Simplification: $(3x^3y^2-1)y'+3x^2y^3=1$ to $y'=-frac{3x^2y^3−1}{3x^3y^2−1}$.












1












$begingroup$


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:



$$y'=frac{1-3x^2y^3}{3x^3y^2−1}$$






algebra-precalculus







share|cite|improve this question











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  • 1




    $begingroup$
    Note that $1 - 3x^2y^3 = -(3x^2 y^3 - 1)$ since minus times minus is plus.
    $endgroup$
    – Winther
    Feb 2 '17 at 22:45












  • $begingroup$
    so my answer was correct?
    $endgroup$
    – Anon
    Feb 2 '17 at 22:46










  • $begingroup$
    Yes the last equation is the same as the next to last equation.
    $endgroup$
    – Winther
    Feb 2 '17 at 22:47
















1












$begingroup$


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:



$$y'=frac{1-3x^2y^3}{3x^3y^2−1}$$






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $1 - 3x^2y^3 = -(3x^2 y^3 - 1)$ since minus times minus is plus.
    $endgroup$
    – Winther
    Feb 2 '17 at 22:45












  • $begingroup$
    so my answer was correct?
    $endgroup$
    – Anon
    Feb 2 '17 at 22:46










  • $begingroup$
    Yes the last equation is the same as the next to last equation.
    $endgroup$
    – Winther
    Feb 2 '17 at 22:47














1












1








1





$begingroup$


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:



$$y'=frac{1-3x^2y^3}{3x^3y^2−1}$$






algebra-precalculus







share|cite|improve this question











$endgroup$




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:



$$y'=frac{1-3x^2y^3}{3x^3y^2−1}$$






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 6:07









Shaun

9,380113684




9,380113684










asked Feb 2 '17 at 22:42









AnonAnon

11211




11211








  • 1




    $begingroup$
    Note that $1 - 3x^2y^3 = -(3x^2 y^3 - 1)$ since minus times minus is plus.
    $endgroup$
    – Winther
    Feb 2 '17 at 22:45












  • $begingroup$
    so my answer was correct?
    $endgroup$
    – Anon
    Feb 2 '17 at 22:46










  • $begingroup$
    Yes the last equation is the same as the next to last equation.
    $endgroup$
    – Winther
    Feb 2 '17 at 22:47














  • 1




    $begingroup$
    Note that $1 - 3x^2y^3 = -(3x^2 y^3 - 1)$ since minus times minus is plus.
    $endgroup$
    – Winther
    Feb 2 '17 at 22:45












  • $begingroup$
    so my answer was correct?
    $endgroup$
    – Anon
    Feb 2 '17 at 22:46










  • $begingroup$
    Yes the last equation is the same as the next to last equation.
    $endgroup$
    – Winther
    Feb 2 '17 at 22:47








1




1




$begingroup$
Note that $1 - 3x^2y^3 = -(3x^2 y^3 - 1)$ since minus times minus is plus.
$endgroup$
– Winther
Feb 2 '17 at 22:45






$begingroup$
Note that $1 - 3x^2y^3 = -(3x^2 y^3 - 1)$ since minus times minus is plus.
$endgroup$
– Winther
Feb 2 '17 at 22:45














$begingroup$
so my answer was correct?
$endgroup$
– Anon
Feb 2 '17 at 22:46




$begingroup$
so my answer was correct?
$endgroup$
– Anon
Feb 2 '17 at 22:46












$begingroup$
Yes the last equation is the same as the next to last equation.
$endgroup$
– Winther
Feb 2 '17 at 22:47




$begingroup$
Yes the last equation is the same as the next to last equation.
$endgroup$
– Winther
Feb 2 '17 at 22:47










1 Answer
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$begingroup$

Your question boils down to showing $$ab+c=diff b=-frac{c-d}{a}$$ for $aneq 0$.



We have



$$begin{align}
ab+c=d &iff ab=d-cquad(text{subtract }c) \
&iff ab=-(c-d)quad(text{factor out }-1) \
&iff b=-frac{c-d}{a}quad(text{dividing by }atext{ assuming } aneq 0).
end{align}$$






share|cite|improve this answer









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    $begingroup$

    Your question boils down to showing $$ab+c=diff b=-frac{c-d}{a}$$ for $aneq 0$.



    We have



    $$begin{align}
    ab+c=d &iff ab=d-cquad(text{subtract }c) \
    &iff ab=-(c-d)quad(text{factor out }-1) \
    &iff b=-frac{c-d}{a}quad(text{dividing by }atext{ assuming } aneq 0).
    end{align}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Your question boils down to showing $$ab+c=diff b=-frac{c-d}{a}$$ for $aneq 0$.



      We have



      $$begin{align}
      ab+c=d &iff ab=d-cquad(text{subtract }c) \
      &iff ab=-(c-d)quad(text{factor out }-1) \
      &iff b=-frac{c-d}{a}quad(text{dividing by }atext{ assuming } aneq 0).
      end{align}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Your question boils down to showing $$ab+c=diff b=-frac{c-d}{a}$$ for $aneq 0$.



        We have



        $$begin{align}
        ab+c=d &iff ab=d-cquad(text{subtract }c) \
        &iff ab=-(c-d)quad(text{factor out }-1) \
        &iff b=-frac{c-d}{a}quad(text{dividing by }atext{ assuming } aneq 0).
        end{align}$$






        share|cite|improve this answer









        $endgroup$



        Your question boils down to showing $$ab+c=diff b=-frac{c-d}{a}$$ for $aneq 0$.



        We have



        $$begin{align}
        ab+c=d &iff ab=d-cquad(text{subtract }c) \
        &iff ab=-(c-d)quad(text{factor out }-1) \
        &iff b=-frac{c-d}{a}quad(text{dividing by }atext{ assuming } aneq 0).
        end{align}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 3:38









        ShaunShaun

        9,380113684




        9,380113684






























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