To find three numbers, on these conditions, that a times 
the first, h times the second, and c times the third, shall 
be equal to a given number p; that d times the first, e 
times the second, and f limes the third, shall be equal 
to another given number q; and that g times the first, 
h times the second, and k times the third, shall be equal 
to a third given number r. 
Let the three required numbers be denoted by x, y, 
and z, and then we shall have 
ax 4 by 4 cz zz p, 
dx + cy 4- fz - q, 
gx 4- hy 4- kz zz r. 
From d times the first of which subtract a times the 
second, and from g times the first subtract a times the 
third, and you will have these two new equations 
$ bdy — aey 4 cdz — afz zz dp — aq, 
1- cgz — akz zz gp — ar\ 
viz. 
t bgy -T- ahy 
or, which are the same, 
x 
X 
bd- 
and, bg — ah 
z zz dp — aq, 
z zz gp — or. 
Multiply the first of these two equations by the coeffi 
cient of y in the second, and vice versa, and let the last of 
the two products be subtracted from the former, and you 
will next have cd 
x z~ bg 
aj x bg- 
■bd—oex eg —ah 
ah x dp — aq—• Id—ae x gp—ar; and 
therefore s = bs ~ ah x ~ M ~ ae y Wi 
cd—af X bg — uh — bd — ae x eg—ak 
whence x and y may also be found. 
Let the given equations be 
x 4- y 4 z zz 12, 
2a' 4 3y 4 4z zz 38, 
3a; 4 tfy 4los = 83; 
Or, which is the same thing, let a — l, b — 1, c = 1, p zz 12, 
dzz°, czz3,fzz4, 9—38, gzz3, hzz6, kzrlO, and r~83: 
then these values being substituted above in that of z,