Class 9 Mathematics Chapter 2: Linear Equations in One Variable — Complete Important Questions with Solutions

Linear Equations in One Variable (Chapter 2) carries 8–12 marks in CBSE Class 9 Mathematics papers, distributed across MCQ (1–2 marks), short-answer (3–4 marks), and long-answer sections (3–5 marks). The board pattern prioritizes three skill levels: (1) Direct solving—isolating the variable using inverse operations; (2) Application—converting word problems into equations and solving; (3) Conceptual depth—justifying steps, verifying solutions, and handling fractional or decimal coefficients. Recent papers (2022–24) emphasize equations where variables appear on both sides (e.g., 3x + 5 = 2x + 9) and multi-step word problems requiring careful equation formation. The new rationalized syllabus removed some extraneous content but strengthened focus on problem-solving over rote methods. Mastering these 18 questions ensures you can tackle any board variant—whether direct numerical solving, comparison of solutions, or word-based scenarios involving ages, speeds, or geometry. Board examiners often repeat patterns within different contexts; recognizing these templates through varied important questions is the fastest way to full marks.

Multiple-choice questions test quick recall and conceptual clarity. Here are five representative 1-mark questions: **Q1.** The solution of the equation 2x + 3 = 9 is: (A) 2 (B) 3 (C) 4 (D) 5 **Answer:** (B) 3 **Explanation:** 2x + 3 = 9 → 2x = 6 → x = 3 **Q2.** Which of the following is a linear equation in one variable? (A) 2x + 3y = 5 (B) x² + 2x = 0 (C) 5x − 7 = 0 (D) x + y + z = 0 **Answer:** (C) 5x − 7 = 0 **Explanation:** A linear equation in one variable contains only one variable with power 1. Only option (C) satisfies this. **Q3.** If 3(x − 2) = 9, then x equals: (A) 3 (B) 5 (C) −1 (D) 6 **Answer:** (B) 5 **Explanation:** 3(x − 2) = 9 → 3x − 6 = 9 → 3x = 15 → x = 5 **Q4.** The solution of (x/2) + 3 = 5 is: (A) 2 (B) 4 (C) 6 (D) 8 **Answer:** (B) 4 **Explanation:** (x/2) + 3 = 5 → (x/2) = 2 → x = 4 **Q5.** If 5x − 4 = 2x + 5, then x equals: (A) 2 (B) 3 (C) 4 (D) 5 **Answer:** (B) 3 **Explanation:** 5x − 4 = 2x + 5 → 5x − 2x = 5 + 4 → 3x = 9 → x = 3

Two-mark questions require clear working and correct algebraic steps. Partial credit is awarded for method, so show every transposition. **Q1.** Solve: 4x − 3 = 2x + 7 **Solution:** 4x − 3 = 2x + 7 4x − 2x = 7 + 3 (transposing variables to left, constants to right) 2x = 10 x = 5 **Verification:** LHS = 4(5) − 3 = 17; RHS = 2(5) + 7 = 17 ✓ **Q2.** Solve: (3x/5) − 2 = (x/5) + 4 **Solution:** (3x/5) − 2 = (x/5) + 4 (3x/5) − (x/5) = 4 + 2 (collecting x terms on left) (2x/5) = 6 2x = 30 x = 15 **Verification:** LHS = (3×15/5) − 2 = 9 − 2 = 7; RHS = (15/5) + 4 = 3 + 4 = 7 ✓ **Q3.** Solve: 2(x − 3) + 3(x − 2) = 8 **Solution:** 2x − 6 + 3x − 6 = 8 (expanding brackets) 5x − 12 = 8 5x = 20 x = 4 **Verification:** 2(4−3) + 3(4−2) = 2(1) + 3(2) = 2 + 6 = 8 ✓ **Q4.** Solve: (2x + 1)/3 = (x + 3)/2 **Solution:** Cross-multiplying: 2(2x + 1) = 3(x + 3) 4x + 2 = 3x + 9 4x − 3x = 9 − 2 x = 7 **Verification:** LHS = (2×7+1)/3 = 15/3 = 5; RHS = (7+3)/2 = 10/2 = 5 ✓ **Q5.** Solve: 5(x + 2) − 2(3x − 1) = 3(x + 1) **Solution:** 5x + 10 − 6x + 2 = 3x + 3 −x + 12 = 3x + 3 12 − 3 = 3x + x 9 = 4x x = 9/4 or 2.25 **Verification:** LHS = 5(2.25+2) − 2(3×2.25−1) = 5(4.25) − 2(5.75) = 21.25 − 11.5 = 9.75; RHS = 3(2.25+1) = 3(3.25) = 9.75 ✓

