Question

(k-3)x+3y=k
kx+ky=12
determine the value of k for which the given system of equations has infinitely many solutions.

Answer 1

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 .

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Answer 2

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 .

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