Five years ago, the age of a man was three years more than four times the age of his daughter. Three years hence, the age of the man will be six years less than thrice the age of his daughter. After how many years from now will their combined age be 80 years?
Solution
Let $f$ be the present age of the man and that of the daughter be $s$. Then,
$f-5=4(s-5)+3 \Rightarrow f-4s=-12$ ...(1)
$f+3=3(s+3)-6 \Rightarrow f-3s=0$ ...(2)
(1)-(2) gives
$-s=-12$
$\therefore s=12$
From equation (2) $f=3s \Rightarrow f=3(12)=36$
Now, let after $x$ years, the sum of their ages be 80 years.
$f+s+2x=80$
$ \Rightarrow 36+12+2x=80$
$\therefore 2x=80-48$ or $ x=16$
$\therefore$ After 16 years their combined age will be 80 years.