Solve Riddle With Algebra 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.

$b$ = 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:

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