How old did the man live to be? On a man's tombstone, it is said that one sixth of his life was spent in childhood and one twelfth as a teenager. One seventh of his life passed between the time he became an adult and the time he married; five years later, his son was born. Alas, the son died four years before he did. He lived to be twice as old as his son did.
Question- How old did the man live to be?
(Assuming the end of teen life as adulthood)

Let the man lived for $x$ years.

Childhood: $\frac{x}{6}$;
Teenager: $\frac{x}{12}$;
From the end of teenage/adulthood to married: $\frac{x}{7}$;
From marriage till his son was born: $5$;
Till his son was dead: $y$;
Till he died: $4$.

Summation of all these is equal to $x$. The second equation comes from the fact that the father lived to be twice as old as his son.

$$\frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + y + 4 = x$$
$$x = 2y$$
Substituting $y$ from the second equation in first,
$$\frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + \frac{x}{2} + 4 = x$$
$$\frac{3x}{4} + \frac{x}{7} + 9 = x$$
$$\frac{x}{4} - \frac{x}{7} = 9$$
$$x = 84, y = 42$$
$84$ years

$\frac{x}{6}+\frac{x}{12}+\frac{x}{7} + 5 + y + 4 = 2y$. Simplifying this equation gives:
$\frac{11}{28} (x) + 9 = y$

Solve this for integer values of $x$ and $y$ using trial and error. Start with $x = 28$, $56$ and finally $84$ till you get an integer solution for $y$.

This gives $x = 84$ and $y = 42$