The slope of the tangent line to the line 2x+3y=5 can be found by differentiating 2x+3y=5.
Differentiating 2x+3y=5 with respect to x, we get
[tex]\begin{gathered} 2+3\frac{dy}{dx}=0 \\ 3\frac{dy}{dx}=-2 \\ \frac{dy}{dx}=\frac{-2}{3} \end{gathered}[/tex]m=dy/dx is the slope of tangent line.
Hence, slope, m=-2/3.
Now, the equation of the tangent line passing through point (x1, y1)=(-2, 3) with slope m=-2/3 can be found as,
[tex]\begin{gathered} m=\frac{y_1-y}{x_1-x} \\ \frac{-2}{3}=\frac{3-y}{-2-x} \\ -2(-2-x)=3(3-y) \\ 4+2x=9-3y \\ 3y+2x=5 \end{gathered}[/tex]Therefore, the equation of the tangent line is 3y+2x=5.
