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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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Verified Answer
The correct answer is:
$\frac{3 y-4 x-1}{2 y-3 x+2}$
We have, $2 x^2-3 x y+y^2+x+2 y-8=0$
On differentiating w.r.t. $x$, we get
$\begin{array}{ll}4 x-3 x \frac{d y}{d x}-3 y+2 y \frac{d y}{d x}+1+\frac{2 d y}{d x}-0=0 \\ \Rightarrow & 4 x+(-3 x+2 y+2) \frac{d y}{d x}-3 y+1=0 \\ \Rightarrow & \frac{d y}{d x}=\frac{3 y-4 x-1}{2 y-3 x+2}\end{array}$
On differentiating w.r.t. $x$, we get
$\begin{array}{ll}4 x-3 x \frac{d y}{d x}-3 y+2 y \frac{d y}{d x}+1+\frac{2 d y}{d x}-0=0 \\ \Rightarrow & 4 x+(-3 x+2 y+2) \frac{d y}{d x}-3 y+1=0 \\ \Rightarrow & \frac{d y}{d x}=\frac{3 y-4 x-1}{2 y-3 x+2}\end{array}$
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