1. **State the problem:** Find the coordinates of the points of intersection of the curve $$4 + \frac{5}{y} + \frac{3}{x} = 0$$ and the line $$y = 15x + 10$$. 2. **Substitute** the line equation into the curve equation: $$4 + \frac{5}{15x + 10} + \frac{3}{x} = 0$$ 3. **Multiply both sides** by the common denominator $$x(15x + 10)$$ to clear fractions: $$4x(15x + 10) + 5x + 3(15x + 10) = 0$$ 4. **Expand terms:** $$4x(15x + 10) = 60x^2 + 40x$$ $$3(15x + 10) = 45x + 30$$ So the equation becomes: $$60x^2 + 40x + 5x + 45x + 30 = 0$$ 5. **Combine like terms:** $$60x^2 + (40x + 5x + 45x) + 30 = 0$$ $$60x^2 + 90x + 30 = 0$$ 6. **Divide entire equation by 6** to simplify: $$\cancel{6}0x^2 + \cancel{6}0x + \cancel{6}0 = 0 \Rightarrow 10x^2 + 15x + 5 = 0$$ 7. **Divide entire equation by 5** to simplify further: $$\frac{10x^2}{5} + \frac{15x}{5} + \frac{5}{5} = 0 \Rightarrow 2x^2 + 3x + 1 = 0$$ 8. **Solve quadratic equation** $$2x^2 + 3x + 1 = 0$$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=2$$, $$b=3$$, $$c=1$$. Calculate discriminant: $$\Delta = 3^2 - 4 \times 2 \times 1 = 9 - 8 = 1$$ So, $$x = \frac{-3 \pm \sqrt{1}}{2 \times 2} = \frac{-3 \pm 1}{4}$$ 9. **Find roots:** - $$x_1 = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2}$$ - $$x_2 = \frac{-3 - 1}{4} = \frac{-4}{4} = -1$$ 10. **Find corresponding y values** using $$y = 15x + 10$$: - For $$x = -\frac{1}{2}$$: $$y = 15 \times \left(-\frac{1}{2}\right) + 10 = -7.5 + 10 = 2.5$$ - For $$x = -1$$: $$y = 15 \times (-1) + 10 = -15 + 10 = -5$$ 11. **Check denominator conditions:** Denominator $$3x^2 + 2x \neq 0$$ to avoid division by zero. Calculate for each root: - For $$x = -\frac{1}{2}$$: $$3 \times \left(-\frac{1}{2}\right)^2 + 2 \times \left(-\frac{1}{2}\right) = 3 \times \frac{1}{4} - 1 = \frac{3}{4} - 1 = -\frac{1}{4} \neq 0$$ - For $$x = -1$$: $$3 \times 1 + 2 \times (-1) = 3 - 2 = 1 \neq 0$$ Both denominators are nonzero, so both points are valid. **Final answer:** The points of intersection are $$\boxed{\left(-\frac{1}{2}, 2.5\right)}$$ and $$\boxed{(-1, -5)}$$.