Given, pair of equations
2x + 3y = 2
and (c + 2)x + (2c + 1)y = 2(c – 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 = 2, b1 = 3 and c1 = – 2
and a2 = c + 2, b2 = 2c + 1 and c2 = – 2(c – 1)
For infinitely many solutions,
⇒ 2(2c + 1) = 3(c + 2)
⇒ 4c + 2 = 3c + 6
⇒ 4c – 3c = 6 – 2
⇒ c = 4
Hence, the required value of c = 4.