Full text: A treatise of algebra

1S4 
• THL RESOLUTION OF 
corresponding value of x. Hence it follows, that, if 
the greatest value of x be divided by the co-eflicient of //, 
the remainder will be the least value of x, and that the 
quotient 4- 1 will give the number of all the answers. 
But it is to be observed, that the equations here spoken 
of, are such, wherein the said co-efficients are prime 
to each other; if this should not be the case, let the 
equation given be, first of all, reduced to one of this 
form, by dividing by the greatest common measure. 
PROBLEM III* 
To find how many different ways it is possible to pay 
look in guineas and pistoles, only ; reckoning guineas at 
21 shillings each, and pistoles at 17. 
Let x represent the number of guineas, and y that 
of the pistoles; then the number of shillings in the 
guineas being 2Lr, and in the pistoles, 17//, we shall 
therefore have 21a? f- 17// — 2000, and consequently x — 
2000— 17y , 5—17// 
— = 95 f 
which being a whole 
number, by the question, it is manifest that 
must also be an integer: now the least value of ?/, in 
whole numbers, to answer this condition, will be found 
= 4, and the expression itself rr 3; the corresponding, 
or greatest value of a? being = 92; which being divided 
by 17, the co-efficient of y (according to the preceding 
note) the quotient comes out 5, and the remainder 7 : 
therefore the least value of .t is 7, and tlir number of 
answers (- 5 + 1) r 6: and these are as follow, 
II 
fO 
75 ! 58 
41 
24 
11 
>> 
25 J 46 
67 
68 
PROBLEM IV. 
To determine whether it be possible to pay 100I. in gui 
neas and moidores only; the former being reckoned ut 21 
shillings each, and the latter at 27. 
Here, by proceeding as in the last question, we have
	        
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