# What is the slope of the tangent line of  xy-y^2+x =C , where C is an arbitrary constant, at (-3,1)?

Jun 22, 2016

$\frac{4}{7}$

#### Explanation:

$x y - {y}^{2} + x = C$

$\left(x y\right) ' - \left({y}^{2}\right) ' + \left(x\right) ' = 0$

$\left(y + x y '\right) - \left(2 y y '\right) + 1 = 0$

$y ' \left(x - 2 y\right) = - \left(1 + y\right)$

$y ' = \frac{1 + y}{2 y - x}$

$y {'}_{\left(- 1 , 3\right)} = \frac{1 + 3}{2 \left(3\right) - \left(- 1\right)} = \frac{4}{7}$