A simple system of equations

Question

I'm trying to refresh my school math knowlegde and have trouble solving a simple system of equations:

$\begin{cases} x + xy + y = -3,\\ x - xy + y = 1. \end{cases}$

I derive $y$ from the second:

$y - xy = 1 - x$

$y(1-x)=(1-x)$

Hence,

$y = \frac {(1-x)}{(1-x)} = 1$, provided that x ≠ 1

Next, I substitute $y=1$ in the first equasion,

$x + x + 1 = -3$; $2x = -4$, $x=-2$.

The answer seems to be $(-2; 1)$.

The problem book, however, also lists a second answer, $(1; -2)$.

I feel that I've done something wrong. Give me a hint, please.

Answer 1

By saying "provided that $x \neq 1$", after getting a solution you have to go back to try out $x=1$.

Answer 2

Note that $$y(1-x)=(1-x)\iff (1-x)(y-1)=0\iff x=1\ \text{or}\ y=1.$$ Since you've already considered the case when $y=1$, you need to consider the case when $x=1$.

Answer 3

Add the two together to get $~2x+2y=-2\iff x+y=-1$. Now replace this value in either one of the initial two equations to get $xy$. And when you know both the Sum and the Product of two numbers, you can determine their values by solving the quadratic $~u^2-su+p=0$, where $s=x+y$ and $p=xy$, since this is what you get when expanding $~(u-x)(u-y)=0.~$ So all that's left to do now is applying the well-known quadratic formula. :-)