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like manner, it may be shewn that three equations must be given to determine three unknown quantities ; and so on. It mig!.t be further remarked, that these equations must be independent one upon the other. Thus, ar + ly = c, and dr + ey = f are independent, but ar + ly =1, and ax =( by, are not independent, for the one may be derived from the other; and the two, though of a different

rm, amount, in effect, to no more than one equation. What we have now said will lead us to

PROBLEM I.

To exterminate two unknown quantities, or to reduce the two simple

equations containing them, to a single one. (67.) Rule 1.-1. Find the value of one of the unknown quantities in terms of the other, from each of the two given equations.

2. Make the two values thus found equal to each other, and there will arise a new equation containing only one unknown quantity, whose alue may be found as before.

Example 1. Given {3x +2y=10} it is required to find r and

y. 23-34

10+ 2y From the first equation r=

from the second x= 2

5 23-3y_10+2y Consequently

2

5 Or 115– 15y=20+4y, Or 19y=115-20=95 ; 95

23-3y 23—15 That is y =5, And r =

=4. 19

2

2

x+y=a 2. Given r-1

x . From the first equation r=a-y, and from the second x=b+y.

a-6 Therefore a-y=+y, or 2y=2–6, consequently y

2 and x = a - y.

a-l Orr=a

2

2

's and

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3. Given

18+ y || 18

it is required to find x and y.

From the first equation x = 14

2y 3

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And from the second r=24

2 2y

3y T'here ore 14 = 24 -, And 42 – 2y = 72

3

2

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and y.

2y

24 And I=14

S 14

: 6. 3

3 4. Given 4:1 +y = 34, and 4y + x = 16, it is required to find a

Ans. 2= = 8, and y=2. Зу 9

3.1 2y

61 5. Given + and +

it is required 5 4

4 5 120

2r

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20

to find x and

y.

=

Ans. I =

2'

and y

3

6. Given x+y=s, and x-yo=d, it is required to find x and y.

6 td sa-d 2s

2s

Ans.

and y=

7. Given t-y= -d, and x:y

nim, it is required to find x

and y.

{

Rule 2.1. Consider which of the unknown quantities must be first terminated, and find its value in that equation where it is least involved.

2. Substitute the value, thus found, for its equal in the other equation, and a new equation will arise with only one unknown quantity, whose value may be found as before

*+2y=177 Ex. 1. Given

3r— ý= 2 it is required to find z and y. From the first equation x=17-2y. And this value substituted for r in the second, gives 3(17-2y)-y=2, Or 51-6y - y=2, or 51—7y=2; that is 7y = 51-2=49;

49 Whence y= = 7, and x = 17-2y =

= 17-14 = 3. 7

x+y=131 2. Given

*-y= 3 From the first equation r=13-y. And this value, substituted for x in the 2d, gives 13-y-y=3, ON 13—2y=3, That is, 2y=13-3=10,

10 Whence

y=

=5, and x=13-y=13-5=8, 2

Sa:0:: X:y! 3. Given ro+y=0

it is required to find x and y. The first . analogy turned into an equation is bx=ay, or x=7

6

ay an

C, or

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cl9 Or aʻy? +boy' =.cl", or y'=

a? + 12 ch?

ca?
y=

a
and

)!

a? +62 4. Given 2x+3y=16, and 3x-2y=11, it is required to find x and y.

Ans. x=5 and y=2. 5. Given +*

7

Ans. x=7 and y=14. 6. Given 2–12=X+8, and

2y-2
+
-8=

+ 27, it is re

4 quired to find and y.

Ans. x =60 and y=40. 7. Given a :b::I: y, and x-yo=d, it is required to find x and y.

da?

01:3 Ans.

and

bs Rule 3.-1. Multiply or divide the given equations by such numbers or quantities as will make the term which contains one of the unknowr. quantities the same in both equations.

2. Then, by adding or subtracting the equations, according as may oe required, a new equation will arise with only one unknown quantity as before.

$3x + 5y=402 Ex. 1. Given 1 x+2y=14) it is required to find x and

y. First, multiply the 2d equation by 3, and it will give 3x+by = 42.

Then from the last equation subtract the first, and it will give by-5y=42-40, or y=2, :: x=14–2y=1444=10.

2. Given { 20+ 5y = 18$ it is required to find x and y.

Let the first equation be multiplied by 2, and the 2d by 5, and we shall have 10r by=18, 10x +25y=80; And if the former of these be subtracted from the latter, it will give

62 Sly=62, or y= 2,

31

9+3y And consequently r = s by the first equation,

15 Or x=

3.

5} y

5

9+6

5

5

Another method.

Let the first equation be multiplied by 5, and the second by 3,

and we shall have { 2577159=48 Now, let these two equations be added together,

and the sun will be 311=93,

16-21 And consequently y = s by the second equation,

5 16-6 10 Or y =

-2 as before. 5

Any of the three foregoing rules may be applied to the solution of the following

1. Given *# 2 +8y=31, and

Miscellaneous Examples. +

y + 5

+ 10x=192, it is required to 3 Índ I and y.

Ans. x=19 and y=3 27y

2y + 2. Given

+14=18, and + 16 = 19, it is required to 3

3 find x and y.

Ans. x=5 and y=2 3. Given

77–3.2

--y=11, it is required to find ... 6%.

2

Ans. r=6, and y=8. 4. Given ar +by=c, and dx +ey=f, it is required to find x and y.

Ans, x=
ce-of

ae-bd

2x+3y +

=8,and

and y.

andy=

af-dc

ac-ud

PROBLEM II.

To exterminate three unknown quantities, or to reduce three simple equations, containing them, to a single one.

Rule.-1. Let x, y, and %, be the three unknown quantities whicb are to be exterminated.

2. From each of the three given equations, find the value of x. 3. Compare the first value of with the second, and there wil arise an equation involving only y and z.

4. Compare, in like manner, the first value of x with the third, and there will arise another equation involving only y and %.

5. From these two equations, find the values of y and %, according to the former rules, and x, y, and %, will be exterminated as was required.

Note. In nearly the same manner, any pumber of unknown quantities may de exterminated; but practice will often suggest much shorter methods for performing the operation.

Eramples. x + y + x=zy x+2y+3x=62 it is required to find x, y, and z. 4r+ sy+x=10)

1. Given

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2. Given

fr+by+tx = 62
5x + ty+$47 it is required to find x, y, and 2.
Ir + sy + z = 38)

}

First, the given equations, cleared of fractions, become

12r+ 8y + 6z=1489
20x + 15y + 12x=2820
30.3 +244 +20%=4560

And, if the second of these equations be subtracted from double the first, and three times the third from five times the second, we shall have

4x+y=156 10x+3y=420

And again, if the second of these be subtracted from three times the first, it will give

48
12x – 10x=468-420, or r= -=24;

2

Therefore y=156–4x=60, and x

1488-87- 12.x

6

120.

3. Given x+y+z=53, x+2y+32=105, and x+3y + 4z=134, it is required to find x, y, and %.

Ans. x=24, y=6, and z=23.

4. Given x+y=a, x+x=1, and y+x=c, it is required to find 1, y, and %.

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