Infinitely Many Solutions for System of Linear Equations

For a system of linear equations $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$ to have infinitely many solutions: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Here, $\frac{a+5}{1} = \frac{b^2-15}{1} = \frac{8b}{4} = 2b$.

From $\frac{b^2-15}{1} = 2b$: $b^2 - 2b - 15 = 0 \implies (b - 5)(b + 3) = 0$ So $b = 5$ or $b = -3$.

Case 1: $b = 5$ $a + 5 = 2(5) = 10 \implies a = 5$ $ab = 5 \times 5 = 25$

Case 2: $b = -3$ $a + 5 = 2(-3) = -6 \implies a = -11$ $ab = (-11) \times (-3) = 33$

Thus, the maximum possible value of $ab$ is $33$.