Simplification: $(3x^3y^2-1)y'+3x^2y^3=1$ to $y'=-frac{3x^2y^3−1}{3x^3y^2−1}$.
$begingroup$
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
edited Dec 22 '18 at 6:07
Shaun
9,241113684
asked Feb 2 '17 at 22:42
AnonAnon
11211
$endgroup$
add a comment |
$begingroup$
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
edited Dec 22 '18 at 6:07
Shaun
9,241113684
asked Feb 2 '17 at 22:42
AnonAnon
11211
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 22 '18 at 6:07
Shaun
9,241113684
asked Feb 2 '17 at 22:42
AnonAnon
11211
$endgroup$
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
edited Dec 22 '18 at 6:07
Shaun
9,241113684
asked Feb 2 '17 at 22:42
AnonAnon
11211
edited Dec 22 '18 at 6:07
Shaun
9,241113684
asked Feb 2 '17 at 22:42
AnonAnon
11211
edited Dec 22 '18 at 6:07
Shaun
9,241113684
edited Dec 22 '18 at 6:07
Shaun
9,241113684
edited Dec 22 '18 at 6:07
Shaun
9,241113684
9,241113684
asked Feb 2 '17 at 22:42
AnonAnon
11211
asked Feb 2 '17 at 22:42
AnonAnon
11211
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your question boils down to showing $$ab+c=diff b=-frac{c-d}{a}$$ for $aneq 0$.
We have
$$begin{align}
ab+c=d &iff ab=d-cquad(text{subtract }c) \
&iff ab=-(c-d)quad(text{factor out }-1) \
&iff b=-frac{c-d}{a}quad(text{dividing by }atext{ assuming } aneq 0).
end{align}$$
answered Dec 22 '18 at 3:38
ShaunShaun
9,241113684
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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$begingroup$
Your question boils down to showing $$ab+c=diff b=-frac{c-d}{a}$$ for $aneq 0$.
We have
$$begin{align}
ab+c=d &iff ab=d-cquad(text{subtract }c) \
&iff ab=-(c-d)quad(text{factor out }-1) \
&iff b=-frac{c-d}{a}quad(text{dividing by }atext{ assuming } aneq 0).
end{align}$$
answered Dec 22 '18 at 3:38
ShaunShaun
9,241113684
$endgroup$
add a comment |
$begingroup$
Your question boils down to showing $$ab+c=diff b=-frac{c-d}{a}$$ for $aneq 0$.
We have
$$begin{align}
ab+c=d &iff ab=d-cquad(text{subtract }c) \
&iff ab=-(c-d)quad(text{factor out }-1) \
&iff b=-frac{c-d}{a}quad(text{dividing by }atext{ assuming } aneq 0).
end{align}$$
answered Dec 22 '18 at 3:38
ShaunShaun
9,241113684
$endgroup$
add a comment |
$begingroup$
Your question boils down to showing $$ab+c=diff b=-frac{c-d}{a}$$ for $aneq 0$.
We have
$$begin{align}
ab+c=d &iff ab=d-cquad(text{subtract }c) \
&iff ab=-(c-d)quad(text{factor out }-1) \
&iff b=-frac{c-d}{a}quad(text{dividing by }atext{ assuming } aneq 0).
end{align}$$
answered Dec 22 '18 at 3:38
ShaunShaun
9,241113684
$endgroup$
Your question boils down to showing $$ab+c=diff b=-frac{c-d}{a}$$ for $aneq 0$.
We have
$$begin{align}
ab+c=d &iff ab=d-cquad(text{subtract }c) \
&iff ab=-(c-d)quad(text{factor out }-1) \
&iff b=-frac{c-d}{a}quad(text{dividing by }atext{ assuming } aneq 0).
end{align}$$
answered Dec 22 '18 at 3:38
ShaunShaun
9,241113684
answered Dec 22 '18 at 3:38
ShaunShaun
9,241113684
answered Dec 22 '18 at 3:38
ShaunShaun
9,241113684
answered Dec 22 '18 at 3:38
ShaunShaun
9,241113684
9,241113684
add a comment |
add a comment |
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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