CBSE • Class 10 • Mathematics • Chapter 3 (Pair of Linear Equations in Two Variables) • Exercise 3.2

Exercise 3.2: Pair of Linear Equations in Two Variables — NCERT Solutions

Elimination method for simultaneous linear equations and consistency by ratios of coefficients.

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What this exercise covers

Elimination methodConsistencya₁/a₂ vs b₁/b₂ vs c₁/c₂

Step-by-step solutions — Exercise 3.2

4 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 3.2 Q1 • 3 marks

Solve by the elimination method: 2x+3y=122x + 3y = 12 and x−y=1x - y = 1.
Hint (Socratic — try this first)▾
Which variable could you eliminate by multiplying one equation by a suitable number so that the coefficients match?
Step-by-step solution▾

Equations: 2x+3y=12(1)2x + 3y = 12 \quad (1) x−y=1(2)x - y = 1 \quad (2)

Multiply equation (2) by 22 to match the coefficient of xx:

2x−2y=2(3)2x - 2y = 2 \quad (3)

Subtract (3) from (1):

(2x+3y)−(2x−2y)=12−2(2x + 3y) - (2x - 2y) = 12 - 2 5y=105y = 10 y=2y = 2

Substitute y=2y = 2 into equation (2):

x−2=1⇒x=3x - 2 = 1 \Rightarrow x = 3

Solution: x=3, y=2x = 3,\ y = 2.

Common mistake:
Forgetting to multiply the right-hand side when scaling an equation, e.g. writing 2x−2y=12x - 2y = 1 instead of 2x−2y=22x - 2y = 2.
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Exercise 3.2 Q2 • 2 marks

Without solving, determine whether the pair 2x+3y=72x + 3y = 7 and 4x+6y=124x + 6y = 12 is consistent or inconsistent.
Hint (Socratic — try this first)▾
How do the ratios a1a2\frac{a_1}{a_2}, b1b2\frac{b_1}{b_2} and c1c2\frac{c_1}{c_2} compare?
Step-by-step solution▾

Write both equations in the form ax+by+c=0a x + b y + c = 0:

  • Equation 1: 2x+3y−7=02x + 3y - 7 = 0, so a1=2, b1=3, c1=−7a_1 = 2,\ b_1 = 3,\ c_1 = -7.
  • Equation 2: 4x+6y−12=04x + 6y - 12 = 0, so a2=4, b2=6, c2=−12a_2 = 4,\ b_2 = 6,\ c_2 = -12.

Compute the ratios:

a1a2=24=12,b1b2=36=12,c1c2=−7−12=712\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{-7}{-12} = \frac{7}{12}

Here a1a2=b1b2≠c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}.

Conclusion: The lines are parallel, so the pair is inconsistent (no solution).

Common mistake:
Ignoring the signs of c1c_1 and c2c_2 when forming the ratio, or comparing ratios before writing the equations in the standard =0=0 form.
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Exercise 3.2 Q3 • 3 marks

Find the value of kk for which the pair 2x+3y=52x + 3y = 5 and 4x+ky=104x + ky = 10 has infinitely many solutions.
Hint (Socratic — try this first)▾
For infinitely many solutions, all three coefficient ratios must be equal — what does that force kk to be?
Step-by-step solution▾

In standard form:

  • 2x+3y−5=02x + 3y - 5 = 0: a1=2, b1=3, c1=−5a_1 = 2,\ b_1 = 3,\ c_1 = -5.
  • 4x+ky−10=04x + ky - 10 = 0: a2=4, b2=k, c2=−10a_2 = 4,\ b_2 = k,\ c_2 = -10.

For infinitely many solutions:

a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} 24=3k=−5−10\frac{2}{4} = \frac{3}{k} = \frac{-5}{-10}

All must equal 12\frac{1}{2}. From 3k=12\frac{3}{k} = \frac{1}{2}:

k=6k = 6

Check: 24=36=510=12\frac{2}{4} = \frac{3}{6} = \frac{5}{10} = \frac{1}{2} ✓

Answer: k=6k = 6.

Common mistake:
Only using two of the three ratios and getting a value of kk that satisfies dependence but not verifying the cc-ratio, or confusing the condition with that for a unique solution.
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Exercise 3.2 Q4 • 3 marks

Solve by elimination: x2+y3=2\frac{x}{2} + \frac{y}{3} = 2 and x−y=1x - y = 1.
Hint (Socratic — try this first)▾
Can you clear the fractions in the first equation before eliminating a variable?
Step-by-step solution▾

Clear fractions in the first equation by multiplying through by 66:

6⋅x2+6⋅y3=6⋅26 \cdot \frac{x}{2} + 6 \cdot \frac{y}{3} = 6 \cdot 2 3x+2y=12(1)3x + 2y = 12 \quad (1)

Second equation: x−y=1(2)x - y = 1 \quad (2)

Multiply (2) by 22: 2x−2y=2(3)2x - 2y = 2 \quad (3)

Add (1) and (3): (3x+2y)+(2x−2y)=12+2(3x + 2y) + (2x - 2y) = 12 + 2 5x=145x = 14 x=145x = \frac{14}{5}

Substitute into (2): 145−y=1⇒y=145−1=95\frac{14}{5} - y = 1 \Rightarrow y = \frac{14}{5} - 1 = \frac{9}{5}.

Solution: x=145, y=95x = \frac{14}{5},\ y = \frac{9}{5}.

Common mistake:
Multiplying only some terms by the LCM (e.g. forgetting to multiply the constant term 2), leading to an incorrect cleared equation.
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How to approach Exercise 3.2

  1. Re-read the chapter summary first. Open Pair of Linear Equations in Two Variables and refresh the key concepts: Substitution method, Elimination method, Cross-multiplication, Consistent and inconsistent systems.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Pair of Linear Equations in Two Variables

  1. Exercise 3.1 — Algebraic methods — solving simultaneous linear equations by substitution.
  2. Exercise 3.2 — Elimination method for simultaneous linear equations and consistency by ratios of coefficients.
  3. Exercise 3.3 — Word problems — translating real-life situations into pairs of linear equations.

← Exercise 3.1

Algebraic methods — solving simultaneous linear equations by substitution.

Exercise 3.3 →

Word problems — translating real-life situations into pairs of linear equations.

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