Solve the replacement system

Question

$ \begin{cases} 2x^2+xy^2-2x^2-xy=8 \\ xy+x+y=9 \\ \end{cases}$

''solution'':

Answer 1

Here's a cleaner approach: we restructure as follows:$$(x+1)(y+1)=10$$$$(2(x+1)+(y+1)-3)\cdot((x+1)(y+1)+(y+1))=8$$Let $v = x+1$, $w=y+1$. We get $$vw=10$$$$w(2v+w-3)(v+1)=8$$We know that $w\neq0$ by the previous equation, so let $v=\frac{10}w$. The second equation becomes $$w(\frac{20}w+w-3)(\frac{10}w+1)=8$$Solve this for $w$, and then solve for $v$ to finish.

Answer 2

From the second equation we get $$x=-\frac{y}{y+1}$$ (The case $y=-1$ must also be considered.) so we get $$2\left(\frac{-y}{y+1}\right)^2+\left(\frac{-y}{y+1}\right)^2y^2-2\left(\frac{-y}{y+1}\right)^2+\frac{y^2}{y+1}=8$$ For $y$ we get $$-{\frac {{y}^{3}-{y}^{2}+8\,y+8}{y+1}}=0$$

Answer 3

$ y=\frac{9-x}{1+x} $ We got from the second equation let's substitute it in the first equality $$\Rightarrow $$ $ x(\frac{x-9}{1-x})^2-x\frac{x-9}{1-x}=8$ $$2x^3-34x^2+56x-8=0 $$ but this equation has 2 solution $\Rightarrow$ we have 2 pair of solution $(x_1;y_1),(x_2,y_2)$