Three-mark questions often combine equation-solving with word problem translation or verification of multiple solutions. **Q1.** A number is 8 more than half of itself. Find the number. **Solution:** Let the number be x. According to the problem: x = (x/2) + 8 Multiplying both sides by 2: 2x = x + 16 2x − x = 16 x = 16 **Answer:** The number is 16. **Verification:** Half of 16 is 8, and 8 + 8 = 16 ✓ **Q2.** The sum of three consecutive integers is 45. Find the integers. **Solution:** Let the three consecutive integers be x, x+1, and x+2. According to the problem: x + (x+1) + (x+2) = 45 3x + 3 = 45 3x = 42 x = 14 **Answer:** The three integers are 14, 15, and 16. **Verification:** 14 + 15 + 16 = 45 ✓ **Q3.** If I add 10 to a number and divide by 5, I get 4. What is the number? **Solution:** Let the number be x. According to the problem: (x + 10)/5 = 4 Multiplying both sides by 5: x + 10 = 20 x = 10 **Answer:** The number is 10. **Verification:** (10 + 10)/5 = 20/5 = 4 ✓ **Q4.** Solve and verify: (5x − 4)/8 = (x + 1)/4 **Solution:** Cross-multiplying: 4(5x − 4) = 8(x + 1) 20x − 16 = 8x + 8 20x − 8x = 8 + 16 12x = 24 x = 2 **Verification:** LHS = (5×2−4)/8 = 6/8 = 3/4; RHS = (2+1)/4 = 3/4 ✓ **Answer:** x = 2

Five-mark questions require detailed working, careful algebraic steps, and sometimes a brief explanation of the method used. These often appear as Part B or Section II in board papers. **Q1.** A man's age is 4 times his son's age. 5 years ago, the man's age was 7 times his son's age. Find their present ages. **Solution:** Let the son's present age be x years. Then the man's present age is 4x years. 5 years ago: Son's age = (x − 5) years Man's age = (4x − 5) years According to the given condition: 4x − 5 = 7(x − 5) 4x − 5 = 7x − 35 −5 + 35 = 7x − 4x 30 = 3x x = 10 Son's present age = 10 years Man's present age = 4 × 10 = 40 years **Verification:** 5 years ago, son was 5 and man was 35. Is 35 = 7 × 5? Yes, 35 = 35 ✓ **Answer:** Son is 10 years old; Man is 40 years old. **Q2.** A boat travels x km upstream and y km downstream. Its speed in still water is 8 km/h and the speed of current is 2 km/h. If the time taken upstream equals the time taken downstream, express y in terms of x. **Solution:** Speed upstream = 8 − 2 = 6 km/h Speed downstream = 8 + 2 = 10 km/h Time upstream = x/6 hours Time downstream = y/10 hours Given: Time upstream = Time downstream x/6 = y/10 Cross-multiplying: 10x = 6y y = 10x/6 y = 5x/3 **Answer:** y = 5x/3 or y = (5/3)x **Q3.** Solve: 3(2x − 1) − 2(3x − 5) = 1 − (x + 2). Also verify the solution. **Solution:** Expanding the left side: 6x − 3 − 6x + 10 = 1 − x − 2 Combining like terms: (6x − 6x) + (−3 + 10) = −1 − x 0 + 7 = −1 − x 7 = −1 − x Transposing: x = −1 − 7 x = −8 **Verification:** LHS = 3(2×(−8)−1) − 2(3×(−8)−5) = 3(−16−1) − 2(−24−5) = 3(−17) − 2(−29) = −51 + 58 = 7 RHS = 1 − (−8 + 2) = 1 − (−6) = 1 + 6 = 7 LHS = RHS ✓ **Answer:** x = −8

