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 + πy = 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 = 7

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 = 6y

Substituting 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 =  4  is the solution of x + πy = 4
                                            π

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

(iv) 2 x − y = 4        
       3
Substituting x = 0
We get,
0 − y = 4
y = −4
Therefore, if x = 0, then y = −4 is the solution of  2  x − 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 of  2 x − 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 = 10

Substituting 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 = 12

Substituting 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 = 24

Substituting 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

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 =  1  and 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

Solution:

Given, 
(-λ,  5 ) is a solution of equation 3x + 4y = k
       2
Substituting x = – λ and y =  5  in 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 + a² = 0, find the values of a.

Solution:

Given, 
(1 , 6) is a solution of equation 8x – ay + a² = 0
Substituting x = 1 and y = 6 in 8x – ay + a² = 0.
We get,
8 × 1 – a × 6 + a² = 0
a² – 6a + 8 = 0 (quadratic equation)
Using quadratic factorization
a² – 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.