Solve for $x$ and $y$ given $xy + x^2 + y^2 = 109$ and $x^2-y^2=24$

Question

I was able to solve this graphically but algebraically I'm lost. The two equations are:

1. $xy + x^2 + y^2 = 109$
2. $x^2-y^2=24$

What I've tried: substitute equation (2) into (1) to get:

$$ 2y^2+xy-85=0 $$ $$ y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, a=2, b=x, c=-85 $$ $$ y=\frac{-x\pm\sqrt{y^2+704}}{4} $$

at which point I'm lost. Plugging $x=\pm\sqrt{y^2+24}$ into the equation above just leaves a big mess.

Answer 1

Walkthrough Solution : Assuming $x,y$ to be integers observe that $$x^2-y^2=24$$ $$\Longrightarrow (x+y)(x-y)=2^3.3$$ Now equating each of the factors of $RHS$ one with $(x+y)$ and another with $(x-y)$ we will get $$(x,y)=(\pm 7,\pm 5), (\pm 5, \pm 1)$$ Now by Trial and Error Method with another equation we get $$\boxed{(x,y)=(+7,+5),(-7,-5)}$$ Assuming to be only real solutions we can write $$xy+x^2+y^2=109 \Longrightarrow 2y^2+xy=85 $$$$\Longrightarrow x = \frac{85-2y^2}{y}$$ Putting it in another equation we get $$3y^4-364y^2+7225=0$$ which has solutions $y=\pm 5, \pm \frac{17}{\sqrt{3}}$ . Easy to find the correspondent values of $x$ . Therefore $$(x,y)= \boxed{(+7,+5),(-7,-5),(-\frac{19}{\sqrt{3}},+\frac{17}{\sqrt{3}}) , (+\frac{19}{\sqrt{3}},-\frac{17}{\sqrt{3}}) }$$

Answer 2

Without assuming $x,y$ to be integers, from $2y^2+xy-85=0$ we solve for $x$ to get $x=(85-2y^2)/y$. Plug that into the second of the original equations for $x$ to get $$ \left({85-2y^2\over y}\right)^2-y^2=24 $$ Algebraic manipulations bring this to the form $3y^4-364y^2+7225=0$, if I haven't made any silly mistakes. Now that's a quadratic equation for $y^2$, so apply the quadratic formula (or factor the quartic, if you can), etc., etc.

Answer 3

First it is clear from the second equation that $|x|>|y|$.

Note that the first equation means $$xy+(x+y)^2-2xy=(x+y)^2-xy=109$$ and the second equation means $$(x+y)(x-y)=24$$

Substitute $s_1=x+y$ and $s_0 = xy$, so that the equations become $s_1^2-s_0=109$ and $s_1(x-y)=24$. But now note that $(x-y)^2=s_1^2-4s_0$.

Upon squaring the second equation you have $$24^2=s_1^2(s_1^2-4s_0)=s_1^2(436-3s_1^2)$$

Now consider the monic quadratic equation whose roots are $x$ and $y$, $f(t):= (t-x)(t-y) = t^2 -(x+y)t + xy = t^2-s_1t+s_0$.

Now if we solve for $t$ from $f(t)=0$, we have via the quadratic formula $$t=\frac{s_1 \pm \sqrt{s_1^2-4s_0}}{2}=\frac{s_1^2\pm 24}{2s_1}$$

So if we solve for $s_1$, we are done. This can be done via the quadratic equation we made earlier. We know that $$3s_1^4-436s_1^2+24^2=(3s^2-4)(s^2-144)=0$$ And therefore $s_1^2 = 144 \lor s_1^2 = \frac{4}{3}$. So $s_1\in \{-12,12, -\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\}$.

Putting $s_1=12$, we see that $t=7\lor t=5$, meaning $x=7$ and $y=5$. With $s_1=-12$, we get the same values with the the signs reversed ie $x=-7$ and $y=-5$.

For $s_1=\frac{2}{\sqrt3}$ we get that $x = \frac{19}{\sqrt 3}$ and $y= \frac{-17}{\sqrt 3}$. For $s_1=\frac{-2}{\sqrt 3}$, we get the same $x,y$ with their signs reversed.