TO THE RESOLUTION OP PROBLEMS. 
89 
x —— 3 
dered the second time: in like manner, if from —-— 
o 
X 3 '•it 
(the half of — } you, again, take 4, there will re 
main — — 4 or -• ~~ the number of sheep re? 
8 8 
x —- 7 
plaining at last. Hence we have —-—'= 5; therefore 
8 
p — 7 = 40, and x — 47- 
PROBLEM XXVIII. 
The difference of tico numbers being given, equal to 4, 
and the difference of their squares, equal to 40; to find 
the numbers. 
Let the lesser number be x; then, the difference be 
ing 4, the greater must consequently be x f 4, and its 
square xx + Sp t- 16, from which xx, the square of the 
lesser being taken away, the difference is 8x -f 16: 
therefore 8p {16 — 40; which, reduced, gives x — 3; 
whence x f 4 — 7; therefore the two required num 
bers are 3 and 7. 
Ali the problems hitherto delivered are resolved by a 
numeral exegesis, wherein the unknown quantities,only, 
are represented by letters of the alphabet; w r hich seem 
ed necessary, in order to strengthen the Beginner’s 
idea, at setting out, and lead him on by proper gra 
dations: but it is not only more masterly and elegant, 
but also more useful, to represent the known, as well 
as the unknown quantities, by algebraic symbols; since 
from thence a general theorem is- derived, whereby 
all other questions of the same kind may be resolved. 
As an instance hereof, let the last problem be again 
resumed; then the given difference of the required 
numbers being denoted by a, the difference of their 
squares by b, and the lesser number by p; the greater 
will be p + a, and its square p 2 + 2xa + a 2 ; from 
which, p% the square of the lesser number, being de 
ducted, there remains 2xa + a 2 — b: whence if aa 
be subtracted from both sides, there will remain 2ax — 
— aa-, this, divided by 2a, gives x — and