930. How many solutions in integers $x$ and $y$ does the inequality

$$
|x|+|y|<10 ?
$$

have?
$\triangle$ From the condition, it is clear that

$$
|x|<10, \text { i.e. }-10<x<10
$$

Let's consider all cases.

If $x= \pm 9$, then $y=0$. We have two solutions: $(9 ; 0)$ and $(-9 ; 0)$.

If $x= \pm 8$, then $y=0, \pm 1$. This gives us six solutions: $(8 ; 0),(-8 ; 0),(8 ; 1),(-8 ; 1)$, $(8 ;-1)$ and $(-8 ;-1)$.

When $x= \pm 7$, the unknown $y$ takes values $0, \pm 1, \pm 2$. This gives us another 10 solutions.

Similarly, for $x= \pm 6, \pm 5, \pm 4, \pm 3, \pm 2, \pm 1$, we get 14, 18, 22, 26, 30, and 34 solutions, respectively.

Finally, if $x=0$, then $y$ takes values $0, \pm 1, \pm 2, \ldots, \pm 9$. This gives us another 19 solutions to the inequality.

It remains to count the total number of solutions:

$$
(2+6+10+14+18+22+26+30+34)+19=162+19=181
$$

Answer: 181.