Hatcher 2.2.26 Show that if $A$ is contractile in $X$ then $H_n(X,A) =H_n(X) \oplus H_{n-1} (A)$ 
Show that if $A$ is contractible in $X$ then $H_n(X,A) \approx \tilde H_n(X) \oplus \tilde H_{n-1}(A)$

I know that $\tilde H_n(X \cup CA) \approx H_n(X \cup CA, CA) \approx H_n(X,A)$.
And $(X \cup CA)/X = SA$, where $SA$ is the suspension of $A$. So
$H_n((X \cup CA)/X) = H_n(SA)$, where $SA$ is the suspension of $A$. But $SA \simeq A$, and homology is homotopic invariant, we have $H_n((X \cup CA)/X) = H_n(A)$.
I have seen this discussion in one of the post in mathstack but having no clue how to use this in the problem. The long exact sequence I can calculate but why $H_n(X \cup CA, CA) \approx H_n(X,A)$. And where suspension is used and how to split the long exact sequence in the direct sum?

Being contractible in $X$ means that the inclusion $i: A \hookrightarrow X$ is nullhomotopic.
Homology being a homotopy invariant functor gives that the long exact sequence of that pair splits up into short exact sequences:
$$
0\to H_iX  \to H_i(X,A) \to  H_{i-1}A \to 0
$$
Now we only need to show that this sequence splits. A splitting might be constructed by using that an element $\alpha$ in $C_{i-1}A$ is nullhomologous in $X$. Hence there is an element in $C_iX$ which gives $\alpha$ as boundary. This will define an element in $H_i(X,A)$ which maps to $\alpha$.
(it will also define an element in $H_iX$ but this will be zero, but we would expect that by exactness)
