Math, asked by rushilgaba98, 11 months ago

determine the value of k for which the following pair of linear equations has infinite solutions.(k-3)X+3y=k, kx+ KY=12​

Answers

Answered by siddhartharao77
11

Answer:

k = 6

Step-by-step explanation:

Given Equations are:

(i)

(k - 3)x + 3y = k

Here, a₁ = (k - 3), b₁ = 3, c₁ = -k

(ii)

kx + ky = 12

Here, a₂ = k, b₂ = k, c₂ = -12

For infinite solution, we have

a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ (k - 3)/k = 3/k = k/12

Case 1:

(k - 3)/k = 3/k

⇒ k(k - 3) = 3k

⇒ k² - 3k = 3k

⇒ k² = 6k

⇒ k = 6

Case 2:

3/k = 6/12

⇒ 36 = 6k

⇒ k = 6

Case 3:

(k - 3)/k = k/12

⇒ 12k - 36 = k²

⇒ k² - 12k + 36 = 0

⇒ (k - 6)² = 0

⇒ k = 6

For k = 6, there are infinitely many solutions.

Hope it helps!

Answered by Siddharta7
3

The Given pair of linear equation is :

(k - 3 ) x + 3y = k

kx + ky = 12

We can write these Equations as :

(k - 3 ) x + 3y - k = 0……….(1)

kx + ky - 12 = 0 ………….(2)

On comparing with General form of a pair of linear equations in two variables x & y is:

a1x + b1y + c1 = 0

and a2x + b2y + c2= 0

a1= k-3 , b1= -3, c1= -k

a2= k , b2= k , c2= - 12

a1/a2= k-3 /k , b1/b2= 3/k , c1/c2= -k/-12= k/12

Given: A pair of linear equations has a infinite solution, if

a1/a2 = b1/b2 = c1/c2

k-3 /k =3/k= k/12

I II III

Taking the first two terms

a1/a2 = b1/b2

k-3 /k =3/k

k - 3 = 3

k = 3 + 3

k = 6

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