How can we find the two numbers when their sum and the sum of their HCF and LCM are given?

$$a + b = 48$$ $$l + h = 96$$ $$ab = lh$$

So $$l = ab/h$$

Let $a = ph$, $b = qh$

Therefore $$ph + qh = 48$$ $$pqh + h = 96$$ Thus $$pqh + h = 2(ph + qh)$$ $$pq + 1 = 2p + 2q$$ $$pq -2p -2q = -1$$ By adding 4 to both sides we can factor the LHS $$pq -2p -2q + 4 = 3$$ $$(p-2)(q-2) = 3$$ 3 is prime, so one of $p-2$ and $q-2$ must be 3 and the other must be 1 (since we're working with positive integers).

WLOG, let $p-2 = 3$ and $q-2 = 1$ Therefore $p=5$ and $q=3$

$p+q=8$ and since $(p + q)h = 48$

$h = 48/8 = 6$

So $a = 5.6 = 30$ and $b = 3.6 = 18$

and $l = LCM(5.6, 3.6) = 5.18 = 30.3 = 90$

Hence $l + h = 96$