Solve $x^2+4y^2+80 = 15x+30y \\ xy=6 $

Question

Solve for $x$ and $y$:

$$x^2+4y^2+80 = 15x+30y \\ xy=6 $$

I have no idea how to solve this. I tried setting $x=\frac{6}{y}$ and plugging in, but all I get is this:

$(\frac{6}{y})^2+4y^2+80 = 15(\frac{6}{y})+30y$

$\frac{36}{y^2}+4y^2+80 = \frac{90}{y}+30y $

$\dfrac{36+4y^4+80y^2}{y^2}=\dfrac{90+30y^2}{y}$

$36+4y^4+80y^2= 90y+30y^3$

$4y^4-30y^3+80y^2-90y+36=0$

This is probably even harder to solve than the original equation. I think there's some trick to this type of question, but I don't see/know it.

I also tried to simplify this way: $x(x-15)+2y(2y-15)+80=0$

But alas the brackets don't equate each other so I can't go on. Thanks for the help.

Answer 1

Well, you can write the first equation as $$(x+2y)^2+(80-4xy)=15(x+2y)$$ $$\iff (x+2y)^2-15(x+2y) + 56=0$$ $$\implies x+2y \in \{7, 8\}$$ Now for each of those cases, use $y = \dfrac6x$ to solve the resulting quadratic equations.

Answer 2

$$x^2+4\left(\frac{6}{x}\right)^2+80=15x+30\left(\frac{6}{x}\right)$$ $$\implies x^4-15x^3+80x^2-180x+144=0$$ $$\implies (x-2)(x-3)(x-4)(x-6)=0$$

Answer 3

$x^2+4y^2+80=15x+30y\Rightarrow (x+2y)^2-4xy+80=15(x+2y)\Rightarrow x+2y=7 \enspace\text{or}\enspace 8$

When $x+2y=7$, $x=3\enspace\text{or}\enspace 4$; When $x+2y=8$, $x=2\enspace\text{or}\enspace 6$

So the solution is $$ \begin{cases} x=3\\ y=2 \end{cases} \enspace\text{or}\enspace \begin{cases} x=4\\ y=\dfrac{3}{2} \end{cases} \enspace\text{or}\enspace \begin{cases} x=2\\ y=3 \end{cases} \enspace\text{or}\enspace \begin{cases} x=6\\ y=1 \end{cases} $$

Answer 4

Just for interest (I cannot put a figure in the comment), the first equation is that of an ellipse, the second a hyperbola. They seem to be touching beautifully, but they are not. They have 4 points in common.

Answer 5

$4y^4-30y^3+80y^2-90y+36=0$

This is a "nice" quartic which can actually be solved quite easily in a couple of different ways.

• Use the rational root theorem and you'll find the integer roots $1,2,3\,$, giving:

$$4y^4-30y^3+80y^2-90y+36 = 2 (y - 3) (y - 2) (y - 1) (2 y - 3)$$

• Notice that symmetric coefficients are in a ratio of powers of $3$ i.e. $\dfrac{-90}{-30} = 3, \dfrac{36}{4} = 3^2\,$. This suggests the substitution $z = y + \dfrac{3}{y}$ then $z^2 = y^2 + \dfrac{9}{y^2} + 6$. After dividing the original equation by $y^2 \ne 0\,$ this gives a simple quadratic in $z\,$: $$ 0 = 4\left(y^2 + \frac{9}{y^2}\right) - 30 \left(y+\frac{3}{y}\right)+80 = 4\left(z^2 - 6\right) - 30 z + 80 = 4 z^2 -30 z + 56 $$ Solving the quadratic $y + \dfrac{3}{y} = z_k$ for each of the roots $z_{1,2} = \dfrac{7}{2}, 4$ gives the roots $y_{1,2,3,4}$.

