Question

(k - 3)x + 3y = k; kx + ky = 12 find the value of k for which the equation has infinite number of solutions​

Answer 1

Answer:

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

Taking the II and III terms

3/k= k/12

k² = 36

k =√36

k = 6

Hence, the value of k is 6 .

Step-by-step explanation:

Answer 2

Answer:

k = 6, y = 2 - x

Step-by-step explanation:

Expand the following:

(x (k - 3) + 3 y = k, k x + k y = 12)

x (k - 3) = x k + x (-3):

(x k - 3 x + 3 y = k, k x + k y = 12)

Subtract k from both sides of x (-3) + x k + 3 y = k:

(-3 x + x k + 3 y - k = k - k, k x + k y = 12)

k - k = 0:

(-3 x + x k + 3 y - k = 0, k x + k y = 12)

Subtract 12 from both sides of k x + k y = 12:

(-3 x + x k + 3 y - k = 0, k x + k y - 12 = 12 - 12)

12 - 12 = 0:

Answer: |

| (-3 x + x k + 3 y - k = 0, k x + k y - 12 = 0)