# $\begin{cases} x+xy=3 \\ xy^2+xy^3=12 \end{cases}$

I should solve the following system: $$\begin{cases} x+xy=3 \\ xy^2+xy^3=12 \end{cases}$$ I should solve by reducing the system to a system of second degree. I am not sure if the term in English is "reduce" a system to a lower degree.

We can factor the equations in this way: $$\begin{cases} x(1+y)=3 \\ xy^2(1+y)=12 \end{cases}.$$

What can I do next?

• ....Divide these equations ? $($Assuming $y \ne -1$$.)$ – The Demonix _ Hermit Jan 7 at 17:08
• I should solve by reducing the system to a system of second degree. I am not sure if the term in English is "reduce" a system to a lower degree. – Nikol Dimitrova Jan 7 at 17:09
• After dividing , you get $y^2 = 4$ , which is in Second degree. – The Demonix _ Hermit Jan 7 at 17:10
• Notice that $x$ may not be $0$. Assuming $y\neq -1$, you may divide the second by the first to obtain that $y^2 = 4$. – Fimpellizieri Jan 7 at 17:10
• Rewrite the second equation as $(x + xy)y^2 = 12$. – anomaly Jan 7 at 17:11

Take 4(1) - (2) to get

$$4x(1+y)-xy^2(1+y)=0$$

and then factorize

$$x(1+y)(4-y^2)=0$$

So, three cases to examine:

Case 1) $$x= 0$$ leads to no solutions.

Case 2) $$1+y = 0$$ does not lead to valid solutions, either.

Case 3) $$4=y^2$$. Substitute $$y=\pm2$$ it into $$x+xy=3$$ to obtain $$x=1,-3$$.

Thus, the valid solutions are

$$(1,2),\>\>\>\>\>(-3,-2)$$

You can assume $$y \neq -1$$ and $$x \neq 0$$.(Because, otherwise, you get a false equality from the first equation). Now you can write
$$\frac{{xy^2 \left( {1 + y} \right)}} {{x\left( {1 + y} \right)}} = \frac{{12}} {3} = 4$$ This means $$y^2=4$$ so that $$y=\pm 2$$. With $$y=2$$, from the first equation you get $$x=1$$, while with $$y=-2$$ you get $$x=-3$$