Show that -this- equals -this-

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We were given a problem to do over the summer that is proving extremely difficult as of right now and I can't figure out how to get started. It's sort of hard to type so bare with me! (And yes, there is three levels to the first side...) We have to decide how to get the right side using the left side equation.

((1 / x+h) - (1 / x)) / h = -1 / (x (x + h))

The one's are the first level or first numerator, the x's and h the second, and the last h is the third or the last denominator.

I've written this problem down a few times now hopnig that if I just write it down then I can help myself through that way but it seems to just get me even more flustered. One of the only things that I've found that helps me get close to the answer is by taking the right side out so it doesn't confuse me and then multiplying the left side by h/1 to get rid of the denominator, but once I get to that I'm stuck again. Can anyone help out in any way?
 
You have \(\displaystyle \L\,\frac{1}{x+h} - \frac{1}{x}\)

Can you not find the Least Common Denominator?

\(\displaystyle \L\,\frac{1}{x+h}*\frac{x}{x} - \frac{1}{x}*\frac{x+h}{x+h}\)

Now what?
 
You would end up getting x/x(x+h) - x+h/x(x+h)

But I don't see how that would help any because they would just end up cancelling each other out once more. And sorry, I'm sort of still in summer mode and not to mention none of my teachers really went over how to work with LCD in problems xD
 
Sadly, your thinking is a product of incorrect teaching. Let's back up a hair. Wherever you learned this rule, "YOU MUST REDUCE ALL FRACTIONS AT ALL TIMES OR YOU WILL FAIL YOUR MATH CLASS!!!!!", simply forget it.

OK, now back to the problem.

We made the fractions more complicated for a reason. Don't undo what was just done. When adding fractions, one establishes a common denominator then adds the numerators. Try adding the numerators.

Note: Please add parentheses to clarify meaning. This: x/x(x+h) - x+h/x(x+h) should be x/x(x+h) - (x+h)/x(x+h)
 
Merlin, I can't believe I went back to that form. My fifth grade teacher set us up on that rule and it was only last year that my teacher told me to forget about it.

So anyways, if you subract the numerators it turns out:

h / (x(x+h)) but then that's still all over h, correct? Is it possible to cancel out the h from the upmost numerator and the h from the lowest denominator to recieve the correct form? And if so, how come it should be -1? o_O;; You see what summer does? x.x Ruins everything I thought I knew.
 
Centara said:
...if you subract the numerators it turns out:

h / (x(x+h)) but then that's still all over h, correct?
Click to expand...
The fraction will be "over h" (or, for simplicity, over h/1, which you then can flip and multiply), but how did you get the fraction you did?

The numerator, once combined, was x - (x + h). How did you get that x - x - h = +h?

Thank you.

Eliz.
 
Lmao, good question xD I think I disregarded the parenthesis... which is a quite common mistake by me.

But that still gets me stuck, even though I've been working through this packet all day with my friend helping me out over IM's... Sorry to be such a burden =/
 
Hello, Centara!

Do you how to "clear" complex fractions?

\(\displaystyle \L\;\frac{\frac{1}{x+h}\,-\,\frac{1}{x}}{h} \;= \;\frac{-1}{x(x+h)}\)
Click to expand...

The LCD of all the denominators is: \(\displaystyle x(x+h)\)
Multiply top and bottom by the LCD . . . and reduce (cancel):

\(\displaystyle \L\;\;\frac{x(x+h)}{x(x+h)}\,\cdot\,\frac{\frac{1}{x+h}\,-\,\frac{1}{x}}{h} \;= \;\frac{x(\sout{x+h})\cdot\left(\frac{1}{\sout{x+h}}\right) \,-\,\not{x}(x+h)\cdot\left(\frac{1}{\not{x}}\right)}{xh(x+h)} \;=\;\frac{x\,-\,(x\,+\,h)}{xh(x\,+\,h)}\)

\(\displaystyle \L\;\;\;= \;\frac{x\,-\,x\,-\,h}{xh(x\,+\,h)}\;=\; \frac{-\not{h}}{x\not{h}(x\,+\,h)} \;=\;\frac{-1}{x(x\,+\,h)}\)

 
Thanks for all the help guys! I attempted the first way and got to the point where soroban came at and then attempted their way and worked it out that way by myself and got the correct answer, and I know how to do it now so it works out wonderfully! I just have one question though, how did you come up with the LCD being x(x+h)? From the final form of the equation?
 
Hello, Centara!

I assume that you know how to find an LCD . . .


We have: \(\displaystyle \L\;\frac{\frac{1}{x+h}\,-\,\frac{1}{x}}{\frac{h}{1}}\)

The denominators are: \(\displaystyle \,x+h,\;x,\) and \(\displaystyle 1.\)

The LCD of all the denominators is: \(\displaystyle x(x+h)\)

Got it?

 
Yes ^^ That makes sense! Thanks a lot guys, and you might be hearing a bit more from me in the future. Stupid AP calc summer work is more tricky then I originally imagined.
 
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