Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can I simplify this expression, perhaps into two expressions $\frac{x}{x+y}$ or is that already simplified as much as possible?

share|cite|improve this question
Simplification usually has a particular purpose. A simplification that is optimal for one purpose may be non-optimal for another purpose. – André Nicolas Aug 12 '13 at 16:39
Oddly, if you try to expand this expression using long division, then you end up with the series $1 - 1 + 1 - 1 + 1 ...$ – Nick R Aug 12 '13 at 16:56
@Sean Mill: If you got the answer, please determine it by clicking on Tick sign. (I've another one but I'm in doubt to say!) – RSh Aug 12 '13 at 17:14
Hi @Sean Mill , don't hesitate to post your answer . We shall learn from all the answers . – Harish Kayarohanam Aug 12 '13 at 17:51
up vote 4 down vote accepted

It's simplified as much as possible, though sometimes it's more convenient to write it as


share|cite|improve this answer

Form 1: way of giving as sum of 2 quantities $$ \frac{x+y -y}{x+y} = 1 - \frac{y}{x+y} $$

Form 2 : express as bringing x to denominator , to see expression in terms of y divided by x $$ \displaystyle\frac{1}{1 + \displaystyle\frac{y}{x}} $$

Form 3 : take y out , so that you can view the expression in terms of only x divided by y

$$ \frac{\displaystyle\frac{x}{y}}{\displaystyle\frac{x}{y} + 1 } $$

share|cite|improve this answer

You might want $$\frac{x}{x+y}=\frac{(x+y)-y}{x+y}=1-\frac{y}{x+y}.$$

share|cite|improve this answer

The given expression uses two operations (one division and one addition). If we judge simplicity by the number of operations, only an expression with one operation would be simpler, but the expression equals none of $x+y$, $x-y$, $y-x$, $xy$, $\frac xy$, $\frac yx$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.