RD Sharma 2020 solution class 9 chapter 7 Linear Equations In Two Variables Exercise 7.2

Exercise 7.2

Page-7.6

Question 1:

Write two solutions for each of the following equations:

(i) 3x + 4y = 7

(ii) x = 6y

(iii) x + xy = 4

(iv) $\frac{2}{3}x-y=4$

Answer 1:

(i) We are given,

Substituting x = 1 in the given equation, we get

Substituting x = 2 in the given equation, we get

(ii) We are given,

Substituting in the given equation, we get

Substituting in the given equation, we get

(iii) We are given,

Substituting x = 0 in the given equation, we get

Substituting x = 4 in the given equation, we get

(iv) We are given,

Substituting x = 0 in the given equation, we get

Substituting x = 3 in the given equation, we get

Question 2:

Check which of the following are solutions of the equations 2x − y = 6 and which are not:

(i) (3,0)

(ii) (0,6)

(iii) (2, −2)

(iv) $\left(\sqrt{3}, 0\right)$

(v) $\left(\frac{1}{2}, -5\right)$

Answer 2:

We are given,

2x – y = 6

(i) In the equation 2x – y = 6,we have

Substituting x = 3 and y = 0 in 2x – y, we get

is the solution of 2x – y = 6.
 

(ii) In the equation 2x – y = 6, we have

Substituting x = 0 and y = 6 in 2x – y,we get

is not the solution of 2x – y = 6.
 

(iii) In the equation 2x – y = 6,we have

Substituting x = 2 and y = –2  in 2x – y, we get

is the solution of 2x – y = 6.
 

(iv) In the equation 2x – y = 6, we have

Substituting and y = 0 in 2x – y, we get

is not the solution of 2x – y = 6.
 

(v) In the equation 2x – y = 6, we have

Substituting and y = –5 in 2x – y, we get

is the solution of 2x – y = 6.

Question 3:

If x = −1, y = 2 is a solution of the equation 3x + 4y = k, find the value of k.

Answer 3:

We are given,

is the solution of equation .

Substituting and in ,we get

Page-7.7

Question 4:

Find the value of λ, if x  = −λ and y = $\frac{5}{2}$ is a solution of the equation x + 4y − 7 = 0.

Answer 4:

We are given,

is the solution of equation .

Substituting and in ,we get

Question 5:

If x = 2α + 1 and y = α − 1 is a solution of the equation 2x − 3y + 5 = 0, find the value of α.

Answer 5:

We are given,

is the solution of equation .

Substituting and in ,we get

$$\begin{gathered} 2\times \left(2a+1\right)-3\times \left(a-1\right)+5=0 \\ \Rightarrow 4a+2-3a+3+5=0 \\ \Rightarrow a+10=0 \\ \Rightarrow a=-10 \left(\mathrm{answer}\right) \end{gathered}$$

Question 6:

If x = 1 and y = 6 is a solution of the equation 8x − ay + a² = 0, find the value of a.

Answer 6:

We are given,

is the solution of equation .

Substituting and in ,we get

Using quadratic factorization

Question 7:

Write two solutions of the form x = 0, y = a and x = b, y = 0 for each of the following equations:

(i) 5x − 2y = 10

(ii) −4x + 3y = 12

(iii) 2x + 3y =24

Answer 7:

(i) We are given,

Substituting x = 0 in the given equation, we get

Substituting y = 0 in the given equation, we get

(ii) We are given,

Substituting in the given equation, we get
$$\begin{gathered} -4\times 0+3y=12 \\ 3y=12 \\ y=4 \end{gathered}$$
Thus x = 0 and y = 4 is the solution of the −4x + 3y = 12

Substituting y = 0 in the given equation, we get

(iii) We are given,

Substituting x = 0 in the given equation, we get

Substituting y = 0 in the given equation, we get