So, the question is that the sum of HCF and LCM is $96$ and the sum of the numbers is $48$. We need to find the numbers.
Here is my attempt to this question:
Let the numbers be $a$ and $b$ and their LCM and HCF be $l$ and $h$ respectively. So, the 3 equations that we have are
$$a + b = 48$$ $$l + h = 96$$ $$ab = lh$$
Substituting the value of $a$ as $48 - b$ and $l$ as $96 - h$ in $ab = lh$ we get $$(48 - b)b = (96 - h)h \\ \implies 48b - b^2 = 96h - h^2 \\ \implies 48b - 96h = b^2 - h^2$$
On comparing LHS with RHS we get $b$ as $48$ and $h$ as $96$. However, this would mean that LCM and $a$ are $0$ which is not true as LCM can't be less the HCF or the numbers. Is there some other way of doing it?