How to get the man's age. I am very much in need of solution for this problem. I can't figure out the answer for this. does anybody know about this problem below? 
thanks,,
Problem:
A man's boyhood lasted for 1/6 of his life, he played soccer for the next 1/12 of his life, and he married after 1/7 more of his life. a daughter was born 5 years after his marriage & the daughter lived 1/2 as many years as her father did. If the man died four years after his daughter did, how old was the man when he died? 

We can set up this problem by adding all the fractions of his life so they equal $1$ (his whole life). Also, let's say that he lived to $x$ years old.
Now,
$\frac{1}{6}$ of his life was boyhood.
$\frac{1}{12}$ of his life was soccer.
$\frac{1}{7}$ of his life was end of soccer to marriage.
After $5$ years, his daughter was born so
$\frac{5}{x}$ of his life was marriage to daughter's birth.
His daughter lived half the amount of years that he will have lived so
$\frac{1}{2}$ of his life was his daughter's life.
Then he died $4$ years later so
$\frac{4}{x}$ of his life was from his daughter's death to his death.
And all of these should equal to $1$. Thus
$\frac{1}{6}+\frac{1}{12}+\frac{1}{7}+\frac{5}{x}+\frac{1}{2}+\frac{4}{x}=1$
Now we simply solve for $x$. We get
$\frac{9}{x}=1-\frac{1}{6}-\frac{1}{12}-\frac{1}{7}-\frac{1}{2}$
then $9=x(1-\frac{1}{6}-\frac{1}{12}-\frac{1}{7}-\frac{1}{2})$
then $x=\frac{9}{1-\frac{1}{6}-\frac{1}{12}-\frac{1}{7}-\frac{1}{2}}$.
By simplifying, we get $x=84$. So he lived to $84$ years old.
