#### Chapter 13Â Linear Equations in Two VariablesÂ R.D. Sharma Solutions for Class 9th Exercise 13.2

**Exercise 13.2**

(i) 5xâ€“2y = 7

(ii) x = 6y

(iii) x + Ï€y = 4

(iv) 2/3x â€“ y = 4.

**Solution**

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

(i) 5x âˆ’ 2y = 10

(ii) -4x + 3y = 12

(iii) 2x + 3y =24

**Solution**

(i) We are given,

5x â€“ 2y = 10

Substituting x = 0 in the given equation,

We get;

5Ã—0 â€“ 2y = 10

â€“ 2y = 10

â€“ y = 10/2

y = â€“ 5

Thus x =0 and y = -5 is the solution of 5x-2y = 10

Substituting y = 0 in the given equation, we get 5x â€”2 x 0 = 10

5x = 10

x = 10/2

x = 2

Thus x =2 and y = 0 is a solution of 5x-2y = 10

(ii) We are given, â€“ 4x + 3y = 12

Substituting x = 0 in the given equation,

we get;

-4 Ã— 0 + 3y = 12

3y = 12

y = 4

Thus x = 0 and y = 4 is a solution of the -4x + 3y = 12

Substituting y = 0 in the given equation, we get;

- 4 x + 3Â Ã—Â 0 = 12

- 4x = 12

x = -12/4

x = -3

Thus x = -3 and y =0 is a solution of -4x + 3y = 12

(iii) We are given, 2x + 3y = 24

Substituting x = 0 in the given equation, we get;

2Â Ã—Â 0 + 3y = 24

3y =24

y = 24/3

y = 8

Thus x = 0 and y = 8 is a solution of 2x+ 3y = 24

Substituting y = 0 in the given equation, we get;

2x +3Â Ã—Â 0 = 24

2x = 24

x = 24/2

x = 12

Thus x =12 and y = 0 is a solution of 2x + 3y = 24

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

**Solution**

We are given, 2x â€“ y = 6

(i) In the equation 2 x â€“ y = 6,

We have L.H.S = 2x - y and R.H.S = 6

Substituting x = 3 and y = 0 in 2x - y ,

We get L.H.S = 2 x 3 â€“ 0 = 6

â‡’Â L.H.S = R.H.S

â‡’Â (3,0) is a solution of 2x - y = 6.

(ii) In the equation 2x - y = 6,

We have L.H.S = 2x- y and R.H.S = 6

Substituting x = 0 and y = 6 in 2x - y

We get L.H.S = 2Â Ã—Â 0 - 6 = -6

â‡’Â L.H.Sâ‰ Â R.H.S

â‡’Â (0,6) is not a solution of 2x - y = 6.

(iii) In the equation 2x - y = 6,

We have L.H.S = 2x - y and R.H.S = 6

Substituting x = 2 and y = - 2 in 2x - y,

We get L.H.S = 2Â Ã—Â 2 - (-2) = 6

â‡’ L.H.S = R.H.S

â‡’Â (2,-2) is a solution of 2x - y = 6.

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

LHS = 2x-y and RHS = 6

Substituting x = âˆš3 and y = 0 in 2x - y, we get

LHS = 2Ã—âˆš3-0 = 2âˆš3

LHS â‰ RHS

(âˆš3, 0) is not the solution of 2x - y = 6.

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

LHS = 2x-y and RHS = 6

Substituting x = 1/2 and y = -5 in 2x-y, we get

LHS = 2Ã— 1/2 and y = -5 in 2x-y, we get

LHS = RHS

(1/2, -5) is the solution of 2x-y = 6.

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

**Solution**

We are given, 3 x + 4 y = k

Given that, (-1,2) is the solution of equation 3x + 4y = k.

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

We get; 3x â€“ 1 + 4Â Ã—Â 2 = k

k = -3 + 8

k = 5

5. Find the value ofÂ Î», if x =-Î» and y=5/2 is a solution of the equation x+4y-7=0.Â

Â

**Â Solution**

6. If x = 2Î± + 1 and y = Î± âˆ’ 1 is a solution of the equation 2x âˆ’ 3y + 5 = 0, find the value of Î±.

**Solution**

We are given, 2x -3y +5 = 0

(2a + 1, a â€“ 1 ) is the solution of equation 2x â€“ 3y + 5 = 0.

Substituting x = 2a + 1 and y = a â€“ 1 in 2x â€“ 3y + 5 = 0,

We get 2 x 2a + (1- 3) x a â€“ 1 + 5 = 0

â‡’Â 4a + 2 â€“ 3a + 3 + 5 = 0

â‡’ a + 10 = 0

â‡’ a = â€“ 10

7.Â If x=1 and y=6 is a solution of the equation 8x-ay+a

^{2}=0. Find the value of a.

**Solution**

Given,

8x - ay + a

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

^{2}Â = 0
Substituting, x = 1 and y = 6 inÂ 8x-ay+a

^{2}Â = 0, we get
8Ã—1 - aÃ—6 +a

^{2}Â = 0
â‡’ a

^{2}Â - 6a + 8 = 0
Using quadratic factorization,

a

^{2}Â -4a -2a + 8 = 0
â‡’ a(a-4) - 2(a-4) = 0

â‡’ (a-2) (a-4) = 0

Therefore, a = 2, 4