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

Q1 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

Answer:

Given equation: $y = 3x + 5$

Let's put different values of x into the equation and find the corresponding values of y.

Let x = 0, then

y = 3(0) + 5 = 0 + 5 = 5

Let x = 1, then

y = 3(1) + 5 = 3 + 5 = 8

Let x = -1, then

y = 3(-1) + 5 = -3 + 5 = 2

Let x = 2, then

y = 3(2) + 5 = 6 + 5 = 11

Therefore, we can see for each value of x, we get unique value of y.

Thus, the equation has infinitely many solutions.

Q2 Write four solutions for each of the following equations:

(i) $2x + y = 7$ (ii) $\pi x +y = 9$ (iii) $x = 4y$

Answer:

(i) Given equation: $2x + y = 7$

Let's put 4 different values of x.

Putting x = 0, we have , $y=7-2\times 0=7$ means $(0,7)$ is a solution.

Putting x = 1, we have , $y=7-2\times 1=5$ means $(1,5)$ is a solution.

Putting x = 2, we have , $y=7-2\times 2=3$ means $(2,3)$ is a solution.

Putting x = 3, we have , $y=7-2\times 3=1$ means $(3,1)$ is a solution.

Therefore, the four solutions are : $(0,7),(1,5),(2,3),(3,1)$ .

(ii) Given : $\pi x +y = 9$

Let's put 4 different values of x.

Putting x = 0, we have , $y=9-\pi \times 0=9$ means $(0,9)$ is a solution.

Putting x = 1, we have , $y=9-\pi \times 1=9-\pi$ means $(1,9-\pi )$ is a solution.

Putting x = 2, we have , $y=9-\pi \times 2=9-2\pi$ means $(2,9-2\pi )$ is a solution.

Putting x = 3, we have , $y=9-\pi \times 3=9-3\pi$ means $(3,9-3\pi )$ is a solution.

Therefore, the four solutions are : $(0,9),(1,9-\pi ),(2,9-2\pi ),(3,9-3\pi )$.

(iii) Given : $x = 4y$

Let's put 4 different values of x.

Putting x = 0, we have , $y=\frac{0}{4}=0$ means $(0,0)$ is a solution.

Putting x = 1, we have , $y=\frac{1}{4}$ means $(1,\frac{1}{4})$ is a solution.

Putting x = 2, we have , $y=\frac{2}{4}=\frac{1}{2}$ means $(2,\frac{1}{2})$ is a solution.

Putting x = 3, we have , $y=\frac{3}{4}$ means $(3,\frac{3}{4})$ is a solution.

Therefore, the four solutions are : $(0,0)$ , $(1,\frac{1}{4})$ , $(2,\frac{1}{2})$ and $(3,\frac{3}{4})$ .

Q3 (i) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: ((0,2)

Answer:

(i) Given equation: $x - 2y = 4$

According to the question, put (0,2), which means x as 0 and y as 2.

Therefore we get , $x - 2y = 0-2(2)=-4\neq 4$

Thus, $(0,2)$ is not a solution of $x - 2y = 4$.

Q3 (ii) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (2,0)

Answer:

Given equation: $x - 2y = 4$

According to the question, put (2,0), which means x as 2 and y as 0.

Therefore we get , $x - 2y = 2-2(0)=2\neq 4$

Thus, (2,0) is not a solution of $x - 2y = 4$.

Q3 (iii) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (4,0)

Answer:

Given equation: $x - 2y = 4$

According to the question, put (4,0), which means x as 4 and y as 0.

Therefore we get, $x - 2y = 4-2(0)=4=4$

Thus, (4,0) is a solution of $x - 2y = 4$ .

Q3 (iv) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: $(\sqrt2 , 4\sqrt2)$

Answer:

Given equation: $x - 2y = 4$

According to the question, put $(\sqrt2 , 4\sqrt2)$, which means x as $\sqrt2$ and y as $4\sqrt2$.

Therefore we get, $x - 2y =\sqrt{2}-2(4\sqrt{2})=\sqrt{2}-8\sqrt{2}=-7\sqrt{2}\neq 4$

Thus, $(\sqrt2 , 4\sqrt2)$ is not a solution of $x - 2y = 4$ .

Q3 (v) Check which of the following are solutions of the equation $x - 2y = 4$ and which are not: (1,1)

Answer:

Given equation: $x - 2y = 4$

According to the question, put (1,1), which means both x and y as 1.

Therefore we get, $x - 2y = 1-2(1)=-1\neq 4$

Thus, (1,1) is not a solution of $x - 2y = 4$ .

Q4 Find the value of k , if $x = 2$ , $y = 1$ is a solution of the equation $2x + 3y = k$ .

Answer:

Given equation: $2x + 3y = k$

According to the question, put x as 2 and y as 1.

Therefore we get, $k=2x + 3y = 2(2)+3(1)=4+3=7$

Thus, k = 7 for $2x + 3y = k$ putting x = 2 and y = 1.