TITLE: Solve Riddle With Algebra
QUESTION [5 upvotes]: There is a riddle and I believe it can be solved by algebra - please assist

A boy has as many sisters as brothers, but each sister has only half as many sisters as brothers. How many brothers and sisters are there in the family

Here is algebra, but I am stuck
$b=brother$
$s=sister$
$t=total$

boy has as many brothers as sisters

$b + b + s = t$

each sister has only half as many sisters as brothers

$s + s + b = t$
$s + \frac{1}{b} + b = t$
Hence
$b + b + s = s + \frac{1}{b} + b$
$2b + s = s + \frac{3b}{2}$
$2b = \frac{3b}{2}$
$4b = 3b$
Please assist.

REPLY [2 votes]: I think it'd be easiest to do
$b = $ number of boys
$g = $ number of girls and the essential insight is to notice:
A boy has $b-1$ brothers and $g$ sisters, and a girl has $b$ brother and $g-1$ sisters.
So "A boy has as many sisters as brothers" means $b-1 = g$.
And "each sister has only half as many sisters as brothers" means $ g-1 = \frac 12 b$
So solve the two equations two unknowns:
$b-1 = g$
$g-1 =\frac 12 b$
......
If you want to use variables for brothers and sisters you can do
$b_b = $ number of brothers a boy has
$b_g = $ number of brothers a girl has
$s_b = $ number of sisters a boy has and 
$s_g = $ number of sisters a girl has.
Then our riddle is $b_b = s_b$ and $s_g = \frac 12 b_s$.  
But to solve it we need to not the unstated but essential a girl has $1$ fewer sister than a boy, and a boy has one fewer brother than a girl.  So
$b_b = b_g -1$ and $s_g = s_b - 1$.
So solve the four equations four unknowns
$b_b = s_b$
$s_g = \frac 12b_s$
$b_b = b_s -1$
$s_g = s_b - 1$.