$$q ON THE EARLY 



s£ First case of Multarindt. Number equal to Mdl and Shai. Com- 

 plete the MM. to unity if it is deficient, and reduce it to the same if it ex- 

 ceeds, and reduce the numbers and Shai in the same ratio, by dividing all 

 by the co-efficient of the Mai Then square pne half the co-efficient of 

 the Shai, and add this square to the numbers. Subtract from the root of 

 the sum half the co-efficient of the Shai, and the unknown will remain. 

 Example, A person admitted that he. owed.ZAi© ,a sum less than 10, so 

 much that if the square of it was add^d toils product by f what it wants 

 qf 10, the sum would be .12. Suppose the number ^i, its square is Mai; 

 half the remainder from 10 is 5 all but half oj Stia-L- The product of Shai 

 by this is 5 Shai all but | of Mai. 'Therefore ;£ & Mai and £ Shai are 

 equal to 12. One Mdl and ao Shai -are equal Xq- 34,. Subtract half the 

 co-efficient of the Shai from the coot of the sum of the square of J the co- 

 efficient of the Shai and the -numbers. There remains 2, which is the 

 number required, 



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'Second ocase. -Shai equal to numbers and Mdl.. After completing or 

 rejecting, subtract the numbers from the square, of half the co-efficient of 

 the [Shai, and add the root of the remainder to half the co-efficient of the 



Shai; or subtract the former from the latter ; the result is the unknown 

 quantity. Example. What number is that which being multiplied by half 

 of itself and the product increased by 12, the result is, five times the 

 original number. Multiply Shai by half itself, then half of Mdl added to 

 12 is equal to 5 Shai. One Mdl and 24 is equal to 10 Shai. Subtract 24 

 from the square of 5, there remains one, and the root of one is one. The 

 sum or difference of 1 and 5 is the number required. 



Third Case. Mdl equal to number and Shai. After completion or 



rejection, add the square of half the co-efficient of the Shai to the numbers, 



