 < 11
➡️ 11x < 66
➡️ x < 6
✔️ Solution: x ∈ (-∞, 6)
🔵 Question 10:
(x/3) – (x/2) + 1 > 0
🟢 Answer:
➡️ LCM 6 → (2x – 3x)/6 + 1 > 0
➡️ (-x)/6 + 1 > 0
➡️ -x + 6 > 0
➡️ x < 6
✔️ Solution: x ∈ (-∞, 6)
🔵 Question 11:
(3/5)(x – 2) ≤ (5/3)(2 – x)
🟢 Answer:
➡️ Multiply by 15 → 9(x – 2) ≤ 25(2 – x)
➡️ 9x – 18 ≤ 50 – 25x
➡️ 34x ≤ 68
➡️ x ≤ 2
✔️ Solution: x ∈ (-∞, 2]
🔵 Question 12:
(1/2)(3x/5 + 4) ≤ (1/3)(x – 6)
🟢 Answer:
➡️ Multiply by 6 → 3(3x/5 + 4) ≤ 2(x – 6)
➡️ (9x/5) + 12 ≤ 2x – 12
➡️ Multiply by 5 → 9x + 60 ≤ 10x – 60
➡️ -x ≤ -120
➡️ x ≥ 120
✔️ Solution: x ∈ [120, ∞)
🔵 Question 13:
2(2x + 3) – 10 < 6(x – 2)
🟢 Answer:
➡️ 4x + 6 – 10 < 6x – 12
➡️ 4x – 4 < 6x – 12
➡️ -4 + 12 < 2x
➡️ 8 < 2x
➡️ x > 4
✔️ Solution: x ∈ (4, ∞)
🔵 Question 14:
37 – (3x + 5) ≥ 9x – 8(x – 3)
🟢 Answer:
➡️ 37 – 3x – 5 ≥ 9x – 8x + 24
➡️ 32 – 3x ≥ x + 24
➡️ 8 ≥ 4x
➡️ x ≤ 2
✔️ Solution: x ∈ (-∞, 2]
🔵 Question 15:
(x/4) ≤ (5x – 2)/3 – (7x – 3)/5
🟢 Answer:
➡️ Multiply by 60 → 15x ≤ 20(5x – 2) – 12(7x – 3)
➡️ 15x ≤ 100x – 40 – 84x + 36
➡️ 15x ≤ 16x – 4
➡️ -x ≤ -4
➡️ x ≥ 4
✔️ Solution: x ∈ [4, ∞)
🔵 Question 16:
(2x – 1)/3 ≥ (3x – 2)/4 ≥ (2 – x)/5
🟢 Answer:
🧩 Part 1: (2x – 1)/3 ≥ (3x – 2)/4
➡️ Multiply by 12 → 8x – 4 ≥ 9x – 6
➡️ -x ≥ -2
➡️ x ≤ 2
🧩 Part 2: (3x – 2)/4 ≥ (2 – x)/5
➡️ Multiply by 20 → 15x – 10 ≥ 8 – 4x
➡️ 19x ≥ 18
➡️ x ≥ 18/19
✔️ Combined Solution: x ∈ [18/19, 2]
🔵 Question 17
3x − 2 < 2x + 1
🟢 Answer
➡️ 3x − 2x < 1 + 2
➡️ x < 3
✔️ Solution set: (−∞, 3)
📌 Number line: open circle at 3, arrow to the left (all numbers less than 3).
🔵 Question 18
5x − 3 ≥ 3x − 5
🟢 Answer
➡️ 5x − 3x ≥ −5 + 3
➡️ 2x ≥ −2
➡️ x ≥ −1
✔️ Solution set: [−1, ∞)
📌 Number line: closed (filled) circle at −1, arrow to the right (all numbers ≥ −1).
🔵 Question 19
3 (1 − x) < 2 (x + 4)
🟢 Answer
➡️ 3 − 3x < 2x + 8
➡️ −5x < 5
➡️ x > −1
✔️ Solution set: (−1, ∞)
📌 Number line: open circle at −1, arrow to the right (all numbers greater than −1).
🔵 Question 20
x/2 ≥ (5x − 2)/3 − (7x − 3)/5
🟢 Answer
➡️ LCM = 30 → multiply through
➡️ 15x ≥ 10(5x − 2) − 6(7x − 3)
➡️ 15x ≥ 50x − 20 − 42x + 18
➡️ 15x ≥ 8x − 2
➡️ 7x ≥ −2
➡️ x ≥ −2/7
✔️ Solution set: [−2/7, ∞)
📌 Number line: closed (filled) circle at −2/7, arrow to the right (all numbers ≥ −2/7).
