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