Question

Find the values of k for which the pair of linear equation kx+y =k^2 and x+ky =1 have infinitely many solutions.

Answer 1

Given kx + y = k^2   --- (1)

           x + ky = 1  ---- (2)

Both the equations are in the form of a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0.

In the equation (1), we have 

a1 = k, b1 = 1,c1 = -k^2.

In the equation (2),

a2 = 1,b2 = k,c2 = -1.


Given that the equation has infinitely many solutions.

So, a1/b1 = a2/b2

       k/1 = 1/k

       k^2 = 1

       k = 1.


Hope this helps!

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Given Equation are

Kx + y - k² = 0    ..... (i)

x + ky - 1 = 0 .....(ii)

Since (i) and (ii) have infinitely many solutions.

⇒ k/1 = 1/k = - k²/- 1

They are 1st = 2nd = 3rd

From 1st and 2nd = k/1 = 1/k

⇒ k² = 1

⇒ k = ± 1 .... (iii)

From 2nd and 3rd = 1/k = k²/1

⇒ k³ = 1

⇒ k = 1 ....(iv)

From (iii) and (iv), we get

⇒ k = 1

Hence, the value of k is 1.