Is \frac{-1}{x_0^2} (x - x_0) Equivalent to \frac{-x}{x_0^2} + \frac{1}{x_0}?

[username12345]

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

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

After that I multiply to get \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}.

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

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

 

[tiny-tim]

Hi username12345!

ooh, that's very complicated!

just write 1/x₀ = x₀/x₀² ​

 

[The Bob]

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 \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}.

The Bob

 

[username12345]

Sorry, that was a typo, should have been + \frac{x_0}{x_0^2}. Updated first post.

 

[The Bob]

Sorry, that was a typo, should have been + \frac{x_0}{x_0^2}. Updated first post.

That's cool. So do you see how the two are equated now?

The Bob