Solve Riddle With Algebra

$b$ = number of boys = number of brothers each girl has

$g$ = number of girls = number of sisters each boy has

$b-1$ = number of brothers each boy has

$g-1$ = number of sisters each girl has

$$b-1 = g, \qquad (\text{Equation 1})$$

$$g-1 = \dfrac{b}{2}, \qquad (\text{Equation 2})$$

Plugging in for $g=b-1$ into the second equation:

$$b-1-1 = \dfrac{b}{2} \Longrightarrow b=4, g=3$$

Although this is probably not intentional, this problem has two different solutions. If you assume there are $s>0$ sisters, the boy also has $s$ brothers making $s+1$ boys in all. Each boy being a brother of any of the girls, each having $s-1$ sisters, the equation $s+1=2(s-1)$ easily solves to $s=3$ (four boys and three girls in all).

However, the question does not say that $s\neq0$; the boy need not have any sisters. In that case each of the sisters has (whatever) is true; there aren't any sisters so that is vacuously true. So $s=0$ is another solution: a family with one boy and no girls. Technically, in this case there are no brothers and sisters at all in the family (which is what the question was asking), as the one child that is there is neither a sister nor a brother. Actually, nowadays this is the more likely solution, I would think.

The problem is you're using $b$ for the number of the boy's brothers and also for the number of the girl's brothers.

Instead say there are $b$ boys and $g$ girls. Then a boy has $b-1$ brothers, so $$g=b-1.$$Similarly $$g-1=b/2.$$