Page:The Algebra of Mohammed Ben Musa (1831).djvu/113

( 97 )

take a number, one-fifth of which may be divided into sevenths, and one-seventh of which may be divided into fifths. Such a number is thirty-five. Add to it two-sevenths of the same, namely, ten. This gives forty-five. Herefrom the legatee receives ten, each son four-teen, and the daughter seven.

If he leaves a mother, three sons, and a daughter, and bequeaths to some one as much as the share of one of his sons less the amount of the share of a second daughter, in case he had one; then you distribute the heritage into such a number of parts as may be divided among the actual heirs, and also among the same, if a second daughter were added to them.^[1] Such a number is three hundred and thirty-six. The share of the second daughter, if there were one, would be thirty-five, and that of a son eighty; their difference is forty-five, and this is the legacy. Add to it three hundred and thirty-six, the sum is three hundred and eighty-one, which is the number of parts of the entire heritage.

---

1. Let $x$ be the stranger’s legacy; $1-x$ is the residue. A widow’s share of the residue is $\scriptstyle {\frac {1}{6}}$th: there remains ${\tfrac {5}{6}}[1-x]$, to be distributed among the children.
   ${\begin{aligned}\left.{\begin{aligned}{\text{Since there are 3 sons, and 1 daughter,}}\\{\text{a son’s share is.......................}}\end{aligned}}\right\}\;{\tfrac {2}{7}}\times {\tfrac {5}{6}}[1-x]\\\left.{\begin{aligned}{\text{Were there 3 sons and 2 daughters, a}}\\{\text{daughter’s share would be...........}}\end{aligned}}\right\}\;{\tfrac {1}{8}}\times {\tfrac {5}{6}}[1-x]\\{\text{The difference}}={\tfrac {9}{56}}\times {\tfrac {5}{6}}[x-1]\\\end{aligned}}$
   ${\begin{aligned}\therefore x={\tfrac {45}{336}}[1-x]\quad \quad \quad \therefore x={\tfrac {45}{381}}\\1-x={\tfrac {336}{381}};\;{\text{the widow’s share}}={\tfrac {56}{381}}\\{\text{the daughter’s share}}={\tfrac {40}{381}}\end{aligned}}$

o