🔵 Question 21:
Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
🟢 Answer:
Let marks in the third test = x
➡️ Average = (70 + 75 + x)/3
➡️ Condition: (70 + 75 + x)/3 ≥ 60
➡️ 145 + x ≥ 180
➡️ x ≥ 35
✔️ Ravi should score at least 35 marks in the third test.
🔵 Question 22:
To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.
🟢 Answer:
Let marks in 5th exam = x
➡️ Average = (87 + 92 + 94 + 95 + x)/5
➡️ Condition: ≥ 90
➡️ 368 + x ≥ 450
➡️ x ≥ 82
✔️ Sunita must score at least 82 marks in the 5th exam.
🔵 Question 23:
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 first odd integer = x
➡️ Next consecutive odd = x + 2
➡️ Condition 1: both < 10 ⇒ x + 2 < 10 ⇒ x < 8
➡️ Condition 2: sum > 11 ⇒ x + (x + 2) > 11 ⇒ 2x + 2 > 11 ⇒ 2x > 9 ⇒ x > 4.5
So x = 5 or 7 (odd and < 8)
✔️ Pairs: (5, 7) and (7, 9)
🔵 Question 24:
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 first even integer = x
➡️ Next consecutive even = x + 2
➡️ Condition 1: x > 5 ⇒ smallest even possible is 6
➡️ Condition 2: sum < 23 ⇒ x + (x + 2) < 23 ⇒ 2x + 2 < 23 ⇒ 2x < 21 ⇒ x < 10.5
Possible even integers satisfying 6 ≤ x ≤ 10
✔️ Pairs: (6, 8), (8, 10)
🔵 Question 25:
The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.
🟢 Answer:
Let the shortest side = x cm
➡️ Then the longest side = 3x cm
➡️ The third side = (3x − 2) cm
🧮 Perimeter = x + 3x + (3x − 2) = 7x − 2
🧭 Condition:
7x − 2 ≥ 61
➡️ 7x ≥ 63
➡️ x ≥ 9
✔️ Minimum length of the shortest side = 9 cm
🧩 Other sides:
Longest side = 27 cm
Third side = 25 cm
📘 Triangle sides = 9 cm, 27 cm, 25 cm
🔵 Question 26:
A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second?
🟢 Answer:
Let the shortest length = x cm
➡️ Second length = (x + 3) cm
➡️ Third length = 2x cm
🧮 Total length ≤ 91 cm
x + (x + 3) + 2x ≤ 91
➡️ 4x + 3 ≤ 91
➡️ 4x ≤ 88
➡️ x ≤ 22
🧭 Also, third ≥ second + 5
2x ≥ (x + 3) + 5
➡️ 2x ≥ x + 8
➡️ x ≥ 8
✔️ Range of x: 8 ≤ x ≤ 22
✅ Possible lengths of shortest board = between 8 cm and 22 cm
📏 Then:
Second length = x + 3 → between 11 cm and 25 cm
Third length = 2x → between 16 cm and 44 cm
📘 Final Answer:
Shortest side x ∈ [8, 22]
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OTHER IMPORTANT QUESTIONS FOR EXAMS
🔷 Response 1 (Q1–Q18: Section A – MCQs)
🔵 Question 1:
Which of the following is a linear inequality?
1️⃣ x² + 2x ≤ 4
2️⃣ 2x + 3 > 7
3️⃣ x³ + 1 < 0
4️⃣ 5/x ≥ 2
🟢 Answer: 2️⃣ 2x + 3 > 7
🔵 Question 2:
The solution of x + 3 < 5 is
1️⃣ x < 8
2️⃣ x < 2
3️⃣ x > 2
4️⃣ x > 8
🟢 Answer: 2️⃣ x < 2
🔵 Question 3:
Which of the following reverses the inequality sign?
