Multiplication of a fraction

  • #1

Main Question or Discussion Point

Can anyone explain why [tex]\frac{-1}{x_0^2} (x - x_0) = \frac{-x}{x_0^2} + \frac{1}{x_0}[/tex]?

Is [tex]\frac{-1}{x_0^2} (x - x_0) = \frac{-1}{x_0^2} . \frac{(x - x_0)}{1}[/tex]?

After that I multiply to get [tex]\frac{-1}{x_0^2} (x - x_0) = \frac{-x + x_0}{x_0^2} = \frac{-x}{x_0^2} + \frac{x_0}{x_0^2}[/tex].

Then divide [tex]x_0[/tex] into [tex]x_0^2[/tex] which gives [tex]x_0^{-1}[/tex] which equals [tex]\frac{1}{x_0}[/tex].

The equation I am following misses all the intermediate steps so I want to make sure I am understanding it correctly.
 
Last edited:

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
249
Hi username12345! :smile:

ooh, that's very complicated! :eek:

just write 1/x0 = x0/x02 :wink:
 
  • #3
1,100
0
Hey there,

Your ideas are right but, without giving too much away, there is one, small mistake in this line:

After that I multiply to get [tex]\frac{-1}{x_0^2} (x - x_0) = \frac{-x + x_0}{x_0^2} = \frac{-x}{x_0^2} + \frac{-x_0}{x_0^2}[/tex].
The Bob
 
  • #4
Sorry, that was a typo, should have been [tex]+ \frac{x_0}{x_0^2}[/tex]. Updated first post.
 
  • #5
1,100
0
Sorry, that was a typo, should have been [tex]+ \frac{x_0}{x_0^2}[/tex]. Updated first post.
That's cool. So do you see how the two are equated now?

The Bob
 

Related Threads for: Multiplication of a fraction

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
16
Views
3K
Replies
14
Views
4K
Replies
11
Views
4K
  • Last Post
Replies
2
Views
907
Top