ALGEBRA. 
Ex, 2. Let c Ito find x and y. 
( m x — nyzzzd ) 
From the first, max -|- mb y ~mc 
from the other, max — nay xxz ad 
by subtraction, mby + nay — me — ad 
, , ~ me — ad 
mb -j -na 
Again, nax -\-nby = n c 
mb x — nby z=:bd 
by addition, na -j- m b . x = nc -f- l d 
,, - nc - 4 - bd 
therefore, x = 
na-\-mb 
If there be three independent simple equa- 
tions, and three unknown quantities, reduce 
two of the equations to one, containing only 
two of the unknown quantities, by the pre- 
ceding rules; then reduce the third equation 
and either of the former to one, containing 
the same two unknown quantities ; and from 
the two equations thus obtained, the unknown 
quantities which they involve may be found. 
The third quantity may be found by sub- 
stituting their values in any of the proposed 
equations. 
r 2i4-3M-|-<i2; = 16q rT , ■ • 
Ex. Let <3x-|-2y — 5z = 8 C Tofin ^' 
(5x-6y + Szz=6 ) V ' and fc 
From the 2 first equat. 6r-|-9y-|-12z = 48 
6.r -|-4y — 10 z = 16 
by subtr. 5 y -j- 22 z = 32 
from the l«and 3 rd , 10 x -j- 15 y -f- 20 z = 80 
lOr? — 12^-f-6z = 12 
by subtr. 27 y -{- 14 z = 68 
and 5 y -f- 22 z = 32 
hence 135 y -j- 70 z — 340 
and 1 35 y -j- 594 a = 864 
by subtr. 524 a = 524 
a = 1 
5 y -J- 22 a = 32 
that is, 5 y -J- 22 = 32 
5y = 32 — 22= 10 
__10_ o 
y ^ 
2a?-j-3y-j-42i = 16 
that is, 2a;-|-6-f-4 = 16 
2 a = 16 — 6 — 4 = 6 
a = 3. 
The same method may be applied to any 
number of simple equations. 
That the unknown quantities may have 
definite values, there must be as many inde- 
pendent equations as unknown quantities. 
Thus, if x -j- y = a, x = a — y ; and as- 
suming y at pleasure, we obtain a value of x, 
such, that x -j- y = a. 
These equations must also be independent, 
that is, not deducible one from another. 
VOL. I. 
Let x -(- y = a, and 2 x 2 y = 2 a ; this 
latter equation being deducible from the for- 
mer, it involves no different supposition, nor 
requires any thing more for its truth, than 
that x -j- y = a should be a just equation. 
PROBLEMS WHICH PRODUCE SIMPLE 
EQUATIONS. 
From certain quantities which are known, 
to investigate others which have a given rela- 
tion to them, is the business of Algebra. 
When a question is proposed to be resolved, 
we must first consider fully its meaning and 
conditions. Then substituting for such un- 
known quantities as appear most convenient, 
we must proceed as if they w'ere already de- 
termined, and we wished to try whether they 
would answer all the proposed conditions or 
not, till as many independent equations arise 
as we have assumed unknown quantities, 
which will always be the case if the question 
be properly limited; and by the solution of 
these equations, the quantities sought will be 
determined. 
Prob. X. To divide a line of 15 inches int. 
two such parts, that one may be three-fourths 
of the other. 
Let 4r = one part, 
then 3.r = the other. 
7 x = 15, by the question, 
15 
x : 
4r: 
3a?: 
: 8 
' 7 
60 
: 7 = 
45 3 
: — = 6 the other. 
7 7 
rj, one part, 
Prob. 2. If A can perform a piece of work 
in 8 days, and B in 10 days, in what time 
will they finish it together ? 
Let x be the time required. 
In one day, A performs - part of the work ; 
therefore in x days, he performs ~ parts of 
8 
it; and in the same time, B performs ~ 
parts of it ; and calling the work 1, 
-4-— = 1. 
8 ' 10 
10x-[- 8 a; = 80 
18 a: = 80 
— 4 ^ days 
a_-_4i a _4- days. 
Prob. 3. A and B play at bowls, and A 
bets B three shillings to two upon every 
game; after a certain number of games it 
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