1️⃣ Addition
2️⃣ Subtraction
3️⃣ Multiplication by positive number
4️⃣ Multiplication by negative number
🟢 Answer: 4️⃣ Multiplication by negative number
🔵 Question 4:
The solution of –2x ≥ 6 is
1️⃣ x ≥ –3
2️⃣ x ≤ –3
3️⃣ x > –3
4️⃣ x < –3
🟢 Answer: 2️⃣ x ≤ –3
🔵 Question 5:
If 3x – 5 ≤ 4, then x ≤
1️⃣ 3
2️⃣ –3
3️⃣ 9
4️⃣ –9
🟢 Answer: 1️⃣ 3
🔵 Question 6:
The solution set of 2x + 1 < 5 is
1️⃣ x < 2
2️⃣ x > 2
3️⃣ x < 3
4️⃣ x > 3
🟢 Answer: 1️⃣ x < 2
🔵 Question 7:
Which symbol represents “less than”?
1️⃣ >
2️⃣ <
3️⃣ ≥
4️⃣ ≤
🟢 Answer: 2️⃣ <
🔵 Question 8:
If x < 4, the interval form of the solution is
1️⃣ (–∞, 4)
2️⃣ [–∞, 4]
3️⃣ (4, ∞)
4️⃣ [4, ∞)
🟢 Answer: 1️⃣ (–∞, 4)
🔵 Question 9:
If x ≥ 3, the interval notation is
1️⃣ [3, ∞)
2️⃣ (3, ∞)
3️⃣ (–∞, 3]
4️⃣ (–∞, 3)
🟢 Answer: 1️⃣ [3, ∞)
🔵 Question 10:
Which of the following is true?
1️⃣ –2 < 1
2️⃣ 5 < 2
3️⃣ 0 > 4
4️⃣ 3 < –1
🟢 Answer: 1️⃣ –2 < 1
🔵 Question 11:
The solution of 3x – 2 > 4 is
1️⃣ x > 2
2️⃣ x < 2
3️⃣ x > 3
4️⃣ x < 3
🟢 Answer: 1️⃣ x > 2
🔵 Question 12:
Which of the following is not a linear inequality?
1️⃣ x + 2 ≤ 5
2️⃣ 2x – 3 > 7
3️⃣ x² + 1 ≥ 0
4️⃣ 4x < 10
🟢 Answer: 3️⃣ x² + 1 ≥ 0
🔵 Question 13:
If –3x > 9, then x
1️⃣ < –3
2️⃣ > –3
3️⃣ < 3
4️⃣ > 3
🟢 Answer: 1️⃣ x < –3
🔵 Question 14:
The graph of x ≤ 2 on number line is
1️⃣ Open circle at 2, right shade
2️⃣ Filled circle at 2, left shade
3️⃣ Open circle at 2, left shade
4️⃣ Filled circle at 2, right shade
🟢 Answer: 2️⃣ Filled circle at 2, left shade
🔵 Question 15:
In which quadrant is the region x + y ≤ 5 and x, y ≥ 0 located?
1️⃣ I quadrant
2️⃣ II quadrant
3️⃣ III quadrant
4️⃣ IV quadrant
🟢 Answer: 1️⃣ I quadrant
🔵 Question 16:
The solution of inequality 2x + 3 ≥ 7 is
1️⃣ x ≥ 2
2️⃣ x ≤ 2
3️⃣ x > 2
4️⃣ x < 2
🟢 Answer: 1️⃣ x ≥ 2
🔵 Question 17:
For inequality ax < b, solution is
1️⃣ x < b/a if a > 0
2️⃣ x > b/a if a > 0
3️⃣ x < b/a if a < 0
4️⃣ None
🟢 Answer: 1️⃣ x < b/a if a > 0
🔵 Question 18:
The set {x | x > –2, x ≤ 3} in interval notation is
1️⃣ [–2, 3)
2️⃣ (–2, 3]
3️⃣ (–2, 3)
4️⃣ [–2, 3]
🟢 Answer: 2️⃣ (–2, 3]
🔷 Section B – Short Answer (2 Marks Each)
🔵 Question 19:
Solve: 2x – 3 < 7
🟢 Answer:
➡️ Step 1: Add 3 on both sides
2x < 10
➡️ Step 2: Divide by 2
x < 5
✔️ Solution: x < 5
✔️ Interval form: (–∞, 5)
🔵 Question 20:
Solve: 3x + 4 ≥ 1
🟢 Answer:
➡️ Step 1: Subtract 4
3x ≥ –3
➡️ Step 2: Divide by 3
x ≥ –1
✔️ Solution: x ≥ –1
✔️ Interval form: [–1, ∞)
🔵 Question 21:
Solve: –2x + 5 > 1
🟢 Answer:
➡️ Step 1: Subtract 5
–2x > –4
➡️ Step 2: Divide by –2 (reverse sign)
x < 2
✔️ Solution: x < 2
✔️ Interval form: (–∞, 2)
🔵 Question 22:
Find the solution set of 5x – 7 ≤ 3x + 1
🟢 Answer:
➡️ Step 1: Bring variables to one side
5x – 3x ≤ 1 + 7
➡️ Step 2: Simplify
2x ≤ 8
➡️ Step 3: Divide by 2
x ≤ 4
✔️ Solution: x ≤ 4
✔️ Interval form: (–∞, 4]
🔵 Question 23:
Write the solution set for 3x + 2 ≥ 11 in set-builder form.
