CBSE • Class 9 • Mathematics • Chapter 4 (Linear Equations in Two Variables) • Exercise 4.2

Exercise 4.2: Linear Equations in Two Variables — NCERT Solutions

Solutions as ordered pairs — listing infinitely many solutions numerically.

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What this exercise covers

Ordered pairsTable of solutionsInfinite solutions

Step-by-step solutions — Exercise 4.2

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

Exercise 4.2 Q1 • 3 marks

Find four different solutions of the equation 2x+y=72x + y = 7.
Hint (Socratic — try this first)▾
If you freely choose a value for xx, how can you find the matching value of yy?
Step-by-step solution▾

For 2x+y=72x + y = 7, choose values of xx and compute y=7−2xy = 7 - 2x.

  • x=0⇒y=7−0=7⇒(0,7)x = 0 \Rightarrow y = 7 - 0 = 7 \Rightarrow (0, 7)
  • x=1⇒y=7−2=5⇒(1,5)x = 1 \Rightarrow y = 7 - 2 = 5 \Rightarrow (1, 5)
  • x=2⇒y=7−4=3⇒(2,3)x = 2 \Rightarrow y = 7 - 4 = 3 \Rightarrow (2, 3)
  • x=3⇒y=7−6=1⇒(3,1)x = 3 \Rightarrow y = 7 - 6 = 1 \Rightarrow (3, 1)

Four solutions are: (0,7), (1,5), (2,3), (3,1)(0,7),\ (1,5),\ (2,3),\ (3,1).

Since infinitely many values of xx are possible, the equation has infinitely many solutions.

Common mistake:
Writing the ordered pair in reverse order as (y,x)(y, x) instead of (x,y)(x, y).
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Exercise 4.2 Q2 • 2 marks

Check whether (2,1)(2, 1) and (−1,5)(-1, 5) are solutions of the equation 4x+3y=114x + 3y = 11.
Hint (Socratic — try this first)▾
What happens when you substitute the coordinates into the left-hand side — does it equal the right-hand side?
Step-by-step solution▾

Substitute each point into 4x+3y4x + 3y and compare with 1111.

For (2,1)(2, 1):

4(2)+3(1)=8+3=114(2) + 3(1) = 8 + 3 = 11

This equals the RHS, so (2,1)(2, 1) is a solution.

For (−1,5)(-1, 5):

4(−1)+3(5)=−4+15=114(-1) + 3(5) = -4 + 15 = 11

This also equals 1111, so (−1,5)(-1, 5) is a solution.

Both ordered pairs are solutions of 4x+3y=114x + 3y = 11.

Common mistake:
Substituting the values into the wrong variables, e.g. putting x=1x = 1 and y=2y = 2 for the point (2,1)(2,1).
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Exercise 4.2 Q3 • 2 marks

If (3,k)(3, k) is a solution of the equation 2x−5y=12x - 5y = 1, find the value of kk.
Hint (Socratic — try this first)▾
Substitute the known coordinate for xx and solve the resulting equation for kk.
Step-by-step solution▾

Since (3,k)(3, k) is a solution, substitute x=3x = 3 and y=ky = k:

2(3)−5k=12(3) - 5k = 1

6−5k=16 - 5k = 1

−5k=1−6=−5-5k = 1 - 6 = -5

k=−5−5=1k = \frac{-5}{-5} = 1

Therefore k=1k = 1, and the solution point is (3,1)(3, 1).

Common mistake:
Dividing incorrectly and losing the negative signs, giving k=−1k = -1 instead of k=1k = 1.
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How to approach Exercise 4.2

  1. Re-read the chapter summary first. Open Linear Equations in Two Variables and refresh the key concepts: Linear equation, Standard form ax + by + c = 0, Ordered pair, Graph as a line.
  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 Linear Equations in Two Variables

  1. Exercise 4.1 — Writing linear equations in standard form ax + by + c = 0.
  2. Exercise 4.2 — Solutions as ordered pairs — listing infinitely many solutions numerically.
  3. Exercise 4.3 — Graphing linear equations — drawing straight lines from intercepts.
  4. Exercise 4.4 — Lines parallel to axes — equations x = k and y = k and special cases.

← Exercise 4.1

Writing linear equations in standard form ax + by + c = 0.

Exercise 4.3 →

Graphing linear equations — drawing straight lines from intercepts.

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