For what value of c, the following system of linear equations has infinite number of solutions:

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

The pair of equations are:

(c – 1) x – y = 5

(c + 1) x + (1 – c) y = 3c + 1

These equations can be written as:

(c – 1) x – y – 5 = 0

(c + 1) x + (1 – c) y –( 3c + 1)

On comparing the given equation with standard form i.e.

a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = c – 1 , b₁ = –1 , c₁ = –5

a₂ = c + 1, b₂ = 1 – c , c₂ = –(3c + 1)

For infinitely many solutions,

So,

So,

From (I) and (II)

⇒ (c–1)(1–c) = – (c+1)

⇒ c – c² –1 + c = –c – 1

⇒ c – c² –1 + c + c + 1 = 0

⇒ 3c – c² = 0

⇒ c ( 3 – c ) = 0

⇒ c = 0 , c = 3

From (II) and (III)

⇒ –(–3c–1) = –5(1–c)

⇒ 3c + 1 = –5 + 5c

⇒ 3c + 1 + 5 – 5c =0

⇒ 6 – 2c = 0

⇒ 6 = 2c

⇒ c = 3

From (I) and (III)

⇒ (c–1)(–3c–1) = –5 (c+1)

⇒ –3c² – c + 3c + 1 = –5c – 5

⇒ –3c² – c + 3c + 1 + 5c + 5 = 0

⇒ –3c² + 7c + 6 = 0

⇒ 3c² – 7c – 6 = 0

⇒ 3c² – 9c + 2c – 6 = 0

⇒ 3c ( c–3) + 2 (c–3) = 0

⇒ (3c+2) ( c–3) = 0

⇒ c = –2/3 and c = 3

Hence the value of c is 3.