2018 AMC 12A Problem #10

Algebra Level 3

How many ordered pairs of real numbers ( x , y ) (x,y) satisfy the following system of equations? x + 3 y = 3 x y = 1 \begin{aligned} x + 3y &= 3 \\ \big | \, |x| - |y| \, \large | &= 1 \end{aligned}


Next problem

Previous problem

All problems

2 4 1 8 3

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Zain Majumder
Feb 11, 2018

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

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

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

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

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

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

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

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

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

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

There are clearly 3 3 solutions to the equation.

Kevin Tran
Feb 11, 2018

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

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...