Question

Let $A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x + y| \geq 3\}$ and $B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x| + |y| \leq 3\}$. If $C = \{(x, y) \in A \cap B : x = 0 \text{ or } y = 0\}$, then $\sum_{(x,y)\in C} |x + y|$ is:

Answer 1

12

Answer 2

Solution Explanation:

For points on the $x$-axis ($y=0$):

• In $B$, $|x| \leq 3$ and in $A$, $|x+0|=|x|\geq 3$. Thus, $x$ must be $\pm3$.

For points on the $y$-axis ($x=0$):

• In $B$, $|y| \leq 3$ and in $A$, $|0+y|=|y|\geq 3$. Thus, $y$ must be $\pm3$.

So,

$$C = \{(3,0),\,(-3,0),\,(0,3),\,(0,-3)\}.$$

Calculating $|x+y|$ for each:

• $(3,0): |3+0|=3$
• $(-3,0): |-3+0|=3$
• $(0,3): |0+3|=3$
• $(0,-3): |0-3|=3$

Thus,

$$\sum_{(x,y)\in C}|x+y| = 3+3+3+3 = 12.$$