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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
- Sorry if this question seems trivial/short
- This is not a duplicate of any of my other questions
- Sorry if the "algebra-precalculus" tag seems a little out of place, I mean I guess it sort of fits but still.
Best Answer
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.