Find the equation of the tangent line to the curve at the given point.\ y = \frac{x - 10}{x - 3}, \left(12, \frac{2}{9}\right)\ y = \boxed{ }
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$$y' = \frac{(x-3)(1)-(x-10)(1)}{(x-3)^2}$$ $$y' = \frac{x-3-x+10}{(x-3)^2}$$ $$y' = \frac{7}{(x-3)^2}$$ Show more…
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