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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?

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

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

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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 $x=-3$

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