HISTORY OF ALGEBRA. 



received 7. How many sons were there, and how many Dinars? Sup- 

 pose the number of sons Shai, and take the sum of the extremes, that is 

 to say, 1 and Shai. Multiply them by half of Shai. This is the number 

 of Dinars. For the' product of the sum of any series of numbers in 

 arithmetical progression-, is equal to the product of the sum of the two 

 extremes, into half the number of terms. Divide the number of the 

 Dinars by Shai, which is the number of the sons; the quotient, according 

 to the terms of the question; will be seven: Mliltiply 7 by Shaii which 

 is the divisor; 7- Shai is the- product, which is equal to \ Mai and % Shah 

 After Jebr and Mukabaldh, one Mi/is equal to 13- Shak Shai then is 13; 

 and this is the number of *he sons.- Multiply this by 7: The number of 

 Dinars will be found 9j,<g£f 



" Questions of this sort may be solved by position. Thus, suppose 

 the number of "sons to be 5 ; the first error is 4 in defect. Then suppose 

 it to be 9, the second error is 2 Jn- defect; The first Mahfiidh is 10 and 

 the secondis 36' ; their difference is-s@;' the difference of the errors is 2. 

 Another method, which is easy and short, is- this; Double the quotient, 

 (the number 7 in the question) subtract one, and the result is the num- 

 ber of sons* 



" Case the third! Number equal to Mdl. Divide the number by the 

 co-efficient of the Mdl ; the root of the quotient is the unknown quantity. 

 For example. A person admitted that he owed Z aid the greater of two 

 sums of money, the sum of which was 20 and the product 96. Suppose 

 one of them to be 10 and Shai, and the other 10 all but Shai. The pro- 

 duct, which is 100 all but Mdl, is equal to 96 ; and after Jebr and Mukd- 

 baldh, one Mdl is equal to 4, and Shai equal 'to 2. One of the sums then 



is 8 and the other is, and is- is thedebt of Z'aid. 



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