Find the center of symmetry $(x_0,y_0)$ of curve $3x^2+y^2-2x+5y+2=0$.
Deduce that $x_0+y_0$ is one of these : -2.167, -2.104, -2.055, -2.011, -1.953.
I tried to solve this question but I couldn't solve I'm confused.Can you help me how to solve this question?
First of all I wrote $3x^2+y^2-2x+5y+2=0$ under the form 3$(x^2-2x/3)+5(y^2/5+y)=-2$, and at last $3(x-1/3)^2-1/9+5(y^2/5+5)=-2$. But I can't finish it.
If you could get the equation to finally look like $$ p(x-h)^2 + q(y-k)^2 + r = 0 \tag1 $$ where $r < 0$ then you would know you had the equation of an ellipse and you would know where its center is. And if $r > 0$ then you would know you had the equation of a hyperbola and you still would know where its center of symmetry is.
To convert $Ax^2 + By^2 + Cx + Dy + F = 0$ into an equation like $(1)$, we can "complete the square" for $x$ to get the $p(x-h)^2$ term and then complete the square for $y$ to get the $q(y-k)^2$ term. While doing this we get some extra constant terms to add to $F$.
You seem to have the idea how to complete the square for $x$, because you managed to write $3\left(x-\frac13\right)^2$, which looks like $p(x-h)^2$, as desired. You made one error while doing this, because $$ 3\left(x^2-\frac23\right) \neq 3\left(x-\frac13\right)^2 - \frac19 $$ What is actually true is that $$ \left(x^2-\frac23\right) = \left(x-\frac13\right)^2 - \frac19 $$ and therefore $$ 3\left(x^2-\frac23\right) = 3\left(\left(x-\frac13\right)^2 - \frac19\right) = 3\left(x-\frac13\right)^2 - \frac13 $$ But that just affects the value of the constant term (the one I showed as $r$ above), which may affect the shape of your curve but will not affect where its center of symmetry is.
The real source of difficulty for you seems to be how to complete the square for $y$. Notice that when you started to complete the square for $x$, you first took the terms $3x^2 - 2x$ and factored the $3$ out of the $x^2$ term to get $3\left(x^2-\frac23\right)$. And now it is not too hard to see how to apply the formula $x^2 + 2b = (x+b)^2 - b^2$ with $b = -\frac13$. But when you looked at $y^2 + 5y$, for some reason you decided to factor the $5$ out of the $y$ term, rather than use the coefficient of the $y^2$ term. Since the coefficient of the $y^2$ term is $1$, completing the square would have been easier if you had not tried to factor out any coefficients at all!
Try and see what happens if you write $y^2 + 5y$ instead of $5\left(\dfrac{y^2}5 + y\right)$ in the first few steps of your work.