🟢 Answer:
➡️ Step 1: Subtract 2
3x ≥ 9
➡️ Step 2: Divide by 3
x ≥ 3
✔️ Set-builder form: {x ∈ ℝ | x ≥ 3}
✔️ Interval form: [3, ∞)
🔷 Section C – Mid-Length (3 Marks Each)
🔵 Question 24:
Solve the inequality –3x + 2 ≤ 8 and represent the solution on a number line.
🟢 Answer:
➡️ Step 1: Subtract 2
–3x ≤ 6
➡️ Step 2: Divide by –3 (reverse sign)
x ≥ –2
✔️ Solution: x ≥ –2
✔️ Interval form: [–2, ∞)
📈 Number line:
Filled circle at –2, shade to right (since ≥).
🔵 Question 25:
Solve: 2 – 3x < 5x + 10
🟢 Answer:
➡️ Step 1: Bring terms
–3x – 5x < 10 – 2
➡️ Step 2: Simplify
–8x < 8
➡️ Step 3: Divide by –8 (reverse sign)
x > –1
✔️ Solution: x > –1
✔️ Interval form: (–1, ∞)
🔵 Question 26:
Solve the inequality: 4x – 3 ≥ 2x + 1
🟢 Answer:
➡️ Step 1: Move variables
4x – 2x ≥ 1 + 3
➡️ Step 2: Simplify
2x ≥ 4
➡️ Step 3: Divide by 2
x ≥ 2
✔️ Solution: x ≥ 2
✔️ Interval form: [2, ∞)
🔵 Question 27:
Solve: –2(3x – 1) ≤ 8
🟢 Answer:
➡️ Step 1: Expand
–6x + 2 ≤ 8
➡️ Step 2: Subtract 2
–6x ≤ 6
➡️ Step 3: Divide by –6 (reverse sign)
x ≥ –1
✔️ Solution: x ≥ –1
✔️ Interval form: [–1, ∞)
🔵 Question 28 (3 marks)
Solve the double inequality and write the solution in interval form: 2 < 3x − 4 ≤ 11
🟢 Answer:
➤ Step 1: Add 4 to all three parts → 6 < 3x ≤ 15
➤ Step 2: Divide all parts by 3 (positive, sign unchanged) → 2 < x ≤ 5
✔ Solution set (set-builder): { x ∈ ℝ | 2 < x ≤ 5 }
✔ Interval form: (2, 5]
📈 Number line: open circle at 2, filled circle at 5, shade between them.
🔵 Question 29 (5 marks, Long Answer)
Solve and represent on number line: 5 − 2(3x − 1) ≥ 3x + 4
🟢 Answer:
➤ Step 1: Expand LHS → 5 − 6x + 2 ≥ 3x + 4
➤ Step 2: Combine like terms → 7 − 6x ≥ 3x + 4
➤ Step 3: Bring variables to one side → −6x − 3x ≥ 4 − 7
➤ Step 4: Simplify → −9x ≥ −3
➤ Step 5: Divide by −9 (reverse sign) → x ≤ 1/3
✔ Solution set: { x ∈ ℝ | x ≤ 1/3 }
✔ Interval form: (−∞, 1/3]
📈 Number line: filled circle at 1/3, shade to the left.
🔵 Question 30 (5 marks, Long Answer)
Solve the system of inequalities and describe the solution region:
x + y ≤ 6, x ≥ 0, y ≥ 0
🟢 Answer:
➤ Step 1: Boundary line for first inequality → x + y = 6
➤ Step 2: Intercepts: (6, 0) and (0, 6)
➤ Step 3: Since ≤, boundary line is solid; test point (0, 0) satisfies 0 ≤ 6 → region includes origin.
