Question

find the value(s) of K for which the pair of linear equation kx+y=k^2 and x+ky = 1 haveinfitely many solutions

Answer 1

for infinate solution
a1/a2 = b1/b2 = c1/c2
so
k/1=1/k=-k^2/-1
so we get
k^2=1
k= -+1

Answer 2

Answer

For pair of equations kx + 1y = k2 and 1x + ky = 1

We have:

a1/a2 = k/1

b1/b2 = 1/k

c1/c2 = k2/1

For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

k/1 = 1/k

k2 = 1

k = 1, -1 ... (i)

1/k = k2/1

k3 = 1

k = 1 ... (ii)

From (i) and (ii),

k = 1




Mark as breinlist please