In this Exercise 4.2 of Chapter 4, Linear Equations in Two Variables, you learn how to find different solutions for a given equation, and how these solutions can be expressed in pairs (x, y). This exercise is important because it demonstrates that linear equations can provide many answers rather than just one.
These NCERT Solutions are presented in simple steps to better understanding and are designed according to the latest NCERT syllabus. This will develop your understanding of algebra by cementing your basics and improves confidence when solving these types of problems.
Practicing this exercise will help you develop the skills needed to solve questions for exams quickly and accurately. It also helps as a foundation for higher-level maths topics and for competitive examinations like the Olympiads.
The NCERT Solutions for Class 9 Maths Chapter 4 Exercise 4.2 explains how to find many solutions for one linear equation using simple number pairs with step by steps solutions. Download the NCERT Solutions free PDF from below:
1. Which one of the following options is true, and why? y=3x+5 has
(i) a unique solution
(ii) Only two solutions
(iii) Infinitely many solutions.
Sol. Option (iii) is true because a linear equation has infinitely many solutions. Moreover when represented graphically a linear equation in two variables is a straight line which has infinite points and hence, it has infinite solutions.
2. Write four solutions for each of the following equations :
(i) 2x+y=7
(ii) πx+y=9
(iii) x=4y
Sol. (i) 2x+y=7
For x=−1, we get −2+y=7, i.e., y=9. Therefore, (−1,9) is a solution.
For x=0, we get y=7. Therefore, (0,7) is a solution.
For x=1, we get 2+y=7, i.e., y=5. Therefore, (1,5) is a solution.
For x=2, we get 4+y=7, i.e., y=3. Therefore, (2,3) is a solution.
Hence, we have four solutions (−1,9), (0,7),(1,5) and (2,3).
(ii) Proceed as in (i) and we can have four solutions as (0,9),(1,9−π),(2,9−2π) and (3,9−3π).
(iii) Proceed as in (i) and we can have four solutions as (0,0),(4,1),(8,2) and (12,3).
3. Check which of the following are solutions of the equation x−2y=4 and which are not
(i) (0,2)
(ii) (2,0)
(iii) (4,0)
(iv) (√2,4√2)
(v) (1,1)
Sol. (i) Substituting x=0, y=2 in the equation x−2y=4,
we get 0−2(2)=4, i.e., −4=4 but −4≠4. Therefore, (0,2) is not a solution.
(ii) 2−2(0)≠4. Therefore, (2,0) is not a solution.
(iii) Substituting x=4 and y=0 in the equation x−2y=4, we get
L.H.S. = 4−2(0)=4−0=4 = R.H.S. Therefore, L.H.S. = R.H.S. Therefore, (4,0) is a solution.
(iv) √2−2(4√2)=4, i.e., √2−8√2=4,
i.e., −7√2=4 but −7√2≠4. Therefore, (√2,4√2) is not a solution.
(v) 1−2(1)≠4. Therefore, (1,1) is not a solution.
4. Find the value of k if x=2,y=1 is a solution of the equation 2x+3y=k.
Sol. (2)(2)+(3)(1)=k, i.e., 4+3=k, i.e., k=7.
(Session 2025 - 26)