Author Topic: Some Calculus (Read 885 times)

Ace · August 30, 2009, 02:35:13 PM

Well, not exactly, it's only part of my Calculus class, but over the summer I had completely forgotten almost everything I knew about mathematics and I could use a bit of a push. I'm stumped with this problem here:

Quote

Solve for a:

(1/a)+(1/b)=(1/c)

It seems that no matter which way I approach this problem, I always end up with cb=0, completely eliminating a due to one side coming down to "a-a" somehow. Clearly I'm doing something wrong here but I cannot pinpoint what it is.

I could be missing something completely obvious that could solve the problem quickly, who knows.

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Ladios · August 30, 2009, 02:43:00 PM

Try looking for the answer in the back of your book and working backwards from there?

Ace · August 30, 2009, 02:45:52 PM

Quote from: ladios on August 30, 2009, 02:43:00 PM

Try looking for the answer in the back of your book and working backwards from there?

It's on a worksheet. If it were in the book I would not have posted it here.

TheGeek · August 30, 2009, 02:46:48 PM

Code: [Select]

1/a + 1/b = 1/c
1/a = 1/c - 1/b = b/bc - c/bc = (b-c)/bc
a = bc/(b-c)

Get 1/a on the left side of the equation
Change the right side to the same denominator
Add up the right side
Invert both sides of the equation

Ace · August 30, 2009, 02:52:56 PM

Ah thank you, it seems surprisingly simple now. I hadn't thought of isolating 1/a first.

kakashi11 · August 30, 2009, 11:05:39 PM

this is hurting my brain