Can someone tell me if I did the following question right and if not what I did wrong?

1-1

x + h x

______________

h

Answer:1 - h

x +h

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- Nov 6th 2007, 11:52 AM #1

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- Nov 6th 2007, 11:58 AM #2

- Nov 6th 2007, 12:01 PM #3

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- Nov 6th 2007, 12:06 PM #4

- Nov 6th 2007, 12:07 PM #5

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- Nov 6th 2007, 12:25 PM #6

- Nov 6th 2007, 12:30 PM #7
Okay, so you have a fraction over a fraction, such as

$\displaystyle \frac{\mbox{ }\frac{a}{b}\mbox{ }}{\frac{c}{d}}$

This is basically saying $\displaystyle \frac{a}{b} \div \frac{c}{d}$

Whenever you divide by a fraction, you are multiplying by it's inverse, so you just rewrite it:

$\displaystyle \frac{a}{b} \times \frac{d}{c}$

Which is then very easy:

$\displaystyle \frac{ad}{bc}$

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So lets say

$\displaystyle \frac{\mbox{ }\frac{1}{2}\mbox{ }}{\frac{3}{4}}$

Then write as division:

$\displaystyle \frac{1}{2} \div \frac{3}{4}$

Then write as multiplying by the inverse:

$\displaystyle \frac{1}{2} \times \frac{4}{3}$

Multiply numerator, multiply denominator

$\displaystyle \frac{1\times4}{2\times3}$

Simplify

$\displaystyle \frac{4}{6}$

Simplify

$\displaystyle \frac{2}{3}$

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So I can't read your equation, but try again using that rule.

- Nov 6th 2007, 12:32 PM #8

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- Nov 6th 2007, 12:44 PM #9
in the post i directed you to, we saw that we can add/subtract fractions in the following way

$\displaystyle \frac ab \pm \frac cd = \frac {ad \pm bc}{bd}$

thus, $\displaystyle \frac 1{x + h} - \frac 1x = \frac {x - (x + h)}{x(x + h)} = \frac {x - x - h}{x(x + h)}$ (since here, a = 1, b = x + h, c = 1 and d = x)