Page:The Algebra of Mohammed Ben Musa (1831).djvu/155

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namely, ten dirhems; then one hundred and ten dirhems remain for the heirs of the husband. From these the (101) gift to the wife is thing, so that there remain one hundred and ten dirhems less thing; and the heirs of the woman obtain twenty dirhems plus thing. She bequeaths one-third of this, namely, six dirhems and two-thirds, and one-third of thing. The moiety of the residue, namely, six dirhems and two-thirds plus one-third of thing, returns to the heirs of the husband: so that one hundred and sixteen and two-thirds, less two-thirds of thing, come into their hands. He has bequeathed one-third of this, which is thing. There remain, therefore, one hundred and sixteen dirhems and two-thirds less one thing and two-thirds, and this is twice as much as the husband’s gift to the wife added to his legacy to the stranger, namely, four things. Reduce this, and you find one hundred and sixteen dirhems and two-thirds, equal to five things and two-thirds. Consequently one thing is equal to

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$s-d-2x+[c+d+x]=4x$, according to the law of inheritance.
$\therefore \;3s+c-2d=17x,\;{\text{and}}\;x={\tfrac {3s+c-2d}{17}}$
${\text{If}}\;s=120,c=10,\;{\text{and}}\;d=10,x={\tfrac {350}{17}}=20{\tfrac {10}{17}}$
$c+d+x=40{\tfrac {10}{17}}\;;{\tfrac {1}{3}}[c+d+x]=13{\tfrac {9}{17}}$
The wife bequeaths $13{\tfrac {9}{17}}$; $13{\tfrac {9}{17}}$ go to her husband, and $13{\tfrac {9}{17}}$ to her other heirs.
The husband bequeaths to the stranger $20{\tfrac {10}{17}}$; he gives the same sum to the wife ; and $4x=82{\tfrac {6}{17}}$ go to his heirs.