1

I need to express y from $\dfrac{y-2x}{y}=1$, I mean get the $y=...$ equation. But I'm getting 0 instead of y:

$\dfrac{y-2x}{y}=1$

$y=y-2x$

$y-y=-2x$

$0=-2x$

Stdugnd4ikbd
  • 205
  • 1
  • 8
  • 2
    Your steps are correct. From your last equation we can see that $x$ has to be 0. Substituting back into the original equation you get $\frac{y}{y} = 1$, which is true for all $y \in \mathbb{R} \setminus {0}$. – ViktorStein Feb 12 '19 at 20:09
  • @ViktorGlombik Actually, the reason I do this is that I need to make a plot y(x). What should I do? Or it is impossible? – Stdugnd4ikbd Feb 12 '19 at 20:14
  • 1
    This equation simply does not determine $y$ as function of $x$. Think about giving it individual values of $x$. Giving it $x=0$, it returns ALL $y \neq 0$. Given any other $x$ you don't even get a $y$ out. But a function has to assign a (unique) $y$ to each $x$. So $y$ certain isn't a function $y(x)$. – AlephNull Feb 12 '19 at 20:16
  • 2
    Regarding the plot: You would essentially get the $y$-axis. – ViktorStein Feb 12 '19 at 20:18
  • It depends upon how the question was asked. If you are asked to graph $\left{(x,y),|,\frac{y-2x}{y}=1\right}$ then that is the $y$-axis minus the origin. But the equation does not define $y$ as a function of $x$. – John Wayland Bales Feb 12 '19 at 20:33
  • @JohnWaylandBales, what do You mean by "y-axis minus the origin"? – Stdugnd4ikbd Feb 12 '19 at 20:44
  • 1
    I mean $\left{(0,y),|,y\ne0\right}$. – John Wayland Bales Feb 12 '19 at 20:46
  • @JohnWaylandBales, seems, I'm too stupid to understand this. This https://www.wolframalpha.com/input/?i=plot+(y+-+2x)%2Fy+%3D+1 plot corresponds the equation? – Stdugnd4ikbd Feb 12 '19 at 20:49
  • 1
    You only have to understand that if $x=0$ and $y\ne0$ then $\frac{y-2x}{y}=\frac{y}{y}=1$ and that there can be no other kind of solution to the equation. We know that $y$ must not be $0$ and that if both $x\ne0$ and $y\ne0$ that leads to a contradiction of a non-zero number equal to $0$. So the only solutions are $\left{(0,y),|,y\ne0\right}$. – John Wayland Bales Feb 12 '19 at 20:56
  • @JohnWaylandBales, ok – Stdugnd4ikbd Feb 12 '19 at 20:59

0 Answers0