Page:The Algebra of Mohammed Ben Musa (1831).djvu/153

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ment of one hundred dirhems, her dowry being ten (100) dirhems, and he bequeaths to some person one-third of his property; then the computation is this:^[1] Pay to the woman her dowry, that is, ten dirhems; there remain ninety dirhems. Herefrom pay the gift to her, thing; then pay likewise to the legatee who is to receive one-third, thing: for the one-third is divided

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1. This case is distinguished from that in page 133 by two circumstances; first, that the woman does not make any bequest; second, that the husband bequeaths one-third of his property.
   Suppose the husband not to make any bequest. Then, since the woman had at her disposal $d+x$, but did not make any bequest, ${\tfrac {1}{2}}[d+x]$ reverts to her husband; and the like amount goes to her other heirs.
   $\therefore \;s-[d+x]+{\tfrac {1}{2}}[d+x]=2x\;\therefore \;x{\tfrac {1}{5}}[2s-d]$
   and since $s=100$, and $d=10$; $x=38$; $d+x=48$; $d+x=48$;${\tfrac {1}{2}}[d+x]=24$ reverts to the husband, and the like sum goes to her other heirs; and $2x=76$, belongs to the husband’s heirs.
   Now suppose the husband to bequeath one-third of his property. The law here interferes with the testator’s right of bequeathing; and provides that whatever sum is at the disposal of the wife, the same sum shall be at the disposal of the husband; and that the sum to be retained by the husband’s heirs shall be twice the sum which the husband and wife together may dispose of.
   

   $\therefore \;s-{\tfrac {1}{2}}[d+x]-x=4$
   

   $\therefore \;{\tfrac {1}{11}}[2s-d]=x\;{\text{if}}\;s=100,\;{\text{and}}\;d=10;x={\tfrac {190}{11}}=17{\tfrac {3}{11}};d+x=27{\tfrac {3}{11}};{\tfrac {1}{2}}[d+x]=13{\tfrac {7}{11}}$ reverts to the husband, and the like sum goes to the other heirs of the woman; $17{\tfrac {3}{11}}$ is what the husband bequeaths; and $69{\tfrac {1}{11}}=4x$ goes to the husband’s heirs.

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