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October 31, 2014

October 31, 2014

Posted by **LEON** on Saturday, November 13, 2010 at 1:16am.

- math calculus -
**Damon**, Saturday, November 13, 2010 at 4:44amx(3y^2dy) + y^3(dx) +2xdy + 2ydx = 0

dy (3xy^2+2x) = -dx (y^3+2y)

dy/dx = -(y^3+2y)/(3xy^2+2x)

I assume you mean the point (3,1)

x = 3

y = 1

so

m = dy/dx = -(1+2)/(9+6) = - 3/15 = -1/5

put in point

1 = m * 3 + b

1 = -3/5 + b

b = 8/5

y = -x/5 + 8/5

5y = 8-x

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