Interesting primary school problem Let say 
$$\frac{a}{b+c}+\frac{d}{e+f}+\frac{g}{h+i}=1$$
Given that $$a,b,c,d,e,f,g,h,i$$ represents number 1,2,3,4,5,6,7,8,9 (we don't know which alphabet represent which digit)
When dealing with this problem, I came up with the following question:
Q1)Is that an algebraic way to solve this problem? If not, does wild guess is the only way we can use to solve this type of problem?
Q2)I find one answer by luck but I am not sure whether it is the unique solution for this problem. How can i prove that is a unique problem?
 A: The one algebraic approach I can think of is to render
$(1/2)+(1/3)+(1/6)=1$
and thereby identify
$a/(b+c)=(1/2)$
$d/(e+f)=(1/3)$
$g/(h+i)=(1/6)$
Now there are only a few possibilitues for the $1/6$ fraction because the denominator has to be less than $9+9=18$ and has to be a multiple of $6$.  Only the following fit both criteria:
$1/(2+4)$
$2/(3+9)$
$2/(4+8)$
$2/(5+7)$
Say we use $2/(3+9)$.  Now we try to form a fraction of $1/3$ with the remaining digits.  The denominator must be a multiple of $3$ less than $18$, so the numerator has to be no greater than $5$.  A numerator of $1$ forces us to repeat digits:  $1/3=1/(1+2)$.  Ditto for $2$ and $3$ because we use those in our trial $1/6$ expression.  We find that only two choices may work for $1/3$:
$1/3=4/(5+7)$
$1/3=5/(7+8)$
In the first case we are left with the digits $1,6,8$ but alas, we cannot make a fraction of $1/2$ with those.  The second case fails similarly.  So we move on; $1/6$ can't be $2/(3+9)$.
The solution I found uses $1/6=1/(2+4)$.  See if you can work out the $1/3$ and $1/2$ fractions from that.
