I am stumped by this. How do I solve algebraically, without graphing?
I've tried to solve for x in the second equation and substitute into the first. I'm not sure how else to do this. It just ends up being circular. xx+yy = 3xy or x/y + y/x = 3 Ultimately I figured it out by sketching a quick graph and noticing there must be two solutions, and there was no option with two solutions. I am suspect that there is a quicker/better way to do this problem which I am expected to know and understand.
Solve the system:
$ x^2 + y^2 = 15 $
$ xy = 5 $
$ y < 0 $
$\;\;$None of these
Notice that you can add and substract $2xy$ from the first equation: $$x^2+y^2 = 15 \Longrightarrow x^2+2xy+y^2-2xy=15 \Longrightarrow (x+y)^2-2xy = 15$$ And since $xy=5 \Longrightarrow (x+y)^2=25 \Longrightarrow (x+y) =5 $ Or, $(x+y)=-5$.
Assume $(x+y)=5,$ then $x=5-y$, we can set it in the second equation to get: $$(5-y)y=5 \Longrightarrow -y^2+5y+5=0$$ There are 2 real positve solutions to this equation, and since $y<0$ it is not a solution to the equation.
Let's try $(x+y)=-5$, in the same way: $$x=-5-y \Longrightarrow-y^2-5y-5=0 \iff y=-\frac{5+\sqrt{5}}{2} \space \space OR\space \space -\frac{5-\sqrt{5}}{2}$$ Both solutions satisfy $y<0$ and thus they are both solutions. We can set this values in the 2nd equation to get that the solutions are: $$\left(-\frac{5-\sqrt{5}}{2},-\frac{5+\sqrt{5}}{2}\right),\left(-\frac{5+\sqrt{5}}{2} , -\frac{5-\sqrt{5}}{2}\right)$$