Question

Find the value of k for which the 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:

`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`

where, `a_1 = (k  –  1), b_1= -1, c_1= -5 and a_2 = (k + 1), b_2 = (1 – k), c_2= -(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)`

`⇒ k^2 + 1  –  2k = k + 1`

`⇒ k^2  –  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)

`⇒ 3k^2 + k – 3k – 1 = 5k + 5`

`⇒ 3k^2 – 2k – 5k – 1 – 5 = 0`

`⇒ 3k^2 – 7k – 6 = 0`

`⇒ 3k^2 – 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.