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Why is $y=\frac{x+y}{x}$ the same as $y=\frac{x}{x-1}$?

It seems to be the same on graphing calculators, but I don't get why. To generalize my statement, I would like to know how to simplify equations where "$y$" is on both sides, but can't be removed through conventional methods.

anonymous
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John
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2 Answers2

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Ah, but $y$ can be removed through conventional methods in this case. We have $$y=\frac{x+y}{x}$$ $$xy = x + y$$ $$xy - y = x$$ $$(x-1)y = x$$ $$y = \frac{x}{x-1}$$

mweiss
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  • Thanks, i can't believe I missed that. – John May 30 '16 at 20:46
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    @John After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. – Tacet May 30 '16 at 21:15
  • Assuming x-1 ne 0 and x ne 0. That is one difference. The second is undefined at x = 1 but defined at x = 0 (y = 0). The first has y undefined at x = 0 and has the impossible statement y = y + 1 at x = 1. At all other values they are the same. Likewise at y= 0 the second is fine (x = 0) but and y = 0 in the first you have the unsolvable 0 = x/x. – fleablood May 30 '16 at 21:41
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To answer the question, they are not the same.

The first one is an implicit expression in $y$. It makes no sense for $x=0$. The second one is explicit in $y$, and it is defined for $x=0$. So the graphs are indeed different, since they differ at least in one point.

Yet for many other $x$, their graphs coincide.

A simpler and similar example is $f_1(x) =\frac{x}{x}$ and $f_2 (x)=1$. Almost the same, except for one little detail at $0$.

Your second question is a bit broad. When $y$ is on both sides, there are several methods, depending on how things are involved: variable re-parametrisation, equation solving, higher dimension embedding...

Laurent Duval
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  • As well as x = 0 making no sense y= 0 makes no sense for the first (0 = x/x). The first is makes no sense at x = 1 (y = y+1) and the second makes sense but is undefined. (y = 1/0). At all other x or all other y both equations are solvable and equivalent. – fleablood May 30 '16 at 21:49