How to solve the system of equations $x^2-5xy+6y^2=0$ and $x^2+y^2=45$

Question

So I was again scrolling through Youtube when I came across this video by Cipher proposing the system of equations

$$ \color{white}{f(x)=} \begin{cases} x^2-5xy+6y^2&=0 \\ x^2+y^2&=45 \end{cases} $$

which I thought that I might be able to solve. Here is my attempt at solving the system of equations:$$x^2+y^2-45=x^2-5xy+6y^2$$$$y^2-45=-5xy+6y^2$$$$-5y^2-45=-5xy$$$$5y^2+45=5xy$$$$y^2+9=xy$$$$\frac{y^2+9}{y}=x$$$$\frac{y^4+18y^2+81}{y^2}+y^2-45=\frac{y^4+18y^2+81}{y^2}-\frac{5y^5+90y^3+485y}{y^2}+6y^2$$$$2y^4-27y^2=-5y^5+7y^4+90y^3+18y^2+485y$$$$-5y^5+5y^4+90y^3-9y^2+485y=0$$$$y^4-y^3-18y^2-1.8y-81=0$$Solving for all roots of this quartic equation gets us:$$\text{Real roots: }y=-4.22813\text{, }5.15294$$$$\text{Complex root: }y=0.037595\pm1.92778i$$And plugging all four of these solutions into the equation $x^2+y^2=45$ gets us$$x=\pm5.20797\text{, }\pm4.29502\text{, }\pm6.97962\pm0.00103838i$$Which means the $4$ solutions to the system of equations

$$ \color{white}{f(x)=} \begin{cases} x^2-5xy+6y^2&=0 \\ x^2+y^2&=45 \end{cases} $$

are$$(\pm5.20797,-4.22813)\text{, }(\pm4.29502,5.15294)\text{, and }(\pm6.97962\pm0.00103838i,0.037595\pm1.92778i)$$

My question

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Is my solution correct, or what could I do to attain the correct solution/attain it more easily?

Answer 1

We have $$x^2-5xy+6y^2=\color{blue}{x^2+y^2}-5xy+5y^2=\color{blue}{45}-5xy+5y^2=0.$$

So we can write $$9-xy+y^2=0$$ and solve the equation for $y$ in terms of $x$: $$y=\frac{x\pm\sqrt{x^2-36}}{2}.\tag1$$

Now, for which $x$ values will we produce a corresponding $y$ value? We have $$x^2-5xy+6y^2=(x-3y)(x-2y)=0\implies x=3y\ \text{or}\ x=2y.$$

For $x=2y$, we have in $(1)$: $$y=\frac{2y\pm\sqrt{4y^2-36}}{2}\implies 0=\sqrt{4y^2-36}\implies y=\pm3\implies x=\pm6,$$ and for $x=3y$, we have in $(1)$: $$y=\frac{3y\pm\sqrt{9y^2-36}}{2}\implies 8y^2=36\implies y=\pm\frac3{\sqrt2}\implies x=\pm\frac9{\sqrt2}.$$

Therefore, we have four real solutions: $(6,3), (-6,-3), \left(\frac9{\sqrt2},\frac3{\sqrt2}\right), \left(-\frac9{\sqrt2},-\frac3{\sqrt2}\right)$.

Here's a link to the two equations plotted on the same graph.

Answer 2

Since $xy\neq 0$, we can use that for $y\neq 0$ from the first equation

$$x^2-5xy+6y^2=0 \implies \left(\frac x y\right)^2-5\,\frac x y+6=0 \implies \frac xy =2\;\lor \;\frac xy =3$$

therefore from the second one using $x=2y\;\lor\; x=3y$ we obtain

$$x^2+y^2=45 \implies 5y^2=45 \;\lor \;10y^2=45$$

from which we can easily conclude.

Answer 3

GEOMETRIC APPROACH

Since $x^2-5xy+6y^2=(x-2y)(x-3y),$ the first equation is a common equation of two lines $$x-2y=0, \quad x-3y=0.$$ Each line meets the circle given by $$x^2+y^2=45 $$ in two points, because the lines pass through origin and the circle is centered in origin.
It suffices to upload $x=2y$ or $x=3y$ in $x^2+y^2=45 $ to obtain corresponding values of $y$ and then compute $x.$

Answer 4

Your approach would work, but your substitution of $ \ x \ = \frac{y^2 \ + \ 9}{y} \ $ into $ \ x^2 + y^2 - 45 \ = \ 0 \ $ should have produced $$ (y^4 + 18y^2 + 81) \ + \ y^4 \ - \ 45y^2 \ \ = \ \ 2y^4 \ - 27y^2 \ + \ 81 $$ $$ = \ \ (2y^2 - 9)·(y^2 - 9) \ \ = \ \ 0 \ \ . $$

Note that the equations $ \ x^2 - 5xy + 6y^2 \ = \ 0 \ $ and $ \ x^2 + y^2 \ = \ 45 \ $ are unaffected (invariant) by changing the signs of both $ \ x \ $ and $ \ y \ . \ $ So if a ordered pair (or "point") $ \ (x \ , \ y) \ $ is a solution to the pair of equations, so is $ \ (-x \ , \ -y) \ \ . \ $ We see this in the even symmetry of the quartic polynomial above. (This provides a hint that something is not right about the numerical results you show.)

user376343 presents one geometric approach to the system of equations ($ \ x^2 - 5xy + 6y^2 \ = \ 0 \ $ being the equation of a "degenerate" conic section). Another would be to use the symmetry of the solutions about the origin to say that any pair of solutions lie on a line $ \ y \ = \ mx \ $ through the origin. We then have $$ x^2 \ - \ 5x·mx \ + \ 6·(mx)^2 \ \ = \ \ (6m^2 \ - \ 5m \ + \ 1)·x^2 $$ $$ = \ \ 6·\left( m \ - \ \frac13 \right)·\left( m \ - \ \frac12 \right)·x^2 \ \ = \ \ 0 \ \ . $$ [As we should expect, this is consistent with the factorization user376343 and Andrew Chin show.]

Since $ \ x \ = \ 0 \ $ does not lead to a consistent result for both equations, we only need to use the two line slopes in the equation for the circle to obtain

$$ \mathbf{m \ = \ \frac13 \ \ : } \quad x^2·\left(1 \ + \ \left[\frac13 \right]^2 \right) \ \ = \ \ 45 \ \ \Rightarrow \ \ x^2 \ \ = \ \ \frac{9·45}{10} $$ $$ \Rightarrow \ \ x \ = \ \pm \frac{9}{\sqrt2} \ \ , \ \ \ y \ = \ \frac13·x \ = \ \pm \frac{3}{\sqrt2} \ \ ; $$

$$ \mathbf{m \ = \ \frac12 \ \ : } \quad x^2·\left(1 \ + \ \left[\frac12 \right]^2 \right) \ \ = \ \ 45 \ \ \Rightarrow \ \ x^2 \ \ = \ \ \frac{4·45}{5} $$ $$ \Rightarrow \ \ x \ = \ \pm 6 \ \ , \ \ \ y \ = \ \frac12·x \ = \ \pm 3 \ \ . $$