A problem by Suraj Mishra

Here we have that $a \times b = 24 \times (a - b)$ and $a + b = 14$

Operating, we have $b = 14 - a$ , and substituting we have $a (14 - a) = 24 [a - (14-a)]$ So... $14a - a^{2} = 24 (2a - 14)$ $14a - a^{2} = 48a - 336$ $a^2 + 34a - 336 = 0$ and there we get that a can only be 8, as the other solution, -42, solves everything right, but is not the larger number of that pair.