Class 10 Mathematics Notes Chapter 3 (Chapter 3) – Examplar Problems (English) Book

Alright class, let's get started with Chapter 3: Pair of Linear Equations in Two Variables from your NCERT Exemplar. This is a crucial chapter, not just for your board exams but also for various government competitive exams where quantitative aptitude is tested. We'll focus on the core concepts and problem-solving techniques highlighted in the Exemplar, which often includes slightly more challenging questions.

Chapter 3: Pair of Linear Equations in Two Variables - Detailed Notes

1. Introduction & General Form:

• An equation of the form ax + by + c = 0, where a, b, and c are real numbers, and a and b are not both zero, is called a linear equation in two variables (x and y).
• A pair of linear equations in two variables involves two such equations, typically written as:
  • a₁x + b₁y + c₁ = 0
  • a₂x + b₂y + c₂ = 0
    Where a₁, b₁, c₁, a₂, b₂, c₂ are real numbers such that a₁² + b₁² ≠ 0 and a₂² + b₂² ≠ 0.
• A solution (x, y) to a pair of linear equations is a pair of values that satisfies both equations simultaneously.

2. Graphical Representation & Interpretation:

• Every linear equation in two variables represents a straight line on the Cartesian plane.

• When we consider a pair of linear equations, we are essentially looking at two straight lines. There are three possibilities for these two lines:

  • (i) Intersecting Lines: The lines cross at exactly one point.

    • Algebraic Interpretation: The pair has a unique solution.
    • Condition (Ratio of Coefficients):
      (a₁/a₂) ≠ (b₁/b₂)
    • Consistency: The system is consistent.

  • (ii) Parallel Lines: The lines never intersect.

    • Algebraic Interpretation: The pair has no solution.
    • Condition (Ratio of Coefficients):
      (a₁/a₂) = (b₁/b₂) ≠ (c₁/c₂)
    • Consistency: The system is inconsistent.

  • (iii) Coincident Lines: The lines overlap completely (they are essentially the same line).

    • Algebraic Interpretation: The pair has infinitely many solutions.
    • Condition (Ratio of Coefficients):
      (a₁/a₂) = (b₁/b₂) = (c₁/c₂)
    • Consistency: The system is consistent (and often called dependent).

  • Important Note: When checking ratios, if any denominator (a₂, b₂, or c₂) is zero, handle it carefully. For instance, if a₂ = 0, the ratio a₁/a₂ is undefined. Think about the equations directly in such cases (e.g., x = k or y = k). Also ensure equations are in standard form (ax+by+c=0) before comparing coefficients.

3. Algebraic Methods for Solving:

• (i) Substitution Method:

  1. Solve one of the equations for one variable in terms of the other (e.g., express y in terms of x from the first equation).
  2. Substitute this expression into the other equation. This results in a linear equation in one variable.
  3. Solve this single-variable equation.
  4. Substitute the value found back into the expression from step 1 to find the value of the other variable.

• (ii) Elimination Method:

  1. Make the coefficients of one variable (either x or y) numerically equal in both equations by multiplying one or both equations by suitable non-zero constants.
  2. Add or subtract the modified equations to eliminate the variable with equal coefficients. This results in a linear equation in one variable.
  3. Solve this single-variable equation.
  4. Substitute the value found back into either of the original equations to find the value of the other variable.

• (iii) Cross-Multiplication Method: (Note: Sometimes excluded from school syllabus but useful for speed in competitive exams)
  For the system:
  a₁x + b₁y + c₁ = 0
  a₂x + b₂y + c₂ = 0
  The solution is given by:
  x / (b₁c₂ - b₂c₁) = y / (c₁a₂ - c₂a₁) = 1 / (a₁b₂ - a₂b₁)

  • Remember the cyclic order: x → y → 1 and coefficients (b, c) → (c, a) → (a, b).
  • This method directly gives the values of x and y, provided (a₁b₂ - a₂b₁) ≠ 0 (which is the condition for a unique solution).

