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I need to solve the system: $$x^2+2xy+y^2-1 = 0$$ where variable is $x$ AND $$x^2 + 2xy = 0$$ where variable is $y$. From the first Ι take discriminant, and end in one solution $x_1 = 1-y$ and another $x_2 = -1-y$.

I am a little confused on the second one. The variable is $y$ so I can't say (?) this is a quadratic equation. So, I end up where $y = -x/2$ or constant $x$ equals $0$? But even if I take $x$ as variable end in the same solution... But even after this, how do I proceed answering the first question? Thanks a lot.

Jimmy R.
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darkchampionz
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  • $y=\pm1$ then use this fact in the other condition and solve for x. If what you said is the correct question. – Chinny84 Oct 17 '14 at 23:15
  • Is the $x$ on the first equation the same as the $x$ in the second equation? And the same for $y$? I mean, if $x=1$ on the first equation does it also equal $1$ on the second one? – Ioannes Oct 17 '14 at 23:18
  • You could also graph it (if your using a graphing calculator, you'll have to isolate $y$). You'll see that the equations intersect at $(-2,1)$ and $(2,-1)$, or $(\pm 2,\pm1)$ – SameOldNick Oct 17 '14 at 23:37

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Just plug $x^2 + 2xy=0$ into your first equation. You get $y= 1$ or $y=-1$.

Then plug those two into the solutions you got for $x$. So you get $x=-2$ or $x=0$ or $x=2$.

Then evaluate your solutions. (You'll notice for instance that $x=2,\ y=1$ is NOT a solution, but $x=2,\ y=-1$ is.)

got it--thanks
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  • nice, thanks for your answer. but what do I do if it is unpluggable? – darkchampionz Oct 17 '14 at 23:23
  • What do you mean by unpluggable? – got it--thanks Oct 17 '14 at 23:24
  • one cannot exactly fit into the first like this example, like if I had x^2+4xy – darkchampionz Oct 17 '14 at 23:28
  • In general, if you have a system of 2 equations in 2 variables, you'll want to try to solve 1 of them for one variable in terms of the other (for instance, get $y=-\frac x{4}$ in your example). Then plug that expression into the other equation (plug this $y$ into the other equation to get an expression only in $x$'s, then solve) to get some solutions for one of your variables. Then plug those solutions back into the first equation to get the possible solutions to the second variable (get an expression only in $y$'s). Then evaluate each combination of solutions. – got it--thanks Oct 17 '14 at 23:30