Question:
The sum of ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man's age at the time. Find their present ages.
Solution:
Let the present age of the man be $x$ years
Then present age of his son is $=(45-x)$ years
Five years ago, man's age $=(x-5)$ years
And his son's age $(45-x-5)=(40-x)$ years
Then according to question,
$(x-5)(40-x)=4(x-5)$
$40 x-x^{2}+5 x-200=4 x-20$
$-x^{2}+45 x-200=4 x-20$
$-x^{2}+45 x-200-4 x+20=0$
$-x^{2}+41 x-180=0$
$x^{2}-41 x+180=0$
$x^{2}-36 x-5 x+180=0$
$x(x-36)-5(x-36)=0$
$(x-36)(x-5)=0$
So, either
$(x-36)=0$
$x=36$
Or
$(x-5)=0$
$x=5$
But, the father's age never be 5 years
Therefore, when $x=36$ then
$45-x=45-36$
$=9$
Hence, man's present age is $=36$ years and his son's age is $=9$ years