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For example, $x+y=x\times y$ is easy to express as $y=\frac{x}{x-1}$, how about $x-y=\frac{x}{y}$?

I tried multiply both sides by $y$ and become

$y^2-xy+x=0$

but up to this step I don't know how to continue, can anyone help?

BLAZE
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ggrr
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2 Answers2

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Solve for y in quadratic equation

$y^2-xy+x=0$

Using quadratic formula.

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It cannot be done, because, as you correctly noticed, the expression (if $y\neq 0$) is equivalent to the expression

$$y^2 - xy + x = 0$$

This expression can have, for a fixed value of $x$, two solutions, one solution, or even no real solution.

For example, if $x=1$, then there is no real value of $y$ that satisfies the equality, but if $x=-1$, there are two such values of $y$.

5xum
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