How to solve this simultaneous equation of implicit equation?

Question

Can someone show me step by step how to solve these simultaneous equations?

$$x^2-y^2-4xy+4x-1=0$$ $$2x^2-2y^2+2xy+4y+2=0$$

Answer 1

Writing your equations with following two functions

$$\begin{gathered} \phi \left( {x,y} \right) = {x^2} - {y^2} - 4 \cdot x \cdot y + 4 \cdot x - 1 \hfill \\ \psi \left( {x,y} \right) = 2 \cdot {x^2} - 2 \cdot {y^2} + 2 \cdot x \cdot y + 4 \cdot y + 2 \hfill \\ \end{gathered} $$

Then we have to found the zeros like so

$$\begin{gathered} \phi \left( {x,y} \right) = 0 \hfill \\ \psi \left( {x,y} \right) = 0 \hfill \\ \end{gathered} $$

Calculating

$$\phi \left( {x,y} \right) - \frac{1}{2} \cdot \psi \left( {x,y} \right) = 0$$

we get

$$ - 5 \cdot x \cdot y + 4 \cdot x - 2 \cdot y - 2 = 0$$

which leads us to

$$y = \frac{{4 \cdot x - 2}}{{5 \cdot x + 2}}$$

Putting this in $$\phi \left( {x,y} \right) = 0$$

we get

$$\phi \left( {x,y} \right) = \phi \left( {x,\frac{{4 \cdot x - 2}}{{5 \cdot x + 2}}} \right) = \frac{{\left( {x + 1} \right)\left( {5x - 1} \right)\left( {5{x^2} + 4x + 8} \right)}}{{{{\left( {5x + 2} \right)}^2}}} = 0$$

Now we see, we have two pairs of real solution

$$\begin{gathered} x = - 1 \Rightarrow y = 2 \hfill \\ x = \frac{1}{5} \Rightarrow y = - \frac{2}{5} \hfill \\ \end{gathered} $$

These coordinates are the intersection points for two hyperboles

And that's all.

Answer 2

A general solution method has been described in various comments: eliminate some terms by fiddling with the terms a bit and taking a linear combination of the equations, solve the resulting equation for one of the variables in terms of the other, then back-substitute into the original equations. With a pair of distinct conic sections as you have here, you’ll get anywhere from 0 to 4 solutions.

Before plunging into this general procedure, it’s often a good idea to look at the specific problem for potential short-cuts or simplifications. In this case, the geometry of the problem gives us a bit of help. We’ll take one of the suggestions immediately and divide the second equation by $2$. The two equations represent hyperbolas, as can be seen by computing their discriminants: $(-4)^2-4\cdot1\cdot(-1)=20$ and $1^2-2\cdot1\cdot(-1)=3$. Now find their centers, which you can do by either remembering the formula for that or computing the gradient and solving for $x$ and $y$: $$\nabla(x^2-y^2-4xy+4x-1)=\langle2x-4y+4,-2y-4x\rangle=0,$$ so $(x,y)=(-2/5,4/5)$. For the second equation, we get $$\nabla(x^2-y^2+xy+2y+1)=\langle2x+y,-2y+x+2\rangle=0$$ and so again $(x,y)=(-2/5,4/5)$. This is a bit of luck as it allows us to eliminate the linear terms in both equations with a translation. Making the substitutions $x\to x'-2/5$ and $y\to y'+4/5$ produces the system $$x'^2-4x'y'-y'^2-\frac95=0 \\ x'^2+x'y'-y'^2+\frac95=0.$$ Note that you don’t have to go though tedious algebraic manipulations to arrive at these equations. Translating the origin to the center of a conic eliminates the linear terms from its general equation while leaving the quadratic terms alone; the new constant term is just the original expression applied to the two offsets. E.g., the constant term of the first equation after translation is $$\left(-\frac25\right)^2-\left(\frac45\right)^2-4\left(-\frac25\right)\left(\frac45\right)+4\left(-\frac25\right)-1=-\frac95.$$ This looks promising. Subtract the first equation from the second to eliminate both $x'^2$ and $y'^2$, giving $$5x'y'+\frac{18}5=0$$ and then solve for either $x'$ or $y'$ as you see fit. Let’s take $y'=-18/25x'$. Substituting this back into either of the translated equations gives $$x'^2+{27\over25}-{324\over625x'^2}=0,$$ a quartic equation in $x'$, but not to worry: all of the exponents are even, so you can treat it as a quadratic equation, which I hope you can solve without further assistance from us (you should get two real and two complex roots).

Once you have solutions to the translated equations, translate them back to get the solutions to the original equations, i.e., add $-2/5$ and $4/5$, respectively, to the values of $x'$ and $y'$ that you’ve computed.