Let K be a 4-dimensional simplicial complex which has 8 0-simplices, 12 1-simplices, 9 2-simplices, 10 3-simplices and 6 4-simplices. Suppose that $H_0(K)= \mathbb{Z}, H_1(K)= \mathbb{Z}+\mathbb{Z}+\mathbb{Z_2}, H_2(K)= \mathbb{Z}+\mathbb{Z_3}, H_3(K)= \mathbb{Z}+\mathbb{Z_4}.\\$ What is $H_4(K)\\$?
Can I do this problem by simply using the fact that $|Im \partial_i|+|Ker\partial_i|= dim C_i$, where $C_i$ is just the free abelian group generated by the $i$ simplices? When I do this I get a bunch of linear equations that gives me $|Ker\partial_4|=2$, and since $K$ is 4-dimensional, $|H_4(K)|=|Ker\partial_4|$, we have $|H_4(K)|=2 \rightarrow H_4(K)=\mathbb{Z}+\mathbb{Z}$.
So for example, $|H_3(K)|=|Ker \partial_3 / Im \partial_4|=1 \rightarrow |Ker \partial_3|=|Im \partial_4|+1$, etc.
The problem with what you describe (if I'm reading your notation correctly) is that the $H_*(K)$ aren't free; they're just $\mathbb{Z}$-modules. On the other hand, $H_4(K) = \ker d_4$ is free, since it's a subgroup of $C_4$. By the universal coefficient theorem, $H_*(K, \mathbb{Q}) = H_*(K) \otimes \mathbb{Q}$. The Euler characteristic of $K$ is thus $\chi(K) = 1 - 2 + 1 - 1 + r = r - 1$, where $r = \dim_{\mathbb{Q}} (H_4(K) \otimes \mathbb{Q})$. But $\chi$ is invariant under passing to homology, so $\chi(K) = 8 - 12 + 9 - 10 + 6 = 1$. Hence $r = 2$ and $H_4(K) = \mathbb{Z}^r = \mathbb{Z}^2$.