You just need to overwrite what you’re deleting with the next value in the array, propagate that change, and then keep in mind where the new end is:
int array[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
// delete 3 (index 2)
for (int i = 2; i < 8; ++i)
array[i] = array[i + 1]; // copy next element left
Now your array is {1, 2, 4, 5, 6, 7, 8, 9, 9}. You cannot delete the extra 9 since this is a statically-sized array, you just have to ignore it. This can be done with std::copy:
std::copy(array + 3, // copy everything starting here
array + 9, // and ending here, not including it,
array + 2) // to this destination
In C++11, use can use std::move (the algorithm overload, not the utility overload) instead.
More generally, use std::remove to remove elements matching a value:
// remove *all* 3's, return new ending (remaining elements unspecified) auto arrayEnd = std::remove(std::begin(array), std::end(array), 3);
Even more generally, there is std::remove_if.
Note that the use of std::vector<int> may be more appropriate here, as its a “true” dynamically-allocated resizing array. (In the sense that asking for its size() reflects removed elements.)