# Is this right?

• November 6th 2007, 11:52 AM
JessicaHoaglund
Is this right?
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
• November 6th 2007, 11:58 AM
Jhevon
Quote:

Originally Posted by JessicaHoaglund
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

no, that is incorrect i'm afraid, and fix your formatting. if you can't use LaTex, type fractions like this: (numerator)/(denominator)

so, for example, you would type your (incorrect) answer like this: (1 - h)/(x + h)
• November 6th 2007, 12:01 PM
JessicaHoaglund
Okay. Sorry I am not trying to make this difficult I just signed up for this an hour ago so I was not aware how to put the answers in correctly. Thank you.
• November 6th 2007, 12:06 PM
Jhevon
Quote:

Originally Posted by JessicaHoaglund
Okay. Sorry I am not trying to make this difficult I just signed up for this an hour ago so I was not aware how to put the answers in correctly. Thank you.

that's fine. did you try again? what's your answer now? begin by combining the fractions in the numerator. do you know how to do that?
• November 6th 2007, 12:07 PM
JessicaHoaglund
Nope. I am honestly really terrible at this!
• November 6th 2007, 12:25 PM
Jhevon
Quote:

Originally Posted by JessicaHoaglund
Nope. I am honestly really terrible at this!

$\frac {\frac 1{x + h} - \frac 1x}h = \frac {\frac {x - x - h}{x(x + h)}}h$

$= \frac {-h}{xh(x + h)}$

$= \frac {-1}{x(x + h)}$

any questions?

post number 5 here may help
• November 6th 2007, 12:30 PM
angel.white
Okay, so you have a fraction over a fraction, such as

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

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

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

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

Which is then very easy:

$\frac{ad}{bc}$

---------

So lets say
$\frac{\mbox{ }\frac{1}{2}\mbox{ }}{\frac{3}{4}}$

Then write as division:
$\frac{1}{2} \div \frac{3}{4}$

Then write as multiplying by the inverse:
$\frac{1}{2} \times \frac{4}{3}$

Multiply numerator, multiply denominator
$\frac{1\times4}{2\times3}$

Simplify
$\frac{4}{6}$

Simplify
$\frac{2}{3}$
-----

So I can't read your equation, but try again using that rule.
• November 6th 2007, 12:32 PM
JessicaHoaglund
I guess I just don't understand the first step:

How did you get

(x-x-h)/x(x+h)

sorry!
• November 6th 2007, 12:44 PM
Jhevon
Quote:

Originally Posted by JessicaHoaglund
I guess I just don't understand the first step:

How did you get

(x-x-h)/x(x+h)

sorry!

in the post i directed you to, we saw that we can add/subtract fractions in the following way

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

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