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If $2x^2-3xy+y^2+x+2y-8=0$, then $\large\frac{dy}{dx}$ is equal to

\[\begin {array} {1 1} (a)\;\frac{3y-4x-1}{2y-3x+2} & \quad (b)\;\frac{3y+4x+1}{2y+3x+2} \\ (c)\;\frac{3y-4x+1}{2y-3x-2} & \quad (d)\;\frac{3y-4x+1}{2y+3x+2} \end {array}\]

1 Answer

$(a)\;\frac{3y-4x-1}{2y-3x+2}$
answered Nov 7, 2013 by pady_1
 
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