Stumpted by Example

[mk43]

I'm reading a book on math and in the first chapter, in a section discussing existence and uniqueness with regard to polynomials, the author presents a formula that looks like this (and although it might be guessed, what this is supposed to illustrate about polynomials and uniqueness isn't important):

[MATH]f(x) = \frac{y_1}{x_1 - x_2}(x - x_2) + \frac{y_2}{x_2 - x_1}(x - x_1)[/MATH]
He follows this immediately with "and simplify with typical algebra to get":

[MATH]f(x) = \frac{x_1 y_2 - x_2 y_1}{x_1 - x_2} + (\frac{y_1 - y_2}{x_1 - x_2})x[/MATH]
Then adding (in order to get on with the point of the discussion there): "Instead of doing all that algebra I..."

Which has me flabbergasted, because I consider myself reasonably well versed in algebra (objective testing backs that up); I even have a small stack of algebra I and II texts and workbooks which at least a couple of times a week I spend a few hours with, cycling through the material to keep it familiar. I've been doing that for a few years, completing one of them at least several times, and there's nothing in those other texts I have been so utterly stumped by -- in fact, as far as I can remember, this is the first time I've ever gone online to look for help.

But when I look at these formulas, I do not see how the second one is a product of "simplifying with typical algebra". I don't see any way to get from one to the other. This resembles it, and is a such a simplification, but I cannot get from there to the second one either:

[MATH]f(x) = \frac{y_1 x - y_1 x_2}{x_1 - x_2} + \frac{y_2 x - y_2 x_1}{x_2 -x_1}[/MATH]

 

[Romsek]

did you notice they flipped the [MATH] x_1[/MATH] and [MATH]x_2[/MATH] in the denominator of the right hand term?

Try simplifying it as

[MATH]f(x) = \dfrac{y_1}{x_1-x_2}(x-x_2) - \dfrac{y_2}{x_1-x_2}(x-x_1)[/MATH]

 

[Steven G]

Can you show us your work trying to go from the first equation to the 2nd equation? That way we can point out your error.
I would 1st get a common denominator and then factor out an x from the terms that have an x.

 

[mk43]

did you notice they flipped the [MATH] x_1[/MATH] and [MATH]x_2[/MATH] in the denominator of the right hand term?

I did not think about flipping it back to get a common denominator, no...had my blinders on pretty well for this. >_<

[MATH]f(x) = \frac{x y_1 - x_2 y_1 - x y_2 + x_1 y_2}{x_1 - x_2} = \frac{x_1 y_2 - x_2 y_1 + x(y_1 - y_2)}{x_1 - x_2} [/MATH]
Thanks much both of you!