Question

Find the value of k for which the following system of linear equations has an infinite number of solutions:

(k – 1)x – y = 5, (k + 1)x + (1 – k)y = (3k + 1)

Answer 1

The given system of equations:

(k – 1)x – y = 5

⇒ (k – 1)x – y – 5 = 0   ...(i)

And (k + 1)x + (1 – k)y = (3k + 1)

⇒ (k + 1)x + (1 – k)y – (3k + 1) = 0   ...(ii)

These equations are of the following form:

a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0

where, a₁ = (k – 1), b₁ = –1, c₁ = –5 and a₂ = (k + 1), b₂ = (1 – k), c₂= –(3k + 1)

For an infinite number of solutions, we must have:

`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`

i.e., `((k - 1))/((k + 1)) = (-1)/(-(k - 1)) = (-5)/(-(3k + 1))`

⇒ `((k - 1))/((k + 1)) = 1/((k - 1)) = 5/((3k + 1))`

Now, we have the following three cases:

Case I:

`((k - 1))/((k + 1)) = 1/((k - 1))`

⇒ (k – 1)² = (k + 1)

⇒ k² + 1 – 2k = k + 1

⇒ k² – 3k = 0

⇒ k(k – 3) = 0

⇒ k = 0 or k = 3

Case II:

`1/((k - 1)) = 5/((3k + 1))`

⇒ 3k + 1 = 5k − 5

⇒ 2k = 6

⇒ k = 3

Case III:

`((k - 1))/((k + 1)) = 5/((3k + 1))`

⇒ (3k + 1) (k – 1) = 5(k + 1)

⇒ 3k² + k – 3k – 1 = 5k + 5

⇒ 3k² – 2k – 5k – 1 – 5 = 0

⇒ 3k² – 7k – 6 = 0

⇒ 3k² – 9k + 2k – 6 = 0

⇒ 3k(k – 3) + 2(k – 3) = 0

⇒ (k – 3) (3k + 2) = 0

⇒ (k – 3) = 0 or (3k + 2) = 0

⇒ k = 3 or k = `(-2)/3`

Hence, the given system of equations has an infinite number of solutions when k is equal to 3.