Inverse of f:A->B exists iff 'f' is a bijection

1) One-One

f(x1) = f(x2)

a(x1) + b = a (x2) + b

x1 = x2

hence it's one-one function

2) Onto

ax + b = c

ax = b-c

**x = b - c / a**

given a !=0

I want to find out for every "C" there exists an element in domain or not

therefore ,we can see that Yes for every value of 'C' we get the pre-image in Domain

hence , it's Onto -function