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

x + (c + 1) y = 5, (c + 1) x + 9y = 8c – 1

The pair of equations are:

x + (c + 1) y = 5

(c + 1) x + 9y = 8c – 1

These equations can be written as:

x + (c + 1) y – 5 = 0

(c + 1) x + 9y – (8c – 1) = 0

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

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

a₁ = 1 , b₁ = c + 1 , c₁ = –5

a₂ = c + 1, b₂ = 9 , c₂ = –(8c – 1)

For infinitely many solutions,

So,

So,

⇒ From (I) and (II)

⇒ 9 = (c+1)²

⇒ 9 = c² + 1 + 2c

⇒ 9 – 1 = c² + 2c

⇒ 8 = c² + 2c

⇒ c² + 2c – 8 = 0

Factorize by splitting the middle term,

c² + 4c – 2c – 8 = 0

⇒ c ( c + 4 ) – 2 ( c + 4) = 0

⇒ (c+4) (c–2) = 0

⇒ c = –4, c = 2

From (II) and (III)

⇒ (c+1)(–8c=1) = –5 × 9

⇒ –8c² + c – 8c + 1 = –45

⇒ –8c² + c – 8c + 1 + 45 = 0

⇒ –8c² – 7c + 46 = 0

⇒ 8c² + 7c – 46 = 0

⇒ 8c² – 16c + 23c –46 = 0

⇒ 8c ( c–2) + 23 ( c–2) = 0

⇒ (8c+23) ( c–2) = 0

⇒ c = –23/8 and c = 2

From (I) and (III)

⇒ –8c+1 = –5(c+1)

⇒ –8c + 1 = –5c – 5

⇒ –8c + 5c = –5 –1

⇒ –3c = –6

⇒ c = 2

So the value of c = 2.