➤ Step 4: Non-negativity constraints x ≥ 0, y ≥ 0 restrict region to first quadrant.
✔ Solution region: triangular area bounded by axes and the line x + y = 6 in the first quadrant.
✔ Vertex points: (0, 0), (6, 0), (0, 6).
✔ Set description: { (x, y) ∈ ℝ² | x ≥ 0, y ≥ 0, x + y ≤ 6 }
🔵 Question 31 (5 marks, Long Answer)
Find all real x satisfying both inequalities simultaneously:
3x − 7 < 2x + 5 and 4 − x ≤ 2x + 10
🟢 Answer:
Part A: 3x − 7 < 2x + 5
➤ Step 1: Subtract 2x → x − 7 < 5
➤ Step 2: Add 7 → x < 12
Part B: 4 − x ≤ 2x + 10
➤ Step 1: Add x → 4 ≤ 3x + 10
➤ Step 2: Subtract 10 → −6 ≤ 3x
➤ Step 3: Divide by 3 → −2 ≤ x
Combine (intersection): −2 ≤ x < 12
✔ Solution set: { x ∈ ℝ | −2 ≤ x < 12 }
✔ Interval form: [−2, 12)
📈 Number line: filled circle at −2, open circle at 12, shade between them.
🔵 Question 32 (5 marks, Case/Application)
A shop offers two plans for buying notebooks x and pens y (both nonnegative integers).
Constraints:
(i) Each notebook costs ₹20 and each pen costs ₹10. Total spend at most ₹300.
(ii) At least twice as many notebooks as pens: x ≥ 2y.
Write the system of linear inequalities and describe the feasible region.
🟢 Answer:
➤ Step 1: Cost constraint → 20x + 10y ≤ 300 → 2x + y ≤ 30
➤ Step 2: Ratio constraint → x ≥ 2y
➤ Step 3: Nonnegativity → x ≥ 0, y ≥ 0
✔ System: { 2x + y ≤ 30, x − 2y ≥ 0, x ≥ 0, y ≥ 0 }
Region description (xy-plane):
• Bounded by the line 2x + y = 30 (solid, includes boundary) in the first quadrant.
• On or to the right of the line x = 2y (since x − 2y ≥ 0).
• Within the first quadrant (x ≥ 0, y ≥ 0).
Key intercepts:
• 2x + y = 30 → (15, 0) and (0, 30).
• x = 2y → passes through (0, 0) and (10, 5) (intersection with 2x + y = 30 found by 2(2y) + y = 30 → 5y = 30 → y = 6, x = 12).
Feasible polygon vertices: (0, 0), (12, 6), (15, 0).
✔ Feasible set: { (x, y) ∈ ℝ² | 2x + y ≤ 30, x ≥ 2y, x ≥ 0, y ≥ 0 }
🔵 Question 33 (5 marks, Case/Application)
A student needs at least 60 minutes of study in Math (M) and English (E) combined, but not more than 120 minutes due to schedule. Also, Math time must be at least 1.5 times English time. Write inequalities and find possible range of E.
🟢 Answer:
Let M ≥ 0, E ≥ 0 (minutes).
➤ Step 1: Total time lower bound → M + E ≥ 60
➤ Step 2: Total time upper bound → M + E ≤ 120
➤ Step 3: Ratio constraint → M ≥ 1.5E → M − 1.5E ≥ 0
Find E-range using bounds and ratio:
From M ≥ 1.5E and M + E ≥ 60 → substitute M = 1.5E (minimum M for given E)
➤ Step 4: 1.5E + E ≥ 60 → 2.5E ≥ 60 → E ≥ 24
From M ≥ 1.5E and M + E ≤ 120 → using minimal M = 1.5E again
➤ Step 5: 1.5E + E ≤ 120 → 2.5E ≤ 120 → E ≤ 48
Also E ≥ 0 by nonnegativity; combined with above gives 24 ≤ E ≤ 48.
Construct corresponding M-range for any feasible E: M ∈ [1.5E, 120 − E] with E ∈ [24, 48].
✔ Inequalities system: { M + E ≥ 60, M + E ≤ 120, M − 1.5E ≥ 0, M ≥ 0, E ≥ 0 }
✔ Final range for E: 24 ≤ E ≤ 48 (minutes).
Example feasible point: E = 30, M = 60 (satisfies all).
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