I am having a hard time understanding simplifying equations. Please could someone show me how you get from:
$$(3x^3y^2-1)y'+3x^2y^3=1$$
to:
$$y'=-\frac{3x^2y^3−1}{3x^3y^2−1}$$
when I work on this I get:
$$y'=\frac{1-3x^2y^3}{3x^3y^2−1}$$
Simplification: $(3x^3y^2-1)y'+3x^2y^3=1$ to $y'=-\frac{3x^2y^3−1}{3x^3y^2−1}$.
1
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algebra-precalculus
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1Note that $1 - 3x^2y^3 = -(3x^2 y^3 - 1)$ since minus times minus is plus. – 2017-02-02
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0so my answer was correct? – 2017-02-02
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0Yes the last equation is the same as the next to last equation. – 2017-02-02
1 Answers
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Your question boils down to showing $$ab+c=d\iff b=-\frac{c-d}{a}$$ for $a\neq 0$.
We have
$$\begin{align} ab+c=d &\iff ab=d-c\quad(\text{subtract }c) \\ &\iff ab=-(c-d)\quad(\text{factor out }-1) \\ &\iff b=-\frac{c-d}{a}\quad(\text{dividing by }a\text{ assuming } a\neq 0). \end{align}$$