NCERT Solutions Class 9 Mathematics

Exercise 4.2

Q.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

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

(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y

Q.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) ($\sqrt{2}$, 4$\sqrt{2}$) (v) (1, 1)

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

NCERT Solutions Class 9 Mathematics

Exercise 4.2

Q.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

Solution:

Any linear equation will have (iii) infinitely many solutions. For each value of x, we will get a value of y. Since there is no limit to taking the values of x, we will have infinite number of solutions.

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

(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y

Solution:

(i) Given, 2x + y = 7

⇒ y = 7 - 2x

Put x = 0 ⇒ y = 7 - 2 × 0 ⇒ y = 7

⇒ (0, 7) is the solution.

We can similarly tabulate the four solutions as follows,

 x 0 1 2 3 y 7 5 3 1

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

(ii) πx + y = 9 ⇒ y = 9 - πx

 x 0 1 2 3 y 9 9 - π 9 - 2π 9 - 3π

The four solutions of the equation πx + y = 9 could be (0, 9), (1, 9-π), (2, 9-2π) and (3, 9-3π).

(iii) x = 4y

 x 0 4 8 12 y 0 1 2 3

The four solutions of the equation πx + y = 9 may be (0, 0), (4, 1), (8, 2) and (12, 3).

Q.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) ($\sqrt{2}$, 4$\sqrt{2}$) (v) (1, 1)

Solution:

(i) Putting x = 0 and y = 2 in the equation x - 2y = 4, we get,

0 - 2×2 = 4 ⇒ -4 = 4

Since LHS and RHS are not equal, (0, 2) is not a solution of the given equation.

(ii) Putting x = 2 and y = 0 in the equation x - 2y = 4, we get,

2 - 2×0 = 4 ⇒ 2 = 4

Since LHS and RHS are not equal, (2, 0) is not a solution of the given equation.

(iii) Putting x = 4 and y = 0 in the equation x - 2y = 4, we get,

4 - 2×0 = 4 ⇒ 4 = 4

Since LHS and RHS are equal, (4, 0) is a solution of the given equation.

(iv) Putting x = $\sqrt{2}$ and y = 4$\sqrt{2}$ in the equation x - 2y = 4, we get,

$\sqrt{2}$ - 2×4$\sqrt{2}$ = 4

$\sqrt{2}$ - 8$\sqrt{2}$ = 4

$\sqrt{2}$(1 - 8) = 4

⇒ -7$\sqrt{2}$  = 4

Since LHS and RHS are not equal, $\sqrt{2}$, 4$\sqrt{2}$) is not a solution of the given equation.

(v) Putting x = 1 and y = 1 in the equation x - 2y = 4, we get,

1 - 2×1 = 4 ⇒ -1 = 4

Since LHS and RHS are not equal, (1, 1) is not a solution of the given equation.

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

Solution:

Given equation is 2x + 3y = k

Now, x = 2, y = 1 is a solution of the given equation.

Putting the value of x and y in the equation, we get

2 × 2 + 3 × 1 = k

⇒ k = 4 + 3

⇒ k = 7