TO THE RESOLUTION OF PROBLEMS. 
I zb — 24a + 120 d— 4c 
” 119 
— 676; whence 2 (c — 
Otherwise. 
Let the first of the required numbers be denoted by x 
(as above); then, the sum of the first and the second, 
being given equal to a, it is manifest that y the second 
must be equal to a minus the first, that is rz a — x, and 
therefore the second number rr 2a — 2a: moreover, the 
sum of the second, and j of the third, being given zr 5; 
it is likewise evident, that y of the third must be equal 
to b minus the second, that is — b — 2a 4- 2-r, and 
consequently the third number itself — 3b — 6a 6x~ 
In the same manner it will appear that y of the fourth 
number — c — 35 + 6a — 6x; and consequently the 
fourth number itself, zr 4c— 12b f 24a — 24c; whenci•„ 
X 
by the question, 4c — 125 f 24a — 24c -f — zr d t 
— öd 4- 20C — 60 6 + 120a 
and therefore x rr 
as above. 
1 iy 
PROBLEM XXXV. 
To divide the number 90 f aJ into four such parts, 
that if the first be increased by 5 fbj, the second de 
creased by 4 (cj, the third multiplied by 3 fdj, and 
the fourth divided by 2 fej, the result in each case, 
shall be exactly the same. 
Let x, y, z, and u be the parts required; then, by the 
question, we shall have these equations, viz. 
x {- y + z -f u — a, and 
. j , * u 
x + 6 — y — c — dz rr- —. 
. e 
Whence, by comparing dz with each of the three 
other equal values, successively, x zr dz — 5, y rr dz 
f c, and u rr dez ; all which, being substituted, for 
their equals, in the first equation, we thence get dz—. 
b i dz 4- c 1- : f dez ~ a; whence dez f Qdz f 2