IF the equations ( k-3)x + 3y = k and kx + k y =12 have infinite number of solutions find k
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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
Now, 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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