Library / CEMC Fermat /524 of 566

CEMC Fermat, 2016

Algebra Difficulty 3.6 AMC 10/12 Find the answer Canada

Suppose that $x$ and $y$ satisfy $\dfrac{x-y}{x+y}=9$ and $\dfrac{xy}{x+y}=-60$ .

The value of $(x+y)+(x-y)+xy$ is

Pick one

Solution

The first equation $\dfrac{x-y}{x+y}=9$ gives $x-y = 9x+9y$ and so $-8x = 10y$ or $-4x = 5y$ .

The second equation $\dfrac{xy}{x+y} = -60$ gives $xy = -60x - 60y$ .

Multiplying this equation by 5 gives $5xy = -300x - 300y$ or $x(5y) = -300x - 60(5y)$ .

Since $5y = -4x$ , then $x(-4x) = -300x - 60(-4x)$ or $-4x^2 = -60x$ .

Rearranging, we obtain $4x^2 - 60x = 0$ or $4x(x-15)=0$ .

Therefore, $x = 0$ or $x=15$ .

Since $y = -\frac{4}{5}x$ , then $y = 0$ or $y = -12$ .

From the first equation, it cannot be the case that $x=y=0$ .

We can check that the pair $(x,y)=(15,-12)$ satisfies both equations.

Therefore, $(x+y)+(x-y)+xy = 3 + 27 + (-180) = -150$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.