Solving inequalities involving the absolute values of $2$ variables

I need to solve $$|x+y| + |x+3| > 10$$

While I am familiar with solving inequalities with multiple absolute values of $$x$$ such as $$|x-7| > |x-3|$$, I do not know how to solve inequalities that involve $$2$$ variables, $$x$$ and $$y$$ in this case

Hint: Consider the cases: $$x\geq -y$$ and $$x\geq 3$$ or $$x\geq -y$$ and $$x<-3$$ or $$x<-y$$ and $$x\geq -3$$ or $$x<-y$$ and $$x<-3$$ so we get in the first case: $$y>7-2x$$ and in the second one $$y>13$$ and in the next case $$y<-7$$ And in the last case we get $$y<-2x-13$$
You have four cases, depending on whether $$x+y$$ is positive or negative and whether $$x+3$$ is positive or negative.
Case 1: $$x+y>0$$ and $$x+3>0$$. Then the inequality becomes $$x+y+x+3>10$$ or $$y>-2x-3$$. The solution in this case is the intersection of the 3 half planes $$y>-x$$, $$x>-3$$ and $$y>-2x-3$$.
The other cases go the same way. The final solution is the union of the solutions of the 4 cases. In the end, the solution is the entire $$xy$$-plane minus some sort of polygon around the origin.