NCERT Solutions Class 9 Chapter 4 Linear Equations in Two Variables Ex 4.2

Linear Equations in Two Variables Class 9 Exercise 4.2 introduces graphing. You will learn how to find solution pairs for a linear equation and plot them to draw the corresponding straight line. This visual step helps students see that every solution pair lies on the same line, and it connects symbolic algebra to geometric visualization.

The NCERT Solutions for linear equations in two variables class 9 exercise 4.2 provide examples of plotting, show how to choose plotting points for accuracy, and explain how to read key features like intercepts and slope from the graph. 

Class 9 Chapter 4 Exercise 4.2 Questions with Answers

Class 9 Chapter 4 Linear Equations in Two Variables Exercise 4.2 questions ask you to calculate solution pairs and graph them to represent the linear equation visually. Questions along with the answers are given here for exam preparation: 

1. Which one of the following options is true, and why?

y = 3x+5 has

1. A unique solution

2. Only two solutions

3. Infinitely many solutions

Solution: Let us substitute different values for x in the linear equation y = 3x+5

From the table, it is clear that x can have infinite values, and for all the infinite values of x, there are infinite values of y as well. Hence, (iii) infinitely many solutions is the only option true.

2. Write four solutions for each of the following equations:

(i) 2x+y = 7

Solution: To find the four solutions of 2x+y =7, we substitute different values for x and y. Let x = 0 Then, 2x+y = 7 (2×0)+y = 7 y = 7 (0,7) Let x = 1 Then, 2x+y = 7 (2×1)+y = 7 2+y = 7 y = 7-2 y = 5 (1,5) Let y = 1 Then, 2x+y = 7 (2x)+1 = 7 2x = 7-1 2x = 6 x = 6/2 x = 3 (3,1) Let x = 2 Then, 2x+y = 7 (2×2)+y = 7 4+y = 7 y =7-4 y = 3 (2,3) The solutions are (0, 7), (1,5), (3,1), (2,3)

(ii) πx+y = 9

Solution: To find the four solutions of πx+y = 9, we substitute different values for x and y. Let x = 0 Then, πx+y = 9 (π×0)+y = 9 y = 9 (0,9) Let x = 1 Then, πx +y = 9 (π×1)+y = 9 π+y = 9 y = 9-π (1, 9-π) Let y = 0 Then, πx+y = 9 πx+0 = 9 πx = 9 x = 9/π (9/π,0) Let x = -1 Then, πx + y = 9 (π×-1) + y = 9 -π+y = 9 y = 9+π (-1,9+π) The solutions are (0,9), (1,9-π), (9/π,0), (-1,9+π)

(iii) x = 4y

Solution: To find the four solutions of x = 4y, we substitute different values for x and y. Let x = 0 Then, x = 4y 0 = 4y 4y= 0 y = 0/4 y = 0 (0,0) Let x = 1 Then, x = 4y 1 = 4y 4y = 1 y = 1/4 (1,1/4) Let y = 4 Then, x = 4y x= 4×4 x = 16 (16,4) Let y = 1 Then, x = 4y x = 4×1 x = 4 (4,1) The solutions are (0,0), (1,1/4), (16,4), (4,1)

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)

Solutions:

(i) (0, 2)

(x,y) = (0,2) Here, x=0 and y=2 Substituting the values of x and y in the equation x–2y = 4, we get, x–2y = 4 ⟹ 0 – (2×2) = 4 But, -4 ≠ 4 (0, 2) is not a solution of the equation x–2y = 4

(ii) (2, 0)

(x,y) = (2, 0) Here, x = 2 and y = 0 Substituting the values of x and y in the equation x -2y = 4, we get, x -2y = 4 ⟹ 2-(2×0) = 4 ⟹ 2 -0 = 4 But, 2 ≠ 4 (2, 0) is not a solution of the equation x-2y = 4

(iii) (4, 0)

Solution: (x,y) = (4, 0) Here, x= 4 and y=0 Substituting the values of x and y in the equation x -2y = 4, we get, x–2y = 4 ⟹ 4 – 2×0 = 4 ⟹ 4-0 = 4 ⟹ 4 = 4 (4, 0) is a solution of the equation x–2y = 4

(iv) (√2,4√2)

Solution: (x,y) = (√2,4√2) Here, x = √2 and y = 4√2 Substituting the values of x and y in the equation x–2y = 4, we get, x –2y = 4 ⟹ √2-(2×4√2) = 4 √2-8√2 = 4 But, -7√2 ≠ 4 (√2,4√2) is not a solution of the equation x–2y = 4

(v) (1, 1)

Solution: (x,y) = (1, 1) Here, x= 1 and y= 1 Substituting the values of x and y in the equation x–2y = 4, we get, x –2y = 4 ⟹ 1 -(2×1) = 4 ⟹ 1-2 = 4 But, -1 ≠ 4 (1, 1) is not a solution of the equation x–2y = 4

4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k.

Solution: The given equation is 2x+3y = k According to the question, x = 2 and y = 1 Now, substituting the values of x and y in the equation2x+3y = k, We get, (2×2)+(3×1) = k ⟹ 4+3 = k ⟹ 7 = k k = 7 The value of k, if x = 2, y = 1 is a solution of the equation 2x+3y = k, is 7.

How to Prepare for Class 9 Chapter 4 Linear Equations in Two Variables Exercise 4.2 Questions?

Begin by choosing two or three convenient x-values (often including 0) and computing the corresponding y-values carefully; these become your plotting points. Use graph paper and a ruler to mark points precisely, label axes, and draw the line through the plotted points.

To prepare efficiently, follow these steps:

• Choose simple integer values for x (like −2, 0, 2) to get clean y-values.

• Plot at least three points to avoid plotting errors; two points define a line, three confirm it.

• Label axes with equal scale units to prevent distortion of slope.

• Use intercepts (set x=0 or y=0) to draw the line quickly when possible.

• Compare your plotted line with the NCERT Solution graph to check for mistakes.

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