TO TïïE RESOLUTION OF PROBLEMS» 
115 
ed together, must, by the conditions of the question, 
be equal to 500 miles, the whole given distance ; which 
we will call b, and then we shall have p f m x x -f 
x x x — l x e— d 
Q 
— b, or fx + 
gx X X— 1 
i 
= by 
writing / := p + m, and g — e — d\ which equation is 
reduced to gx 1 — gx + 2/r = 2b, or x z — x +• 
^ ; whence, by completing the square, See. x 
Í5 O 
comes out 
— ~l— •- + —. Butin 
g 2 I g 2 
the particular case proposed, the answer is more simple, 
and may be more easily derived from the first equation 
—-— , x x x—1 X e—d . „ , . 
p + ?KXrl —— rr b\ tor, e being =: d, 
xxx — 1 x e 
will here entirely vanish out of 
the equation; and therefore x will be barely — — 
n J p -p m 
— —— — 5. The same conclusion is also readily 
40 + bO J 
derived, without algebra, by the help of comipon arith 
metic only: for seeing the sum of the two distances tra 
velled in the first day is too miles, and that the post B 
increases his distance, everyday, by just as much as the 
post A decreases his, it is evident, that between them 
both, they must travel 100 miles every day; therefore, 
if 500 be divided by 100, the quotient 5 will be the 
number of days, in which they travel the whole 500 
miles-. 
PROBLEM LXIV. 
Two persons, A and B, set out together from the same 
place, and travel both the same wag: A goes 8 miles the 
first day, 12 the seco/id, 16 the third, and so on, increasing 
4 miles everg day: but B goes 1 mile the first day, 4 the 
second, g the third, and so on, according to the square of 
the number of days : the question is, to find how many 
days each must travel before B comes up again with A* 
i 2