2018 AMC 10A Problem 12

From the first equation, $x = 3 - 3y$. The second equation says $|x| - |y| = 1$ or $|x| - |y| = -1$.

Along the line, $x \ge 0$ exactly when $y \le 1$, so the signs of $x$ and $y$ split the line into just three pieces.

Region $y < 0$ (so $x > 0$): $|x| - |y| = (3 - 3y) + y = 3 - 2y$. Setting $3 - 2y = \pm 1$ gives $y = 1$ or $y = 2$, neither negative. No solutions.

Region $0 \le y \le 1$ (so $x \ge 0$): $|x| - |y| = 3 - 3y - y = 3 - 4y$. Setting $3 - 4y = \pm 1$ gives $y = \tfrac12$ or $y = 1$, both in range. Solutions $\left(\tfrac32, \tfrac12\right)$ and $(0, 1)$.

Region $y > 1$ (so $x < 0$): $|x| - |y| = (3y - 3) - y = 2y - 3$. Setting $2y - 3 = \pm 1$ gives $y = 2$ or $y = 1$; only $y = 2$ is in range. Solution $(-3, 2)$.

Quick check of $(-3, 2)$: $\big||-3| - |2|\big| = 1$ and $-3 + 6 = 3$. Three ordered pairs in all.

The answer is $\boxed{\textbf{(C)}\ 3}$.

Nested absolute values are handled by peeling one layer at a time, but the case count explodes unless you use the constraint you already have. Substituting the line first ties the sign of $x$ to $y$, cutting the four sign quadrants down to three intervals and making every case a one-variable linear equation. Always verify each candidate against the region that produced it.

Graph $\big||x| - |y|\big| = 1$: in the first quadrant it is the two rays $x - y = 1$ (from $(1,0)$) and $y - x = 1$ (from $(0,1)$), both heading out along the diagonal direction; reflecting into the other quadrants gives eight rays starting at $(\pm1, 0)$ and $(0, \pm1)$. The line through $(3,0)$ and $(0,1)$ passes through the ray endpoint $(0,1)$, crosses the first-quadrant ray from $(1,0)$, and crosses the second-quadrant ray from $(-1,0)$; it misses everything below the $x$-axis. Three points.

• Removing the outer absolute value as |x| - |y| = 1 only, missing the branch |y| - |x| = 1 that produces the third solution (0, 1).
• Accepting a case solution such as $y = 1$ from the region $y < 0$ without checking that it lies in that region, which inflates the count to 4 or more.