Solving inequalities involving the absolute values of $2$ variables

Question

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

Answer 1

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 $$y<-2x-13$$

Answer 2

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.