# Simplify $\frac{\frac{x+1}{2x}}{\frac{x^2-1}{x}}$

$$\color{white}{\require{cancel}{3}}$$

So recently I have been doing math to see if I could still do simple math, mainly focusing on Algebra. So, I decided to see if I was able to simplify$$\frac{\frac{x+1}{2x}}{\frac{x^2-1}{x}}$$The thing is, I am a little iffy on the answer that I got for it, so I want to verify that my solution is correct. Here is how I got my answer:$$\frac{\frac{x+1}{2x}}{\frac{x^2-1}{x}}$$$$\iff\frac{(x+1)(x)}{(2x)(x^2-1)}$$$$\implies\frac{\cancel{x}(x+1)}{\cancel{x}(2x^2-2)}$$$$\iff\frac{x+1}{2x^2-2}$$$$\iff\frac{x+1}{2(x^2-1)}$$$$\iff\frac{x+1}{x^2-1}\cdot\frac{1}{2}$$$$\implies\frac{\cancel{x+1}}{\cancel{(x+1)}(x-1)}\cdot\frac{1}{2}$$$$\iff\frac{1}{x-1}\cdot\frac{1}{2}$$$$\iff\frac{1}{2(x-1)},\text{ }x\neq-1,0,1$$My question

Is my solution correct, or what could I do to attain the correct solution, or if it is correct, what could I do attain the correct solution more easily?

To clarify

1. Sorry if this question seems trivial/short
2. This is not a duplicate of any of my other questions
3. Sorry if the "algebra-precalculus" tag seems a little out of place, I mean I guess it sort of fits but still.
• One small peice of feedback. It didn't help you to expand the 2 term into the denominator in the 2nd step. You undid it very quickly in the next 1-2 steps. In general I find it better to keep constant factors accumulating separately from the polynomial terms you are focused on Commented Apr 24, 2023 at 14:17
• @jameselmore Sorry, I still make that mistake a lot without realizing, I'll work on that :\ Commented Apr 24, 2023 at 14:20
• @CrSb0001, please do not upvote answers just because they’re posted. Besides the confirmed answer, one answer is plain wrong while the other is incomplete. Commented Apr 24, 2023 at 14:50
• The technique is good but the answer is missing the restrictions for x, which should appear at the end. Commented Apr 24, 2023 at 15:22

The answer is mostly correct; you need to include the restriction that $$x \neq 0,-1$$ for the answer to be complete, since you divide by $$x+1$$ and $$x$$ in arriving at the final expression.

As for reaching the solution more easily, you have small things you could do like not multiplying the $$2$$ in the denominator and then factoring it back out, but these won't change your approach too much, and are things you'll improve at with repetition. The heart of the approach is correct.

edit: Also need to include that $$x \neq 1$$ since the final expression is undefined for this value of $$x$$.

edit: Technically, some of the $$\iff$$ symbols are used correctly, and some are not (such as when you divide out $$x$$'s), but since we only care about going in 'one direction' logically, it's good practice to make use of $$\implies$$ symbols here exclusively.

• Do I need to include the restriction that $x\neq0,-1$ because if I allow $x=0$ I get $\frac{\frac{1}{0}}{\frac{-1}{0}}$, which is undefined, and $x=-1$ because I get $\frac{\frac{0}{-2}}{\frac{0}{-1}}$ which simplifies to $\frac{0}{0}$ which is undefined? Commented Apr 24, 2023 at 14:24
• The reason you include it is because at those values you are dividing by zero. That happens to lead to the indeterminate (not undefined) situation, but the division by zero is the reason you include the restriction here Commented Apr 24, 2023 at 14:52
• This answer looks good, but it should say something about the incorrect use of $\iff$ Commented Apr 24, 2023 at 18:28
• @jjagmath edited Commented Apr 24, 2023 at 18:41

$$\frac{\frac{x+1}{2x}}{\frac{x^2-1}{x}}=\frac{\cancel{\frac{x+1}{x}}\frac{1}{2}}{\cancel{\frac{x+1}{x}}(x-1)}=\frac{1}{2(x-1)}$$ where $$x\neq 0,-1$$ or $$1.$$

• How could you simplify a rational expression by dividing by zero? Commented Apr 24, 2023 at 14:51
• @WindSoul $\frac AA=1$ is not correct? It is better to have no As. Commented Apr 24, 2023 at 15:20
• Bob Dobbs, $\frac AA=1$ is not correct unless $A\ne 0$. Your missed that in your answer. Commented Apr 24, 2023 at 18:06
• @WindSoul Ok. It is in details. Commented Apr 24, 2023 at 18:21

$$\iff$$ means "if and only if". Using it here is inappropriate/doesn't make sense. An example of correct usage of "$$\iff$$" is:

$$x\ \text{is an odd integer} \iff x \text{ is an integer which is not divisible by } 2.$$



"Equals" is a lot more appropriate than "if and only if" here.

You should also mention that the above expressions are only legitimate if we are assuming that $$x$$ cannot equal $$-1,0,$$ or $$1.$$

So suppose $$x\neq -1,0,$$ or $$1.$$ Then we have:

$$\frac{\frac{x+1}{2x}}{\frac{x^2-1}{x}}$$

$$=\frac{(x+1)\cancel{x}}{2(x^2-1)\cancel{x}}$$

$$=\frac{\cancel{x+1}}{2(x-1)\cancel{(x+1)}}$$

$$=\frac{1}{2(x-1)}.$$

• Sorry about that, I'll work on using "$\iff$" correctly Commented Apr 24, 2023 at 14:31
• A lot of students mis-use $\iff.$ You need to expose yourself with lots of correct settings in which $\iff$ is used (correctly). Commented Apr 24, 2023 at 14:33