Question

Find the value of k for which the system of equations x+2y-3=0 ky+5x+7=0 has a unique solution​

Answer 1

Answer:

The given system of equations:  

x + 2y = 3  ⇒ x + 2y - 3 = 0 ….(i)

And, 5x + ky + 7 = 0 …(ii)  

These equations are of the following form:  

a1x+b1y+c1 = 0, a2x+b2y+c2 = 0  where, a1 = 1, b1= 2, c1 = -3 and a2 = 5, b2 = k, c2 = 7  

For a unique solution, we must have:

a1/a2 ≠ b1/b2 i.e 1/5 ≠ k/2

k ≠ 10

Thus for all real values of k other than 10, the given system of equations will have a unique solution.