2018 AMC 12A Problem #10

$x=3-3y$ from the first equation. We can substitute this into the second: $||3-3y|-|y||=1 \implies |3-3y|-|y|=\pm1$ . Now we just need to find the number of unique solutions for $y$ .

I decided to do some case testing, though there may be a faster method:

Case 1: $|3-3y|-|y|=1$

$y\le0 \implies 3-3y+y=1 \implies y=1$ . Extraneous since $1>0$ .

$0<y\le1 \implies 3-3y-y=1 \implies \boxed{y=\frac{1}{2}}$ .

$y>1 \implies 3y-3-y=1 \implies \boxed{y=2}$ .

Case 2: $|3-3y|-|y|=-1$

$y\le0 \implies 3-3y+y=-1 \implies y=2$ . Extraneous since $2>0$ .

$0<y\le1 \implies 3-3y-y=-1 \implies \boxed{y=1}$ .

$y>1 \implies 3y-3-y=-1 \implies y=1$ . Extraneous since $1=1$ .

There are clearly $3$ solutions to the equation.

There are 3 ordered pairs of real numbers satisfy the following system of equations.