Solve system of equation

Question

I have an equation which looks like this:

$$\begin{align}3x^2+4x+y^2&=0 \\2xy+2y&=0 \end{align}$$

and I am not able to get all four possible solutions by hand. Maybe someone can help me to solve this problem. Thanks in advance.

Transform the first equation to y=... and then insert it into the second one — mscha, Nov 12 '21 at 17:20

score 1 · Accepted Answer · answered Nov 12 '21 at 17:22

1

Hint; Factor the second equation first: $$2xy+2y=2y(x+1)=0.$$

answered Nov 12 '21 at 17:22

Bernard

175,478

Thank you, got it – mscha Nov 12 '21 at 17:30

score 0 · Answer 2 · answered Nov 12 '21 at 17:25

0

$2xy+2y=0\Rightarrow 2y(x+1)=0$, if $y=0$, then let's try to find a solution from the first equation, $3x^2+4x=0\Rightarrow x(3x+4)=0 \Rightarrow x=0$ or $x = -\frac{4}{3}$.

Now if $x=-1$, we have $3(-1)^2+4(-1)+y^2 = 3-4+y^2=0\Rightarrow y=\pm1$

answered Nov 12 '21 at 17:25

aliberro

408

Thank you, got it – mscha Nov 12 '21 at 17:30

Solve system of equation

2 Answers2