Solve the simultaneous congruences: $x \equiv 71\pmod {112}$ , $x \equiv 111\pmod {189} $

Question

I am a bit stuck on this. i have solved some of it but dont know if it is correct or not

$x - 71 = 112k \ \ \rightarrow \ \ x = 71 + 112k$

$x - 111 = 189c \ \rightarrow \ \ x = 111 + 189c$

$111 + 189c = 71 + 112k$

Then how to find c or k??

Got another question: $x\equiv 10\pmod {11}$ , $x\equiv 5\pmod {12}$ , $x\equiv 1\pmod {13}$

is that similar to the question above?

Answer 1

$$111+189c=71+112k\quad|-111\iff189c=112k-40\iff189c=8(14k-5)\iff$$ $$\iff c=8d\iff189d=14k-5\quad|-14k\iff7\cdot(27d-2k)=-5\iff7\mid5\iff$$ Contradiction!