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? 
 A: Others have explained how to translate to the equation below. Here's a nice way to solve it.
$$\begin{eqnarray} L &\,=\,&\  \,\frac{L}6 + \frac{L}{12} + \frac{L}7\ +\ \frac{L}2\, +\, 9\\ \\ \\
\smash{\overset{\large\times\ \color{#c00}{84}}\iff}\ \ 84 L &\,=\,& 14L + 7L + 12L + 42L + 9\cdot 84\\ \\
 &\,=\,& 75L + 9\cdot 84\\ \\
\iff\ \ \ 9L &\,=\,& 9\cdot 84\\ \\
\iff\ \ \ \ \ L &\,=\,& 84\end{eqnarray}\qquad$$
Note on lcm $= \color{#c00}{84}$: $\rm\,\  \color{#0a0}{2}\mid \color{#0a0}6\mid 12\,\Rightarrow\, lcm(\color{#0a0}{2,6},12, 7) = lcm(12,7) = 12\cdot 7= \color{#c00}{84},\,$ by $\,\gcd(12,7)=1.$
A: 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.
A: Let $L_M$ denote the length of the man's life and $L_D$ denote the length of the daughter's life.
Literally writing down what the problem tells us, we have $L_M = L_M\left(\frac16 + \frac1{12}+\frac17\right) + 5 + \frac12L_M + 4$.
Noting that the LCM of $12$ and $7$ is $84$, we then have $L_M = L_M\left(\frac{14}{84}+\frac{7}{84}+\frac{12}{84}\right) + \frac{42}{84}LM + 9$
So $L_M = \frac{75}{84}L_M + 9$. So $9=L_M - \frac{75}{84}L_M$, i.e. $9 = L_M\left(1-\frac{75}{84}\right)$, i.e. $9=\frac{9}{84}L_M$, so $1 = \frac{1}{84}L_M$, so $\frac{1}{L_M} = \frac{1}{84}$, so $L_M = 84$. As his daughter reached half his age, $L_D = \frac12 \cdot 84 = 42$.
A: Let 
X = Man's Age
M = Man's Age at Marriage.  
B = Man's Age at Daughter's Birth. 
D = Daughter's Age. 
We have 4 equations in 4 unknowns.
M + 5 = B
B + D + 4 = X
D = X / 2
M = 11/28 X
Recasting this as a linear system, we get
M + 5 - B = 0
B + D + 4 - X = 0
D - X / 2 = 0
M - 11/28 X = 0
Solving for this linear system, gives us that X = 84.
A: i will give you a more conceptual answer. let the age of the father when he died be $x$. then his age when he had the child is
$$x(\frac{1}{6}+\frac{1}{12}+\frac{1}{7})+5$$
his age when his daughter died is $x-4$.
it is given that her age when she died is $\frac{x}{2}$.
$$(x-4)-(x( \frac{1}{6}+\frac{1}{12}+\frac{1}{7} )+5)=\frac{x}{2}$$
solving this equation gives the age of the father when he died as $84$.
