How did they go from 3 to 4?

[warwick]

I have a question regarding this worked example:

How did they go from step 3 to step 4?
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Edited by stapel -- Reason for edit: Moving content from subject line to post.

 

[stapel]

Hint: Different of squares. :wink:

Eliz.

 

[Deleted member 4993]

In other words remember

a^2 - b^2 = (a+b)(a-b)

 

[warwick]

So, the x sub 2 minus x sub 1 term was canceled in the denominator and twice in the numerator?

 

[Deleted member 4993]

So, the x sub 2 minus x sub 1 term was canceled in the denominator and twice in the numerator?

No...

First the (x_2 - x_1) was factored out in the numerator.

Then "it" was "canceled" from the numerator and the denominator.

You really need to review your algebra - before you get into this calculus business.

Go to:

http://www.purplemath.com/modules/index.htm

And review thoroughly.

 

[warwick]

So, the x sub 2 minus x sub 1 term was canceled in the denominator and twice in the numerator?

No...

First the (x_2 - x_1) was factored out in the numerator.

Then "it" was "canceled" from the numerator and the denominator.

You really need to review your algebra - before you get into this calculus business.

Go to:

http://www.purplemath.com/modules/index.htm

And review thoroughly.

My algebra is fine. What I said happened AFTER the difference of two squares was factored in the numerator. I just didn't write that it WAS factored first.

 

[galactus]

Let's use x and y instead of subscripts.

\(\displaystyle \L\\\frac{A(x^{2}-y^{2})+B(x-y)}{x-y}\)
.......\(\displaystyle \uparrow\)
\(\displaystyle \L\\\frac{A\overbrace{(x-y)(x+y)}^{\text{difference of\\2 squares}}-B(x-y)}{x-y}\)

\(\displaystyle \L\\A(x+y)-B\)