(i)
kx + y = k – 2; 9x + ky = k
Sol. kx + y = k – 2
Comparing
with a1x
+ b1y
= c1 we
get, a1 =
k, b1 =
1, c1 =
k – 2
9x
+ ky = k
Comparing with
a2x + b2y = c2 we
get, a2 = 9, b2
= k, c2 =
k
∵ The simultaneous equations have
infinitely many solutions.
a1/a2
= b1/b2 = c1/c2
∴ k/9
= 1/k = k-2/k [From eq.
∴ k/9
= 1/k OR 1/k = k-2/k
∴ k2
= 9 OR k/k = k-2
∴ k
= ±√9 OR 1 = k – 2
∴ k
= ± 3 OR
3 = k
k = 3 satisfies both conditions hence k =
3 is the value for which the given simultaneous equations have infinitely many
solutions.
