These NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Ex 4.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise 4.2

Question 1.

Which one of the following statements is true and why?

y = 3x + 5 has

(i) a unique solution.

(ii) only two solutions.

(iii) infinitely many solutions.

Solution:

We have given the equation

y = 3x + 5

Put x = 0, then y = 5

Therefore, (0, 5) is the solution of this equation. Again, put x = 1 then y = 8.

Therefore, (1, 8) is also the solution to this equation. Again, put x = 2, then y = 11.

Therefore, (2, 11) is also the solution of this equation. Again x = 3, x = 4 and so on.

Now, it is clear that this equation has infinitely many solutions.

Note: A linear equation in two variables has infinitely many solutions.

Question 2.

Write four solutions for each of the following equations.

(i) 2x + y = 7

(ii) πx + y = 9

(iii) x = 4y

Solution:

(i) We have 2x + y = 7

Put x = 0 then y = 7

Therefore, (0, 7) is the solution of this equation.

Again put x = 1, then y = 5.

Therefore (1, 5) is also the solution of this equation. Again put x = 2, then y = 3.

Therefore (2, 3) is also the solution of this equation. Again put x = 3 then y = 1.

Therefore (3, 1) is also the solution of this equation.

Therefore, (0, 7), (1, 5), (2, 3) and (3, 1) are the four solutions of the equation 2x + y = 7.

(ii) We have given πx + y = 9

or, \(\frac {22}{7}\) x + y = 9 (∵ π = \(\frac {22}{7}\))

or, y = 9 – \(\frac {22}{7}\) x

Put x = 0 then y = 9

Therefore (0, 9) is the solution of this equation.

Again, put x = 1 then y = \(\frac {41}{7}\)

Therefore (\(\frac {22}{7}\)) is also the solution of this equation.

Again put x = 7, then y = -13

Therefore (7, – 13) is also the solution of this equation.

Again put x = 14, then y = -3

Therefore (14, -35) is also the solution of this equation.

Hence (0, 9) (\(\frac {22}{7}\)) (7, 13) and (14, -35) are the four solutions of equation πx + y = 9

(iii) We have given that x = 4y

Put y = 0, then x = 0

Therefore, (0, 0) is the solution of this equation.

Again put y = 1, then x = 4 Therefore, (4, 1) is also the solution of this equation.

Again, put y = 2 then x = 8 Therefore, (8, 2) is also the solution of this equation.

Again put y = 3, then x = 12 Therefore, (12, 3) is also the solution of this equation.

Hence, (0, 0), (4, 1) (8, 2), and (12, 3) are the four solution of equation x = 4y.

Question 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)

Solution:

(i) we have given that x – 2y = 4

Put x = 0 and y = 2

Then, 0 – 2 × 2 = 4

-4 = 4

Here, L.H.S. ≠ R.H.S.

Therefore, (0, 2) is not the solution of equation x – 2y = 4

(ii) Put x = 2 and y = 0

Then, 2 – 2 × 0 = 4

2 = 4

Here, L.H.S. ≠ R.H.S.

Therefore (2, 0) is not the solution of equation x – 2y = 4

(iii) Put x = 4 and y = 0 in equation x – 2y = 4

Then 4 – 2 × 0 = 4

4 = 4

Here, L.H.S. = R.H.S.

Therefore, (4, 0) is the solution of equation x – 2y = 4

(iv) Put x = √2 and y = 4√2 in equation x – 2y = 4

Then √2 – 2 × 4√2 = 4

√2 – 8√2 = 4

-7√2 = 4

Here, L.H.S. ≠ R.H.S.

Therefore, (√2, 4√2) is not the solution of equation x – 2y = 4.

(v) (1, 1)

Put x = 1 and y = 1 in equation x – 2y = 4

Then 1 – 2 × 1 = 4

1 – 2 = 4

-1 = 4

L.H.S. ≠ R.H.S.

Therefore (1, 1) is not the solution of equation x – 2y = 4.

Question 4.

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

Solution:

We have given that x = 2 and y = 1 is the solution of the equation 2x + 3y = k.

Put x = 2 and y = 1 in equation 2x + 3y = k

2 × 2 + 3 × 1 = k

4 + 3 = k

k = 7

Hence, the value of k in equation 2x + 3y = k is 7.