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I need help ! Pleaseee Solve (x-2y +1) dx + (4x -3y -6) dy = 0

Mathematics
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use green's theorem ( http://en.wikipedia.org/wiki/Green%27s_theorem )
Seriously??? can you tell me how to solve it?
It's been a while since i took calculus 3

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Other answers:

then???
have you any idea, @Cynosure-EPR ?????
I'm sorry =( I do know how to do it but I'm rushed with a programming assignment (worth 40% of grade) due at midnight =(
ok no problem @Cynosure-EPR :) good luck for u :)
@zepp have u any idea??
Nope, I'm only at Calc II, sorry ):
ok np @zepp thank u for coming :)
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\[(x-2y +1) dx + (4x -3y -6) dy = 0\]\[-(x-2y +1) dx = (4x -3y -6) dy \]\[\frac{ dy }{ dx }=\frac{ -(x-2y+1) }{ (4x -3y -6) }\]
Do you know how to do separation of variables? http://en.wikipedia.org/wiki/Separation_of_variables
This looks like a diff eq question
nvm, I don't think we can use that here
then ???? how to solve y?? i really know separation variable, but in this case that method cannot work,
yes.., i think so.., we cannot use that in this case.., :)
http://tutorial.math.lamar.edu/Classes/DE/Exact.aspx
My teacher skipped partial derivatives and that looks like what is needed for this problem. I'm going to tapout. Good luck
i see.., wonderful site for me.., thank u @ChmE,
solve the system of equation \[\large \left(\begin{matrix}x-2y+1=0 \\ 4x-3y-6=0\end{matrix}\right)\]
when you get the solution, say x0 and y0, use u = x - x0 and v = y - y0. try this @gerryliyana
ok.., i'll try
did you get this DE? \[\large (u-2v)du + (4u-3v)dv=0\] x changed to u, y changed to v, the constants +1 and -6 disappeared.
\[\left( \frac{ x - 2y + 1 }{ 4x -3y - 6 } \right) \times \left[ \frac{ 4 }{ 1 } \right]\] \[\left( \frac{ 4x - 8y + 4 }{ 4x - 3y -6 } \right) = 0\] \[\left( \frac{ 4x - 8y = -4 }{ 4x -3y= 6 } \right)\] -------------------------------------------- __ \[-8y -(-3y) = -10 \rightarrow y = 2\] \[4x - 3(2) = 6 \rightarrow x = 3\]
on the right track. now, u = x - 3 and v = y - 2
OMG..., my connection is lost, and I lost all my type, :(
hahaha. enter them a few equations at a time.
don't worry. you are not the first one.

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