Write two solutions for each of the following equations:
(i) 5x - 2y = 7
(ii) x = 6y
(iii) x + πy = 4
(iv) (2/3)x - y = 4.
(i) We are given,
3x + 4y = 7
Substituting x = 1
In the given equation, We get
3 ×1 + 4y = 7
4y = 7- 3 - 4 = 4Y
Y = 1
Thus x = 1 and y = 1 is the solution of 3x + 4y = 7
Substituting x = 2 in the given equation, we get
3× 2 + 4y = 7
4y = 7 - 6
y = 1/4
Thus x = 2 and y = 1/4 is the solution of 3x + 4y = 7
(ii) We are given, x = 6y
Substituting x = 0 in the given equation, we get 0 = 6y
6 y = 0
y = 0
Thus x = 0, ⟹ Solution (0, 0)
Substituting x = 6
6 = 6y
y = 6/6
y = 1
⟹ Solution (6, 1)
(iii) We are given x + πy = 4
Substituting x = 0 in the given equation,
We get 0 + πy = 4
πy = 4
y = 4/π
⟹ Solution = (0, 4/π)
Substituting y = 0 in the given equation, we get x + 0 = 4
x = 4
⟹ Solution = (4, 0)
(iv) We are given (2/3) x – y = 4
Substituting x = 0 in the given equation, we get 0 - y = 4
y = - 4
Thus x = 0 and y = - 4 is a solution
Substituting x = 3 in the given equation, we get (2/3) × 3 − y = 4
2 - y = 4
y = 2 - 4
y = -2
Thus x = 3 and y = – 2 is a solution
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
(i) We are given, 5x - 2y = 10
Substituting x = 0 in the given equation, We get;
5× 0 - 2y = 10 - 2y = 10 - y = 10/2
y = - 5
Thus x = 0 and y = - 5 is the solution of 5x - 2y = 10
Substituting y = 0 in the given equation, we get 5x – 2 × 0 = 10
5x = 10
x = 10/5
x = 2
Thus x = 2 and y = 0 is a solution of 5x - 2y = 10
(ii) We are given, - 4x + 3y = 12
Substituting x = 0 in the given equation, we get;
- 4 × 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;
- 4x + 3 × 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; 2 × 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 + 3 × 0 = 24
2x = 24
x = 24/2
x =12
Thus x = 12 and y = 0 is a solution of 2x + 3y = 24
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) (1/2 , - 5)
We are given, 2x - y = 6
(i) In the equation 2x - 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 × 3 - 0 = 6 ⟹ L.H.S = R.H.S
⟹ (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 = 2 × 0 - 6 = - 6
⟹ L.H.S ≠ R.H.S
⟹ (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 = 2 × 2 - (-2) = 6
⟹ L.H.S = R.H.S
⟹ (2, - 2) is a solution of 2x - y = 6.
(iv) In the equation 2x - 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 × √3 - 0
⟹ L.H.S ≠ R.H.S
⟹ (√3, 0) is not a solution of 2x - y = 6.
(v) In the equation 2x - y = 6,
We have L.H.S = 2x - y and R.H.S = 6
Substituting x = 1/2 and y = in 2x - y, we get L.H.S = 2 × (1/2) - (-5)
⟹ 1 + 5 = 6
⟹ L.H.S = R.H.S
⟹ (12, - 5) is a solution of 2x - y = 6.
If x = – 1, y = 2 is a solution of the equation 3x + 4y = k, find the value of k.
We are given, 3x + 4y = 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 - 1 + 4 × 2 = k
K = - 3 + 8 k = 5
Find the value of λ, if x = -λ and y = 5/2 is a solution of the equation x + 4y – 7 = 0
We are given, x + 4y - 7 = 0 (- λ, - 5) is a solution of equation 3x + 4y = k
Substituting x = - λ and y = 5/2 in x + 4y - 7 = 0,
We get; - λ + 4 × (5/2) - 7 = 0 -λ + 4 × 5/2 - 7 = 0
λ = 10 - 7
λ = 3
If x = 2a + 1 and y = a -1 is a solution of the equation 2x - 3y + 5 = 0, find the value of a.
We are given, 2x - 3y + 5 = 0 (2a + 1, a - 1) is the solution of equation 2x - 3y + 5 = 0.
Substituting x = 2a + 1 and y = a - 1 in 2x - 3y + 5 = 0,
We get 2 × 2a + (1- 3) x a - 1 + 5 = 0
⟹ 4a + 2 - 3a + 3 + 5 = 0
⟹ a + 10 = 0
⟹ a = - 10
If x = 1 and y = 6 is a solution of the equation 8x - ay + a2 = 0, find the values of a.
We are given, 8x - ay + a2 = 0 (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
Using quadratic factorization a2 - 4a - 2a + 8 = 0 a(a - 4 ) - 2(a - 4) = 0 (a - 2)(a - 4) = 0
a = 2, 4