$y^3 = ax^2 + b$

$\text{Point } (1,2) \text{ lies on curve}$

$2^3 = a(1)^2 + b$

$8 = a + b$

$\text{Given tangent line: } 2y = x + 3$

$y = \frac{x+3}{2}$

$\text{slope} = \frac{1}{2}$

$\text{Differentiate: } y^3 = ax^2 + b$

$3y^2 \frac{dy}{dx} = 2ax$

$\frac{dy}{dx} = \frac{2ax}{3y^2}$

$\text{At } (1,2): \frac{dy}{dx} = \frac{2a}{3 \cdot 4} = \frac{a}{6}$

$\frac{a}{6} = \frac{1}{2}$

$a = 3$

$b = 8 - a = 5$

$3b - 5a = 3(5) - 5(3) = 15 - 15 = 0$

Final Answer: 0