Inequality. Find Relationship between x and y

Question

I am trying to determine relationship between $x$ and $y$ based on given equation. $x+y=-\frac{1}{10}$ and $xy=-\frac{1}{5}$.

I want answer in a way means whether $x<y$, $x>y$, $x=y$ or relationship between $x$ and $y$ cannot be determined.

What I did from the above equation $xy=-\frac{1}{5}$ I found out $y=-\frac{1}{5x}$. I place the value of $y$ in $x+y=104$ and got equation $10x^2+x-2=0$. But I got stuck here.

Answer 1

Look closely at your system of equations and you can instantly spot that the relationship cannot be determined. The reason is thus: $x$ and $y$ are transposable in your system. Notice: $$y+x=-\frac{1}{10};yx=-\frac 15$$ To see what I mean, compare with a system like: $$x+3y=5$$ $$x^2-y=2$$ In this system, you can't swap $x$ and $y$ freely because they are being treated in different ways, but in your system, exactly the same thing is happening to each. Thus it is impossible to say that one will be larger then the other.

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Allow me to also showcase this algebraically

You have: $$10x^2+x-2=0$$ Applying the quadratic formula we get: $$x=\frac{-1\pm\sqrt{1+4(10)(2)}}{20}=\frac{-1\pm\sqrt{81}}{20}=\frac{-1\pm9}{20}=\frac 25, -\frac 12$$ Now checking $y=\frac{-1}{5x}$ for each of these we see $y=-\frac 12, \frac 25$ for the respective $x$.

Note the two pairs of solutions $(x,y)$ $$(\frac 25, -\frac 12)$$ $$(-\frac 12, \frac 25)$$ The swapping of the numbers is inherited from the transposability feature mentioned earlier.

Answer 2

HINT

Recall that by the given condition $x$ and $y$ are the solutions of the quadratic equation

$$t^2+\frac1{10}t-\frac15=0$$

indeed

$$(t-x)(t-y)=t^2-(x+y)t+xy=0$$

Answer 3

$$x+y=-\dfrac{1}{10}$$

$$xy = -\dfrac 15$$

This may not be the quickest way to find the solution, but I always liked the symmetry of it.

\begin{align} x^2 + 2xy + y^2 &= \dfrac{1}{100} \\ -4xy &= \dfrac 45 \\ \hline x^2 - 2xy + y^2 &= \dfrac{81}{100} \\ \hline x-y &= \pm \dfrac{9}{10} \\ x+y &= -\dfrac{1}{10} \\ \hline 2x &= \dfrac{-1 \pm 9}{10} \\ x &\in \left\{ \dfrac 25, -\dfrac 12 \right\} \\ (x,y) \in \left\{ \left(\dfrac 25, -\dfrac 12\right), \left(-\dfrac 12, \dfrac 25\right) \right\} \end{align}