# 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 = 7Substituting 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.