Find the value of k for which following system of equations has a unique solution:
4x+ky+8=0
2x+2y+2=0
For the system to have unique solution, lines must intersect each other.
i.e., If the lines a1x+b1y+c1=0 and a2x+b2y+c2=0 have the unique solution,
Then a1a2≠b1b2
Given equations are,
4x+ky+8=0 and 2x+2y+2=0 has a unique solution.
⇒42≠k2 [Since, a1a2≠b1b2]
⇒2k≠8
⇒k≠82
i.e., The value of k is all the real numbers except 4.