\[\begin{array}{1 1}(1)5y+3x=2&(2)5y-3x=2\\(3)3x-5y=2&(4)3x+3y=2\end{array}\]

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$y= \large\frac{x^3}{5}$

$\large\frac{dy}{dx} =\large\frac{3x^2}{5}$

$\large\frac{dy}{dx} (-1 ,-1 \large\frac{1}{5} )=\large\frac{3 \times (-1)^2}{5} =\large\frac{3}{5}$

Slope of the tangent at $(-1,\large\frac{-1}{5})$ having slope $\large\frac{3}{5}$ is

$y+ \large\frac{1}{5} =\frac{3}{5} $$(x+1)$

$\large\frac{5y+1}{5} =\frac{3x+3}{5}$

$3x-5y+2=0$

Hence 2 is the correct answer.

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