2018 AMC 10A Problem 12

Absolute values are hard to work with until you know whether the inside is positive or negative. The line lets us decide that.

1. Use the line. From $x + 3y = 3$ we get $x = 3 - 3y$. So once we know $y$, we know $x$.
2. When is $x$ positive? $x = 3 - 3y$ is $0$ or more exactly when $y \le 1$. And $y$ itself changes sign at $0$. So the line splits into three pieces:
- $y < 0$: here $x > 0$ and $y < 0$;
- $0 \le y \le 1$: here $x \ge 0$ and $y \ge 0$;
- $y > 1$: here $x < 0$ and $y > 0$.
3. Open the outer absolute value. $\big||x| - |y|\big| = 1$ means $|x| - |y|$ is either $1$ or $-1$.
4. Piece $y < 0$. Here $|x| = 3 - 3y$ and $|y| = -y$, so $|x| - |y| = 3 - 3y + y = 3 - 2y$.
- $3 - 2y = 1$ gives $y = 1$; $3 - 2y = -1$ gives $y = 2$. Neither is negative, so no solutions here.
5. Piece $0 \le y \le 1$. Here $|x| = 3 - 3y$ and $|y| = y$, so $|x| - |y| = 3 - 4y$.
- $3 - 4y = 1$ gives $y = \tfrac12$, so $x = \tfrac32$.
- $3 - 4y = -1$ gives $y = 1$, so $x = 0$.
- Both $y$ values are in the range, so we get $\left(\tfrac32, \tfrac12\right)$ and $(0, 1)$.
6. Piece $y > 1$. Here $|x| = 3y - 3$ and $|y| = y$, so $|x| - |y| = 2y - 3$.
- $2y - 3 = 1$ gives $y = 2$, so $x = -3$. This is in range.
- $2y - 3 = -1$ gives $y = 1$, which is not bigger than $1$. Rejected (it is the point we already found).
7. Check the last one. $(-3, 2)$: $-3 + 6 = 3$, and $\big|3 - 2\big| = 1$. Good.

There are $3$ ordered pairs: $\left(\tfrac32, \tfrac12\right)$, $(0,1)$, $(-3,2)$.

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

An absolute value becomes an ordinary expression once you know the sign of what is inside. Plugging in the line first ties the sign of $x$ to $y$, so instead of many sign cases you get just three stretches of the line, each with a simple linear equation. Always check that each answer really falls in the stretch you assumed.

Draw it. In the top-right quarter of the plane, $\big||x| - |y|\big| = 1$ is two rays: $x - y = 1$ starting at $(1, 0)$ and $y - x = 1$ starting at $(0, 1)$, both going up and to the right at slope $1$. Flipping these over both axes gives eight rays. The line $x + 3y = 3$ goes through $(3, 0)$ and $(0, 1)$. It touches the ray start $(0,1)$, crosses the ray from $(1,0)$ at $\left(\tfrac32, \tfrac12\right)$, and crosses a ray in the top-left quarter at $(-3, 2)$. Below the $x$-axis the line moves away from the rays and never meets them. Three points.

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 $\boxed{\textbf{(C)}\ 3}$.