Higher-order thinking questions require conceptual understanding and multi-step reasoning. These are increasingly common in board papers. **Case-Study Question:** A shopkeeper buys pens and notebooks. Let p be the number of pens and n be the number of notebooks. Each pen costs ₹5 and each notebook costs ₹8. The shopkeeper spends ₹100 total. He also buys 5 more notebooks than pens. (i) Form two linear equations representing these conditions. (ii) Solve to find how many pens and notebooks he bought. (iii) If the shopkeeper instead had ₹110 to spend (keeping the condition "5 more notebooks"), how many items would he buy in total? **Solution:** (i) **Equation 1** (Total cost): 5p + 8n = 100 **Equation 2** (Relationship): n = p + 5 (ii) **Solving using substitution:** Substitute n = p + 5 into Equation 1: 5p + 8(p + 5) = 100 5p + 8p + 40 = 100 13p = 60 p = 60/13 ≈ 4.62 Since p must be a whole number, let's check if the problem has a solution in whole numbers. If p = 4, then n = 9. Verification: 5(4) + 8(9) = 20 + 72 = 92 ≠ 100 (close, but not exact) If p = 5, then n = 10: 5(5) + 8(10) = 25 + 80 = 105 ≠ 100 The exact solution is non-integer; in a real context, the shopkeeper would buy approximately 4–5 pens and 9–10 notebooks. (iii) **With ₹110 budget:** 5p + 8n = 110 and n = p + 5 Substituting: 5p + 8(p + 5) = 110 13p + 40 = 110 13p = 70 p = 70/13 ≈ 5.38 Again, this yields a non-integer. In practice, the shopkeeper could buy 5 pens and 10 notebooks (₹105) or 6 pens and 11 notebooks (₹118), depending on rounding preference. **Learning Point:** Real-world problems don't always yield "clean" answers. Understanding this reinforces that linear equations are tools for approximation and decision-making, not just textbook exercises.

Drilling these 18 questions once is insufficient; mastery requires recognizing patterns across varied contexts and building speed under time constraints. CBSETUTOR.ai's adaptive AI tutor personalizes your revision by: **1. Pattern Recognition:** The AI generates 50+ variants of each question type (e.g., "number problems," "age problems," "geometric problems") by modifying numbers and contexts while keeping the underlying algebraic structure identical. You internalize the method, not just one answer. **2. Real-Time Error Feedback:** When you transpose incorrectly, divide by a negative without flipping inequality signs, or forget verification, the AI immediately flags the mistake with a hyper-specific explanation, preventing re-learning of wrong methods. **3. Adaptive Difficulty Ladder:** Start with 1-mark direct solving, progress to 2-mark fractional equations, then 3-mark word problems, and finally 5-mark multi-step scenarios—all within a single learning session tailored to your pace. **4. Board-Pattern Simulation:** The AI generates full-length mock papers mirroring exact 2024–25 CBSE Class 9 Mathematics paper structure (Section A: 1-mark MCQs, Section B: 2-mark Q's, Section C: 3-mark Q's, Section D: 5-mark Q's), with time limits and auto-marking. **5. Spaced Revision:** The AI revisits mastered topics at optimal intervals to prevent forgetting, ensuring Chapter 2 stays fresh right until board day. Start a 3-day free trial at cbsetutor.ai to experience live, AI-driven drilling of these exact patterns with instant step-by-step corrections.