For what values of k will the following pair of linear equations have infinitely many solutions?
kx + 3y - (k – 3) = 0
12x + ky - k = 0

Solution:

In the above equation a₁= k, a₂ = 12, b₁ = 3, b₂ = k, c₁ = -(k - 3) and c₂ = -k

If a solution has infinitely many solutions, then

⇒ a₁/ a₂ = b₁ / b₂ = c₁ / c₂

For the above pair of equations,

⇒ k / 12 = 3 / k = k - 3 / k

⇒ k/ 12 = 3/ k

⇒ k² = 36

⇒ k = ± 6

Also

⇒ 3/ k = (k - 3)/ k

⇒ 3k = k(k - 3)

⇒ 3k = k² -3k

⇒ k² = 6k

⇒ k = 6

The value of k which satisfies both the equations is 6.

Thus, for k = 6, the pair of linear equations have infinitely many solutions

For what values of k will the following pair of linear equations have infinitely many solutions? kx + 3y - (k – 3) = 0, 12x + ky - k = 0

Summary:

For k = 6 the pair of linear equations kx + 3y - (k – 3) = 0 and 12x + ky - k = 0 have infinitely many solutions

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