# A simple system of equations

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.

• You excluded the case $x=1$ in your workings. So you have to go back and check that too. Sep 29, 2014 at 11:58
• Thank you, @MarkBennet! So, when I get an $x≠a$ condition, where $a$ is some number, in one of the equasions of the system, I should plug it into all other equasions, calculate the second variable, then check on all the equations if this $(x, y)$ combination solves them all? Sep 29, 2014 at 12:47

By saying "provided that $x \neq 1$", after getting a solution you have to go back to try out $x=1$.
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 $x=1$.
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. :-)
• Great, I see you're using Vieta's formulas in an imaginative way! (0: I will try to follow this route. As I understand, $u_1, u_2$ in this quadratic equation will equal $x, y$. I'm too dense to understand the line "since this is what you get when expanding $(u-x)(u-y)=0.$" though. (0: Sep 29, 2014 at 14:02
• I solved the quadratic, getting the result $(1; -2)$. How will I get (or logically derive) the second result, $(-2; 1)$, I wonder. Sep 29, 2014 at 14:43
• @CopperKettle: Your two equations are both symmetrical in x and y, so if $(a,b)$ is a solution, then so is $(b,a)$. As for your other remark, just open up the parentheses, and see what you get. Sep 29, 2014 at 15:20