4. Consistency and Inconsistency (Summary):

• Consistent System: A system of linear equations having at least one solution.
  • Unique Solution (Intersecting lines): a₁/a₂ ≠ b₁/b₂
  • Infinitely Many Solutions (Coincident lines): a₁/a₂ = b₁/b₂ = c₁/c₂
• Inconsistent System: A system of linear equations having no solution.
  • No Solution (Parallel lines): a₁/a₂ = b₁/b₂ ≠ c₁/c₂

5. Equations Reducible to a Pair of Linear Equations:

• Some equations are not linear initially but can be transformed into a linear pair by making suitable substitutions.
• Example: Solve:
  • 2/x + 3/y = 13
  • 5/x - 4/y = -2
  • Substitution: Let u = 1/x and v = 1/y. The equations become:
    • 2u + 3v = 13
    • 5u - 4v = -2
  • Now, solve this linear system for u and v using any method (elimination, substitution).
  • Once you find u and v, find x and y using x = 1/u and y = 1/v.
• Be careful with the domain (e.g., x ≠ 0, y ≠ 0 in the example above).

6. Word Problems:

• This is a very important application area for competitive exams.
• Steps:
  1. Read Carefully: Understand the problem and identify the unknowns.
  2. Define Variables: Assign variables (e.g., x and y) to represent the unknowns.
  3. Formulate Equations: Translate the information given in the problem into two linear equations involving the variables. Look for two distinct conditions or pieces of information.
  4. Solve the System: Use any appropriate algebraic method to solve the pair of equations.
  5. Check and Interpret: Verify if the solution satisfies the conditions of the original problem and write the answer in the context of the problem (with units, if applicable).
• Common Types: Problems involving ages, numbers (digits), fractions, speed-distance-time (including upstream/downstream boat problems), time and work, geometry (angles, lengths), fixed charges and per-unit charges.

Key Focus Areas for Competitive Exams (from Exemplar perspective):

• Finding values of constants (like k, p, etc.) for the system to have a unique solution, no solution, or infinitely many solutions. This directly tests your understanding of the coefficient ratio conditions.
• Geometric Interpretation: Questions asking about the nature of lines represented by the equations.
• Word Problems: Especially those involving speed/time/distance (upstream/downstream), ages, and fractions. These require careful equation setup.
• Reducible Equations: Recognizing and solving equations that require substitution.
• Checking for consistency/inconsistency without actually solving.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts discussed, suitable for practice:

1. The pair of equations y = 0 and y = -7 has:
   (A) One solution
   (B) Two solutions
   (C) Infinitely many solutions
   (D) No solution

2. For what value of k, do the equations 3x - y + 8 = 0 and 6x - ky = -16 represent coincident lines?
   (A) 1/2
   (B) -1/2
   (C) 2
   (D) -2

3. If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is:
   (A) -5/4
   (B) 2/5
   (C) 15/4
   (D) 3/2

4. The solution of the equations x - y = 2 and x + y = 4 is:
   (A) x = 3, y = 1
   (B) x = 1, y = 3
   (C) x = -3, y = -1
   (D) x = -1, y = -3

5. A pair of linear equations which has a unique solution x = 2, y = –3 is:
   (A) x + y = –1 ; 2x – 3y = –5
   (B) 2x + 5y = –11 ; 4x + 10y = –22
   (C) 2x – y = 1 ; 3x + 2y = 0
   (D) x – 4y – 14 = 0 ; 5x – y – 13 = 0

6. The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 has:
   (A) A unique solution
   (B) Exactly two solutions
   (C) Infinitely many solutions
   (D) No solution

7. If x = a, y = b is the solution of the equations x – y = 2 and x + y = 4, then the values of a and b are, respectively:
   (A) 3 and 1
   (B) 3 and 5
   (C) 5 and 3
   (D) –1 and –3

8. Aruna has only Re 1 and Rs 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Re 1 and Rs 2 coins are, respectively:
   (A) 35 and 15
   (B) 15 and 35
   (C) 35 and 20
   (D) 25 and 25

9. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively:
   (A) 4 and 24
   (B) 5 and 30
   (C) 6 and 36
   (D) 3 and 24

10. The value of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is:
    (A) 3
    (B) -3
    (C) -12
    (D) No value

Answers to MCQs:

1. (D) - Both lines are horizontal and distinct (y=0 is the x-axis, y=-7 is parallel below it). Parallel lines have no solution.
2. (C) - For coincident lines: a₁/a₂ = b₁/b₂ = c₁/c₂. Here: 3/6 = (-1)/(-k) = 8/16. From 3/6 = 1/k, we get 1/2 = 1/k, so k=2. Check: (-1)/(-2) = 1/2 and 8/16 = 1/2. Condition holds for k=2.
3. (C) - For parallel lines: a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Here: 3/2 = (2k)/5. Cross-multiplying gives 15 = 4k, so k = 15/4. We must also check 3/2 ≠ 2/(-1). Since 3/2 ≠ -2, the condition holds.
4. (A) - Adding the two equations: (x-y) + (x+y) = 2+4 => 2x = 6 => x = 3. Substituting x=3 in x+y=4 gives 3+y=4 => y=1.
5. (D) - Check each option by substituting x=2, y=-3.
   (A) 2 + (-3) = -1 (True); 2(2) - 3(-3) = 4 + 9 = 13 ≠ -5 (False)
   (B) 2(2) + 5(-3) = 4 - 15 = -11 (True); 4(2) + 10(-3) = 8 - 30 = -22 (True). But check ratios: 2/4 = 5/10 = (-11)/(-22) = 1/2. These are coincident lines (infinite solutions), not a unique solution.
   (C) 2(2) - (-3) = 4 + 3 = 7 ≠ 1 (False)
   (D) 2 - 4(-3) - 14 = 2 + 12 - 14 = 0 (True); 5(2) - (-3) - 13 = 10 + 3 - 13 = 0 (True). Check ratios: 1/5 ≠ (-4)/(-1). Unique solution.
6. (D) - Check ratios: a₁/a₂ = 1/(-3); b₁/b₂ = 2/(-6) = -1/3; c₁/c₂ = 5/1. Here, a₁/a₂ = b₁/b₂ (-1/3 = -1/3) but a₁/a₂ ≠ c₁/c₂ (-1/3 ≠ 5). This is the condition for parallel lines (no solution).
7. (A) - This is the same system as Q4. Solution is x=3, y=1. So a=3, b=1.
8. (D) - Let x be the number of Re 1 coins and y be the number of Rs 2 coins.
   Total coins: x + y = 50
   Total amount: 1x + 2y = 75
   From first eq, x = 50 - y. Substitute in second: (50 - y) + 2y = 75 => 50 + y = 75 => y = 25.
   Then x = 50 - 25 = 25. So, 25 Re 1 coins and 25 Rs 2 coins.
9. (C) - Let son's present age be s and father's present age be f.
   f = 6s (Equation 1)
   Four years hence: Son's age = s+4, Father's age = f+4.
   f+4 = 4(s+4) => f+4 = 4s + 16 => f = 4s + 12 (Equation 2)
   Substitute f=6s into Equation 2: 6s = 4s + 12 => 2s = 12 => s = 6.
   Then f = 6s = 6(6) = 36. Ages are 6 and 36.
10. (D) - For infinitely many solutions: a₁/a₂ = b₁/b₂ = c₁/c₂.
    Equations: cx - y - 2 = 0 and 6x - 2y - 3 = 0.
    Ratios: c/6 = (-1)/(-2) = (-2)/(-3)
    From (-1)/(-2) = (-2)/(-3), we get 1/2 = 2/3, which is false.
    Since b₁/b₂ ≠ c₁/c₂, the condition for infinitely many solutions can never be met, regardless of the value of c. Therefore, there is no such value of c.

Make sure you practice solving various types of problems, especially the word problems and those involving finding the value of 'k' based on the nature of solutions. Good luck!