# Class 9 RD Sharma Solutions – Chapter 13 Linear Equation in Two Variable- Exercise 13.2

**Question 1. Write two solutions for each of the following equations:**

**(i) 3x + 4y = 7**

**(ii) x = 6y**

**(iii) x + **π**y = 4**

**(iv) 2/3 x − y = 4**

**Solution:**

(i) 3x + 4y = 7

Substituting x =1

We get,

3×1 + 4y = 7

4y = 7 − 3

4y = 4

y = 1

Therefore, if x = 1, then y =1, is the solution of 3x + 4y = 7Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

Demo Class for First Step to Coding Course,specificallydesigned for students of class 8 to 12.The students will get to learn more about the world of programming in these

free classeswhich will definitely help them in making a wise career choice in the future.Substituting x = 2

We get,

3×2 + 4y = 7

6 + 4y = 7

4y = 7 − 6

y =1

4

Therefore, if x = 2, then y =1 ,is the solution of 3x + 4y = 7

4

(ii) x = 6y

Substituting x =0

We get,

6y = 0

y = 0

Therefore, if x = 0, then y =0, is the solution of x = 6ySubstituting x = 6

We get,

6 = 6y

y =6

6

y = 1

Therefore, if x = 6, then y =1, is the solution of x = 6y

(iii) x + πy = 4

Substituting x = 0

We get,

πy = 4

y =4

π

Therefore, if x = 0, then y =4is the solution of x + πy = 4

πSubstituting y =0

We get,

x + 0 = 4

x = 4

Therefore, if y = 0, then x = 4 is the solution of x + πy = 4

(iv)2x − y = 4

3

Substituting x = 0

We get,

0 − y = 4

y = −4

Therefore, if x = 0, then y = −4 is the solution of2x − y = 4

3

Substituting x =3

We get,2×3 − y = 4_{ }3

2 − y = 4

y = 4 − 2

y = −2

Therefore, if x = 3, then y = −2 is the solution of2x − y = 4

3

**Question 2. 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**

**Solution: **

(i) 5x − 2y = 10

Substituting x =0

We get,

5×0 − 2y = 10

−2y = 10

−y =10

2

y = −5

Therefore, if x = 0, then y = −5 is the solution of 5x − 2y = 10Substituting y = 0

We get,

5x − 2×0 = 10

5x = 10

x =10

5

x = 2

Therefore, if y = 0, then x = 2 is the solution of 5x − 2y = 10

(ii) −4x + 3y = 12

Substituting x = 0

We get,

−4×0 + 3y = 12

3y = 12

y =12

3

y = 4

Therefore, if x = 0, then y = 4 is the solution of −4x + 3y = 12Substituting y = 0

−4x + 3 x 0 = 12

– 4x = 12

x = −3

Therefore, if y = 0 then x = −3 is a solution of −4x + 3y = 12

(iii) 2x + 3y = 24

Substituting x = 0

2×0 + 3y = 24

3y =24

y = 8

Therefore, if x = 0 then y = 8 is a solution of 2x+ 3y = 24Substituting y = 0

2x + 3×0 = 24

2x = 24

x =12

Therefore, if y = 0 then x = 12 is a solution of 2x + 3y = 24

**Question 3. Check which of the following are solutions of the equation 2x – y = 6 and which are not:**

**(i) (3, 0) **

**(ii) (0, 6) **

**(iii) (2, -2) **

**(iv) (√3, 0) **

**(v) (**__1__, -5)

** 2**

__1__, -5)

**Solution:**

(i) (3, 0)

Substitute x = 3 and y = 0 in 2x – y = 6

2×3 – 0 = 6

6 = 6 {L.H.S. = R.H.S.}

Therefore, (3,0) is a solution of 2x – y = 6.

(ii) (0, 6)

Substitute

x = 0 and y = 6 in 2x – y = 6

2×0 – 6 = 6

–6 = 6 {L.H.S. ≠ R.H.S.}

Therefore, (0, 6) is not a solution of 2x – y = 6.

(iii) (2, -2)

Substitute x = 0 and y = 6 in 2x – y = 6

2×2 – (–2) = 6

4 + 2 = 6

6 = 6 {L.H.S. = R.H.S.}

Therefore, (2,-2) is a solution of 2x – y = 6.

(iv) (√3, 0)

Substitute x = √3 and y = 0 in 2x – y = 6

2×√3 – 0 = 6

2 √3 = 6 {L.H.S. ≠ R.H.S.}

Therefore, (√3, 0) is not a solution of 2x – y = 6.

(v) (1/2, -5)

Substitute x =1and y = -5 in 2x – y = 6

2

2 ×1– (– 5) = 6

2

1 + 5 = 6

6 = 6 {L.H.S. = R.H.S.}

Therefore, (1, –5) is a solution of 2x – y = 6.

2

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

**Solution:**

Given,

3 x + 4 y = k

(–1, 2) is the solution of 3x + 4y = k.

Substituting x = –1 and y = 2 in 3x + 4y = k,

We get,

3×(– 1 ) + 4×2 = k

–3 + 8 = k

k = 5

Therefore, k is 5.

**Question 5. Find the value of λ, if x = –**λ** and y = **__ 5 __ is a solution of the equation x + 4y – 7 = 0

** **_{ }2

__5__is a solution of the equation x + 4y – 7 = 0

_{ }2

**Solution:**

Given,

(-λ,5) is a solution of equation 3x + 4y = k

2

Substituting x = – λ and y =5in x + 4y – 7 = 0.

2

We get,

–λ + 4 ×5– 7 = 0

2

–λ + 10 – 7 = 0

–λ = –3

λ = 3

**Question 6. If x = 2 **α** + 1 and y = **α** – 1 is a solution of the equation 2x – 3y + 5 = 0, find the value of **α.

**Solution:**

Given,

(2 α + 1, α – 1 ) is the solution of equation 2x – 3y + 5 = 0.

Substituting x = 2 α + 1 and y = α – 1 in 2x – 3y + 5 = 0.

We get,

2×(2 α + 1) – 3 × (α – 1 ) + 5 = 0

4α + 2 – 3α + 3 + 5 = 0

α + 10 = 0

α = –10

Therefore, the value of α is –10.

**Question 7. If x = 1 and y = 6 is a solution of the equation 8x – ay + a**^{2} = 0, find the values of a.

^{2}= 0, find the values of a.

**Solution:**

Given,

(1 , 6) is a solution of equation 8x – ay + a^{2}= 0

Substituting x = 1 and y = 6 in 8x – ay + a^{2}= 0.

We get,

8 × 1 – a × 6 + a^{2}= 0

a^{2}– 6a + 8 = 0 (quadratic equation)

Using quadratic factorization

a^{2}– 4a – 2a + 8 = 0

a × (a – 4) – 2 × (a – 4) = 0

(a – 2) (a – 4)= 0

a = 2, 4

Therefore, the values of a are 2 and 4.