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| Some Calculus |
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| Ace:
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) --- End quote --- 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. |
| Ladios:
Try looking for the answer in the back of your book and working backwards from there? |
| Ace:
--- Quote from: ladios on August 30, 2009, 03:43:00 PM ---Try looking for the answer in the back of your book and working backwards from there? --- End quote --- It's on a worksheet. If it were in the book I would not have posted it here. |
| TheGeek:
--- Code: ---1/a + 1/b = 1/c 1/a = 1/c - 1/b = b/bc - c/bc = (b-c)/bc a = bc/(b-c) --- End code --- * 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:
Ah thank you, it seems surprisingly simple now. I hadn't thought of isolating 1/a first. |
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