An algebra problem by Nguyễn Hữu Khánh

$\begin {cases} y^2 + x + xy - 6y +1 = 0\\ y^3x - 8y^2 + x^2y + x = 0 \end {cases} \quad \Rightarrow \begin {cases} (y^2 + x) + (xy +1) - 6y = 0 \\ (y^2 + x)( xy + 1) - 9y^2 = 0 \end {cases}$

$\Rightarrow \begin {cases} (y^2 + x) + (xy +1) = 6y \\ (y^2 + x)( xy + 1) = 9y^2 \end {cases} \quad \Rightarrow \begin {cases} u + v = 6y \\ uv = 9y^2 \end {cases} \quad \Rightarrow \begin {cases} v = 6y - u\\ u(6y-u) = 9y^2 \end {cases}$

$\Rightarrow u^2 - 6yu + 9y^2 = (u - 3y)^2 = 0\quad \Rightarrow u = 3y\quad \Rightarrow y^2 + x = 3y$

$\Rightarrow x = 3y-y^2$ ; and substituting in $y^2 + x + xy +1 = 6y$ , we have:

$y^2 + 3y-y^2 + (3y-y^2)y +1 - 6y = - 3y + 3y^2 - y^3 + 1 = 0$

$\Rightarrow y^3 - 3y^2 + 3y - 1 = (y-1)^3 = 0 \quad \Rightarrow y = 1 \quad \Rightarrow x = 3(1) - 1^2 = 2$

$\Rightarrow x + y = 2 + 1 = \boxed {3}$

From the equation $y^{2} + x + xy -6y + 1 = 0$ $x = -y^{2}+6y-1)/(1+y)$

Replacing the above value of 'x' in equation $y^{3}x-8y^{2}+x^{2}y+x=0$ and simplifying, we get an equation completely in y: $-y^{6}+6y^{5}-15y^{4}+20y^{3}-15y^{2}+6y-1=0$

Just by a look at the coefficients (coefficients and constant add up to 0), y =1 is one of the solutions. Replace this value in the first equation to get y = 2. And hence the answer.. 3..

Lazy man's solution:

Equate the two equations and get a 0 on the right side:

$y^2 + x + xy - 6y + 1 = y^3x-8y^2+x^2y+x$

$y^2 + x + xy - 6y + 1 - y^3x+8y^2-x^2y-x = 0$

If you just start randomly plugging in values for $y$ (because doing so with $x$ might lead you to need to solve an ugly cubic function), you'll arrive at that of plugging in $1$ for $y$ .

$1+x+x-6+1-x+8-x^2-x=0$

$x^2 = 4 \rightarrow x=\pm2$

So the answer ( $x+y$ ) must be either $2+1=3$ or $1-2=-1$ . Try both of them and see that the first of them works.

Let eq1 be ${y}^{2}+x+xy-6y+1=0$ and eq2 be ${y}^{3}x-8{y}^{2}-{x}^{2}y+x=0$ .

We begin by multiplying eq 1 by $xy$ and subtract eq 2. With a bit of factoring and rearrangement, we get $(x-2)(x-4){y}^{2}=x(1-y).$

Hence we look at the possibilities when $x=0,2,4$ $y=0,1$

Here we find the integer solutions (0,0), (2,1), (4,1). By substituting these solutions into the original system of equations, only (2,1) works. Therefore, the answer is $2+1 =3.$