If 2 x^2-3 x y+y^2+x+2 y-8=0, then frac{d y}{ | vedclass

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(A) Given the equation: $2 x^2-3 x y+y^2+x+2 y-8=0$Differentiating both sides with respect to $x$ using the chain rule and product rule:$\frac{d}{d x}

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$\frac{3 y-4 x-1}{2 y-3 x+2}$

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(A) Given the equation: $2 x^2-3 x y+y^2+x+2 y-8=0$Differentiating both sides with respect to $x$ using the chain rule and product rule:$\frac{d}{d x}(2 x^2) - \frac{d}{d x}(3 x y) + \frac{d}{d x}(y^2) + \frac{d}{d x}(x) + \frac{d}{d x}(2 y) - \frac{d}{d x}(8) = 0$$4 x - (3 y + 3 x \frac{d y}{d x}) + 2 y \frac{d y}{d x} + 1 + 2 \frac{d y}{d x} = 0$Grouping the terms containing $\frac{d y}{d x}$:$(2 y - 3 x + 2) \frac{d y}{d x} = 3 y - 4 x - 1$Therefore, $\frac{d y}{d x} = \frac{3 y - 4 x - 1}{2 y - 3 x + 2}$

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

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