TO THE RESOLUTION OF PROBLEMS. 
79 
84 
Hence 3a — 60 f 24 — 84, and a ~ —— =28: so 
that 28, 20, and 12, are the three parts required. 
problem v 
The sum of 660/. was raised (for a certain purposeJ 
by four persons A, B, C, and D; ivhereofB advanced 
twice as much as A ; C as much as A and B; and D as 
much as B and C: what did each person contribute ? 
Let the sum or number of pounds advanced \ 
by A be called * 
then will the number of B’s pounds be denoted by 2a, 
that of C’s by 3a, 
and that of D’s by . bx\ 
the sum of all which is given equal to 660/. that is, 
11a = 660: from which x — ~~~ — 60. Therefore, 
60,120, 180, and 300/. are the respective sums that were 
to be determined. 
PROBLEM VI 
A certain sum of money was shared among five persons, 
A, B, C, D, and E; whereof B received 10/. less than, 
A; C 16/. more than B; Do/, less than C; E 15/. 
more than D: moreover it appeared, that the shares of the 
two last together were equal to the sum of the shares of the 
other three: What was the whole sum shared, and how 
much did each receive ? 
Let x denote the share of A.: 
a? — 10 
а b 6 
x Ь 1 
a 1 + 16 
then 
will be the share of 
and therefore 2a + 17 = за — 4, by the question: from 
whence, by transposition, 21 = a; so that 21, 11, 27, 
22, and 37/. are the several required shares; amounting, 
jn the whole, to 118/,