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,
⇒ 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.