Answer 6

A method suggested by user376343's posting:

The solution of this system of simultaneous non-linear equations is equivalent to determining the intersections of the rectangular hyperbola $ \ xy \ = \ 6 \ $ and the ellipse $$ x^2 \ - \ 15x \ + \ 4y^2 \ - \ 30y \ \ = \ \ - 80 $$ $$ \rightarrow \ \ \left(x \ - \ \frac{15}{2} \right)^2 \ + \ \ 4·\left(y \ - \ \frac{15}{4} \right)^2 \ \ = \ \ - 80 \ + \ \frac{225}{4} \ + \ 4·\frac{225}{16} \ \ = \ \ \frac{130}{4} \ \ . $$

Taking advantage of the relationship among the coefficients of the ellipse equation, we can perform a "vertical stretch" on the curves, writing them as $$ x^2 \ - \ 15x \ + \ (2y)^2 \ - \ 15·(2y) \ \ = \ \ - 80 \ \ \rightarrow \ \ x^2 \ - \ 15x \ + \ \mathcal{Y}^2 \ - \ 15\mathcal{Y} \ \ = \ \ - 80 $$ $$ \rightarrow \ \ \left(x \ - \ \frac{15}{2} \right)^2 \ + \ \left(\mathcal{Y} \ - \ \frac{15}{2} \right)^2 \ \ = \ \ \frac{130}{4} \ \ $$
and $ \ x·(2y) \ = \ 2·6 \ \rightarrow \ \mathcal{Y} \ = \ \frac{12}{x} \ \ . \ $

We make this transformation to convert the ellipse into a circle: in this way, we can locate the intersections of transformed ellipse and transformed hyperbola by finding the points on the hyperbola that lie at a distance of $ \ \frac{\sqrt{130}}{2} \ $ from $ \ \left( \ \frac{15}{2} \ , \ \frac{15}{2} \ \right) \ \ . \ $ Thus, the "distance-squared" formula (really, the same as the circle equation) produces

$$ \left(x \ - \ \frac{15}{2} \right)^2 \ + \ \left(\frac{12}{x} \ - \ \frac{15}{2} \right)^2 \ \ = \ \ \frac{130}{4} $$ $$ \rightarrow \ \ x^2 \ - \ 15x \ + \ \frac{225}{4} \ + \ \frac{144}{x^2} \ - \ \frac{180}{x} \ + \ \frac{225}{4} \ - \ \frac{130}{4} \ \ = \ \ 0 $$ $$ \rightarrow \ \ x^4 \ - \ 15x^3 \ + \ 80x^2 \ - \ 180x \ + \ 144 \ \ = \ \ 0 \ \ , \ \ x \ \neq \ 0 \ \ , $$ the quartic we've seen elsewhere. The Rule of Signs tells us that this polynomial has an even number of positive zeroes (four, two, or none) but no negative zeroes (as we would expect since the ellipse lies entirely in the first quadrant). The polynomial proves to have rational zeroes and factors as $$ x^4 \ - \ 15x^3 \ + \ 80x^2 \ - \ 180x \ + \ 144 \ \ = \ \ (x - 2)·(x^3 \ - \ 13x^2 \ + \ 54x \ - \ 72) $$ $$ = \ \ (x - 2)·(x - 3)·(x^2 \ - \ 10x \ + \ 24 ) \ \ = \ \ (x - 2)·(x - 3)·(x - 4)·(x - 6) \ \ , $$ the $ \ x-$coordinates of the four intersections seen in user376343's graph. Returning to the original system of equations, its four solutions are then $ \ (2 \ , \ 3) \ \ , \ \ (3 \ , \ 2) \ \ , \ \ \left( 4 \ , \ \frac32 \right) \ \ , \ \ (6 \ , \ 1 ) \ \ . $

$$ \ \ $$

It may seem surprising that the first-quadrant branch of the hyperbola has so many intersections with the ellipse. In applying the formula for curvature of a single-variable function curve, $ \ \kappa(x) \ \ = \ \ \large{ \frac{| \ y'' \ |}{( 1 \ + \ y'^{ \ 2})^{3/2}} } \ \ , \ $ we find for the transformed hyperbola,
$$ y' \ = \ -\frac{12}{x^2} \ \ \ , \ \ \ y'' \ = \ \frac{24}{x^3} \ \ \Rightarrow \ \ \kappa_{hyp}(x) \ \ = \ \ \frac{ 24·x^3}{( x^4 \ + \ 144)^{3/2}} \ \ $$ while the curvature of the transformed ellipse is just the reciprocal of the circle's radius, $ \ \kappa_{circ} \ \ = \ \ \frac{ 2}{\sqrt{130}} \ \approx \ 0.1754 \ \ . $ We see in the graph below that the curvature of the two curves are fairly close in the interval $ \ 1.5 \ \le \ x \ \le \ 4.5 \ \ , \ $ with that of the hyperbola being larger or smaller than that of the circle over this interval (tending to zero as it approaches its vertical and horizontal asymptotes). A "close-up" of the transformed circle [in blue] and the transformed hyperbola [in red] is also displayed.