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Ok, this is really easy and it's getting me crazy because I forgot nearly everything I knew about maths!

I've been trying to solve this equation and I can't seem to find a way out. I need to find out when the following equation is valid:

$\frac{1}{x} - \frac{1}{y} = \frac{1}{x-y}$

Well, $x \not= 0$, $y \not= 0$, and $x \not= y$ but that's not enough I suppose.

The first thing I did was passing everything to the left side: $\frac{x-y}{x} - \frac{x-y}{y} - 1 = 0$

Removing the fraction: xy - y² - x² + xy - xy = 0xy

But then I get stuck.. - y² - x² + xy = 0

How can I know when the above function is valid?

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    only real numbers2012-02-20

3 Answers 3

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Your work goes a long way towards the answer. I will assume that you are looking for solutions in real numbers $x$ and $y$. You want to solve $x^2-xy+y^2=0$. Note that $x^2-xy+y^2=\left(x-\frac{y}{2}\right)^2 +\frac{3}{4}y^2.\qquad(\ast)$ The above result is easy to verify by expanding the right-hand side. But it was not obtained by magic: It is a standard application of the powerful idea usually called Completing the Square. You have undoubtedly met this idea earlier in other contexts.

On the right-hand side of $(\ast)$ we have a square, namely $\left(x-\frac{y}{2}\right)^2$, plus $3/4$ of $y^2$. The square of the real number $y$ is always $\ge 0$. So the only way we can satisfy the equation $x^2-xy+y^2=0$ is by taking $y=0$ and $x-y/2=0$, meaning that $x=0$. These values are, as you pointed out, forbidden, so the original equation has no real solutions.

Remark: Through unhappy experiences, I have somewhat of an aversion to fractions, so would prefer to say that the equation $x^2-xy+y^2=0$ is equivalent to $4x^2-4xy+4y^2=0$. But $4x^2-4xy+4y^2=(2x-y)^2+3y^2.$ Or else, if we feel bad about breaking symmetry, we can avoid completing the square, and instead note that $2x^2-2xy+2y^2=(x-y)^2+x^2+y^2.$ Again, we have a sum of squares on the right, and this can be $0$ only if $x$, $y$ (and therefore $x-y$) are all $0$.

Much more mechanically, we can use the Quadratic Formula. For any fixed $y$, the solutions of $x^2-xy+y^2=0$ are $x=\frac{y \pm\sqrt{-3y^2}}{2}.$ If $y\ne 0$, the number under the square root sign is negative, and therefore $\sqrt{-3y^2}$ is not a real number, so $x$ is not a real number.

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    @Schiavini: You are welcome. You will see completing the square come up in fancy disguises in various branches of mathematics.2012-02-21
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$y=x\left(\dfrac{1 \pm\sqrt{-3}}{2}\right)$, combined with $0 \not = y\not =x\not = 0$, answers the question "when the following equation is valid". The first statement is equivalent to $x=y\left(\dfrac{1 \mp \sqrt{-3}}{2}\right)$.

It is valid for all pairs of complex numbers with this property; it is not valid for any pair of real numbers.

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$x^2-xy+y^2=(x+jy)(x+j^2y)$ so $x=y(1+\sqrt{-3})/2$

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    Welcome to Math.SE! If you plan to participate, the in-house [MathJax Tutorial an$d$ Quick Reference](http://meta.math.stackexchange.com/questions/5$0$20/mathjax-basic-tutorial-a$n$d-quick-refere$n$ce) may be a ha$n$dy place to get started with posting and editing math formulas. Often a one-line Answer as you've given requires some additional words to make the solution clear for future Readers.2014-10-17