TO THE RESOLUTION OF PROBLEMS. Ill 
Let the first of these equations be subtracted from the 
second, whence y + 2z zz b — a, or y — b — a— 2z; 
also, if the double of the first be subtracted from the 
second, there will come out z.*— x zz b — 2a, or x — z 
4-2a — b: wherefore, if f be put — b — a (zz 144), 
g zz h — 2a {— 10), and for y and x, their equals/ — 2z 
and 2— g, be substituted, our third equation, x i 4- y z 
4- z x — c y will become 22 — 2gz 4- gg 4- ff — 4fz 
+ 4.zz 4- zz zz c; which, ordered, gives z 1 —- 
■ X z zz ; whence, by putting h zz 
<3 0 
p* 098 
(zz ——), and completing the square, &c. 2 is 
found 
4- 
c-P 
P -2 h 2 
-+ /( 
119 + 1) 
3 3' 
2 ' V 6 
50: therefore y (zz f — 2 2) c: 44, and x (zz 2 — g) 
40. 
PROBLEM LIX. 
A traveller sets out from one city B, to go to another C, 
at the same time as another traveller sets out from C for 
B; they both travel uniformly, and in such proportion, 
that the former y four hours after their meeting, arrives at 
C, and the latter at B, in nine hours after: note the ques 
tion is, to find in how many hours each person performed 
the journey ? 
D 
B- {— C 
Let D be the place of meeting, and put a zz 4, h zz 9, 
and x zz the number of hours they travel before they 
meet: then, the distances gone over, with the same uni 
form motion, being always to each other as the times in 
which they are described, we therefore have, BD : 
DC :: x (the time in which the first traveller goes the 
distance BD) : a (the time in which he goes the distance 
DC) : and for the same reason, BD : DC :; b (the time 
in which the second goes the distance BD): x ( the time 
in which he goes the distance DC): wherefore, since it 
appears that x is to a in the ratio of BD to DC, and b to 
x in the same ratio, it follows that a:: a ;: b : x; whence 
x 7 zzab, and x zz \/ab (zz 6); therefore a abzz 10, 
and b 4- \/ab zz 15, are the two numbers required.