I'm trying to solve for $Y$ in this equation: $\frac {(X-Y)}{Y} = Z.$ I tried applying some of the answers from other questions but I'm having special trouble with figuring out how to get the $Y$ out of the parenthesis. Thanks!


2 Answers 2


If $Z\not =-1$, then$$\frac{X-Y}{Y}=Z\Rightarrow X-Y=YZ\Rightarrow X=Y(Z+1)\Rightarrow Y=\frac{X}{Z+1}.$$ Note that $X\not=0$ because $Y\not=0$.

If $Z=-1$, then $$\frac{X-Y}{Y}=Z\iff X=Y(Z+1)=0\ \text{and}\ Y\not=0.$$ Hence, $Y$ can be any real number except $0$.

As a result, since $X=0\iff Z=-1$, $$Y=\begin{cases}\frac{X}{Z+1}&\text{if $X\not=0\ \text{and}\ Z\not=-1$}\\\text{any real number except $0$}&\text{if $X=0\ \text{and}\ Z=-1$}\end{cases}$$

  • 1
    $\begingroup$ If $Z+1 = 0 \Rightarrow X=0$ $\endgroup$
    – Frank Vel
    Sep 30, 2014 at 11:47

If a variable is present more than once (the $Y$ in the proposed eq.) then, as a general rule, you should start by removing any fractions and multiplying out any brackets.

Since $Y$ composes the denominator, start by multiplying both sides by $Y$ so that it cancels out on the left: $$ Y\cdot\frac{(X-Y)}{Y}=Z\cdot Y\;\to\;X-Y=Z\cdot Y $$ The next step is to put all the terms with $Y$ on the same side of the equation; it can be done by adding $Y$ to both sides: $$ X-Y+Y=Z\cdot Y+Y\;\to\;X=Z\cdot Y+Y $$ Now the right hand side can be factorized by the common factor $Y$: $$ X=Z\cdot Y+Y=Z\cdot Y+1\cdot Y=Y\cdot(Z+1) $$ Then, by dividing both sides by the factor $(Z+1)$ we isolate our variable: $$ Y=\frac{X}{Z+1} $$ on a last note, think of $(Z+1)$ as one thing by keeping it in the bracket.

As an advice, always do one operation as you solve an equation and always do it to both sides, because more complex equations might involve squaring and square-rooting too and other operations. Concentrate on one passage at a time, make a note on what you're doing (adding $Y$, dividing by $(Z+1)$ et similia) and you should get the solution eventually.

  • $\begingroup$ Err...much better obviously... $\endgroup$
    – Paul
    Sep 30, 2014 at 19:03
  • $\begingroup$ Much better answer in my opinion, explains the method's steps in detail, using the equation as an example. $\endgroup$
    – Alex Barac
    Sep 30, 2014 at 20:25

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