Page:The Algebra of Mohammed Ben Musa (1831).djvu/114

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If he leaves three sons, and bequeaths to some onę as much as the share of one of his sons, less the share of a daughter, supposing he had one, plus one-third of the remainder of the one-third; the computation will be this:^[1] distribute the heritage into such a number of parts as may be divided among the actual heirs, and also among them if a daughter were added to them. Such a number is twenty-one. Were a daughter among the heirs, her share would be three, and that of a son seven. The testator has therefore bequeathed to, the (73) legatee four-sevenths of the share of a son, and one-third of what remains from one-third. Take therefore one-third, and remove from it four-sevenths of the share of a son. There remains one-third of the capital less four-sevenths of the share of a son. Subtract now one-third of what remains of the one-third, that is to say, one-ninth of the capital less one-seventh and one-third of the seventh of the share of a son; the remainder

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1. Since there are 3 sons, each son’s share of the residue=${\tfrac {1}{3}}$. Were there 3 sons and a daughter, the daughter’s share would be ${\tfrac {1}{7}}$.
   

   ${\tfrac {1}{3}}-{\tfrac {1}{7}}={\tfrac {4}{7}}$
   Let $x$ be the stranger’s legacy, and $v$ a son’s share
   Then $1-x=3v$
   but $x={\tfrac {4}{7}}v+{\tfrac {1}{3}}[{\tfrac {1}{3}}-{\tfrac {4}{7}}v]$
   and $1-x={\tfrac {2}{3}}+{\tfrac {1}{3}}-{\tfrac {4}{7}}v-{\tfrac {1}{3}}[{\tfrac {1}{3}}-{\tfrac {4}{7}}v]=3v$
   $\therefore \;{\tfrac {2}{3}}+{\tfrac {2}{3}}[{\tfrac {1}{3}}-{\tfrac {4}{7}}v]=3v$
   $\therefore \;{\tfrac {2}{3}}+{\tfrac {2}{9}}=3{\tfrac {8}{21}}\times v,\;{\text{or}}\;{\tfrac {8}{9}}={\tfrac {71}{21}}v$
   $\therefore \;{\tfrac {8}{3}}={\tfrac {71}{7}}v\;\therefore \;v={\tfrac {56}{213}}=$ a son’s share
   $x={\tfrac {45}{213}}=$ the stranger’s legacy.