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I'm trying to see why my textbook's solution is correct and mine isn't.

"Find an expression in terms of $x$ and $y$ for $\displaystyle \frac{dy}{dx}$, given that $x^2+6x-8y+5y^2=13$

First, the textbook's solution, which I understand and agree with fully: enter image description here

Now my similar solution, for which I don't see my error:

Differential: $$2x+6-8\frac{dy}{dx}+10y\frac{dy}{dx}=0$$ $$\frac{dy}{dx}(10y-8)=-2x-6$$ $$\frac{dy}{dx}=\frac{-2x-6}{10y-8}=\frac{-x-3}{5y-4}$$

So I end up with the negative of the correct solution, because I moved the $(2x+6)$ to the RHS and the textbook author moved the other part instead. I would have thought it would produce an equivalent answer?


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HINT $\rm\displaystyle\ \ \frac{-A}{-B}\ =\ \frac{A}B\qquad$ – Bill Dubuque Jun 4 '11 at 18:14
As you now know, your answer is correct. But there is kind of a convention in writing mathematical expressions, that fewer minus signs is generally better. – André Nicolas Jun 4 '11 at 18:22
@user6312 duly noted, thanks! :) – Danny King Jun 4 '11 at 18:24
up vote 6 down vote accepted

Fractions can be written multiple ways.

$$\frac{-x-3}{(5y-4)} = \frac{x+3}{4-5y}$$

In general $$\frac{a-b}{c-d}=\frac{b-a}{d-c}$$

This is just multiplying both the top and the bottom by −1. In other words, your answer and the books differ by multiplication of $$\frac{-1}{-1} = 1$$

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Oh! When you put it like that, it becomes very clear. Thank you! – Danny King Jun 4 '11 at 18:05

Your answers agree. Note that: $$\frac{-x-3}{5y-4} = \frac{-(x+3)}{-(-5y+4)} = \frac{x+3}{4-5y}.$$

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I don't know, what's wrong with your solution. Notice, that


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