Question

find the value of K for which the following pair of linear equation have infinitely many solution. 2x+3y=7,(k+1)x+(2k-1)y=4k+1​

Answer 1

GIVEN :

Pair of Linear equations :

i) 2x +3y=7

ii) (k+1)x+(2k-1)y=4k+1

We know that,

For infinite solutions :

a1/a2 = b1/b2 = c1/c2

From the above equations :

a1 = 2 a2 = 3 a3 = 7

b1 = k + 1 b2 = 2k - 1 b3 = 4k + 1

2/K+1 = 3/2k-1 = 7/4k+1

2/k + 1 = 3/2k - 1

4k - 2 = 3k + 3

k = 5

Therefore, the value of k is 5.

Answer 2

Answer:

Step-by-step explanation:

Given :-

2x - 3y = 7

Solution :-

2x - 3y - 7 = 0

a₁ = 2 , b₁ = -3 , c₁ = -7

⇒ (K+1) x + (1 - 2 k) y = (5 k-4)

⇒ (K+1) x + (1 - 2 k) y - (5 k-4) = 0

a₁ = k+1 , b₁ = 1-2k , c₁ = -(5k + 4)

If the equations have infinitely may solutions :-

a₁/a₂ = b₁/b₂ = c₁/c₂

2/k + 1 = -3/1 - 2k = -7/ - (5k + 4)

2/k + 1 = -3/1 - 2k

Cross multiplying both, we get :-

⇒ -3 (k + 1) = 2 (1 - 2k)

⇒ -3k -3 = 2 - 4k

⇒ -3k+4k = 2+3

⇒ k = 5

Hence, the value of k is 5.