Solve Fractions Problem: Slope at (1,3)

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The discussion centers on calculating the slope of the derivative y' = [3(y-x)^2 - 2x]/[3(y-x)^2] at the point (1,3). The initial calculation yields a slope of 5/6. A participant suggests simplifying the derivative to y' = -2x/[3(y-x)]^2, which would result in a slope of -1/6. The confusion arises from the incorrect assumption that the term [3(y-x)^2] can be canceled out, which is clarified by the hint that $$\frac{a}{a}=1$$ where $$a\ne0$$.

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I have this derivative and I need the slope at (1,3).

y' = [3(y-x)^2 -2x]/[3(y-x)^2]

With this equation I plug in x and y and the slope equals 5/6.

However, can't y' be simplified further to:

y' = [3(y-x)^2]/[3(y-x)^2] -2x/[3(y-x)^2] ?

Thus can't it be simplified to:

y'= -2x/[3(y-x)]^2

thus the slope would be -1/6 when I plug in x and y.What am I doing wrong?

Thanks for all your help!
 
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Hint: $$\frac{a}{a}=1$$ where $$a\ne0$$

You are trying to say $$\frac{a}{a}=0$$.
 
Hopefully this helps.

EDIT: Fixed Image
9wew.png

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