Find real points of intersection of the equations algebraically

Question

Can you help me to find pints of intersection of given system of equesions? Can't do it by myself.

$xy + x - 2y + 3 = 0$

$x^2 + 4y^2 - 9 = 0$

Answer 1

The first equation becomes $$x−2y=-3-xy$$ so $$(x-2y)^2=(-3-xy)^2$$ or $$x^2-4xy+4y^2=9+x^2y^2+6xy$$ and thus $$x^2+4y^2-9=xy(xy+10)$$ and the left hand side is equal to $0$, so $$xy(xy+10)=0$$ So either $x=0$, $y=0$, or $xy=-10$.

If $x=0$, then $y=\frac 32$ (by simply solving the first equation)

If $y=0$, then $x=-3$.

If $xy=-10$, then $x=2y+7$ (by the first equation) so, when substituting this into the second equation, we get $$(2y+7)^2+4y^2-9=0$$ so $$8y^2+28y+40=0$$ which has no real solutions, so $xy=-10$ yields no solutions. Thus, the solutions are $$(x,y)\in\{(0,\tfrac 32),(-3,0)\}$$

Answer 2

Rewrite 1st Equation in terms of $y$:

$$y=\frac{-x-3}{x-2}$$

Rewrite 2nd Equation in terms of $y$:

$$y=\sqrt{\frac{9-x^2}{4}}$$

Setting them equal to one another:

$$\frac{-x-3}{x-2}=\sqrt{\frac{9-x^2}{4}}$$

Simplifying:

$$\frac{x^2+6x+9}{x^2-4x+4}={\frac{9-x^2}{4}}$$

$$4x^2+24x+36=9x^2-36x+36-x^4+4x^3-4x^2$$

$$4x^2+24x=9x^2-36x-x^4+4x^3-4x^2$$

$$x^4-4x^3-x^2+54x=0$$

Real solutions from using synthetic division and rational root theorem:

$$x=0, x=-3$$

Case #1: Plugging $x=0$ into the original equation:

$$-2y+3=0$$

$$y=\frac{3}{2}$$

Intersection point #1: $(0,\frac{3}{2})$

Case #2: Plugging in $x=-3$ to the original equation:

$$-3y-2y=0$$

$$y=0$$

Intersection point #2: $(-3,0)$