Consider the curve given by the equation xy - 3x + 2y = -2. Find the equation of the tangent in the point (2,1). So through implicit differentiation I've found that (if I'm correct) y' = (1-y)/(x-2) How do I proceed to find the tangent line since there are still 2 variables in the function?

Consider the curve given by the equation xy - 3x + 2y = -2. Find the equation of the tangent in the point (2,1).
So through implicit differentiation I've found that (if I'm correct) y' = (1-y)/(x-2)
How do I proceed to find the tangent line since there are still 2 variables in the function?
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Marcel Mccullough
Note that if we have
$xy-3x+2y=-2$
then after differentiation we get
$y+x{y}^{\prime }-3+2{y}^{\prime }=0\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x{y}^{\prime }+2{y}^{\prime }=3-y\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime }\left(x+2\right)=3-y\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime }=\frac{3-y}{x+2}$
Edit: So to find the tangent line, first you need the slope of the tangent line at the point (2,1). You do that by plugging in the coordinates in the derivative expression we found, so ${y}^{\prime }$ at (2,1) is:
${y}^{\prime }=\frac{3-1}{2+2}=\frac{1}{2}$
You know the slope, and you know one point on the tangent line (namely, (2,1)) so you can find the equation of the line from here.