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bx = dy,

Here the analogy in the first, turned into an equation,

ay
gives bx
ay, or x =

b?
And this value, substituted for x in the second,

a'ya
C, or

+ y = , 6

82

bc. Whence we have aʻy+ b*y= boc, or y* = m2 +62

c

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And, consequently, y=bV

and x=0V a' + 622

a2 + 72

EXAMPLES FOR PRACTICE.

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+

= 40.

1. Given

+ 730 51, to find the values 17 of x and y.

Ans. x=7, and y 14. Y x + y

2y 2. Given 12 =

+ +
8

+
2
4

5
3

4 27, to find the values of x and y. Ans. x = 60, y

3. Given x+y=s, and x2 - y2 = d, to find the values of w and y.

82 + d s2d Ans. x =

Y 2s

2s 4. Given 50 3y = 150, and 10x + 154 825, to find x.

Ans. * = 45, and 5. Given x +y=16, and ix:y::3:1, to find & and y.

Ans. x = 12, and 4. Y x 12, and y + 9, to find x and y. 2

y =

4. 7. Given X: Y:: 3:2, and 23 - ya

- y2 = 20, to find x and

Y.

and y.

Y = 25.

y

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Ans. *

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10, and

Ans. * x + y

6, and

y=4.

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2x

Y

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Y 8. Given

+13 and
+57-16

+ 2 27, to find u and y.

Ans. 20

and

? RULE 3.—Let one or both of the given equations be multiplied, or divided, by such numbers, or quantities, as will make the term that contains one of the unknown quantities the same in each of them; then, by adding, or subtracting, the two equations thus obtained, as the case may require, there will arise a new equation, with only one unknown quantity in it, which may be resolved as before.*

* The values of the unknown quantities in the two literal equations ax+by=c, add a'z-_by=c', may be found in general terms, by multiplying the first by a', and the second by a, and then working according

EXAMPLES.

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40 1. Given a +- 2y = 14 to find the values of x and y.

First, multiply the second equation by 3, and it will give 3х + 6 = 42.

Then, subtract the first equation from this and it will give 6y - 5y = 42

. 40, or y

2.
Whence, also, w I 14 – 2y = 14 -4 = 10.
58 Зу 92

and
2x - 5

16 Multiply the first equation by 2, and the second by 5; then 10x by = 18, and 10x + 254 And if the former of these be subtracted from the latter,

62 there will arise 31y = 62, or y=

= 2,
31
9 + 3y

15
Whence, by the first equation, a =

5

5

2. Given {

80.

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3.

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x and

2. Given 3x +-7y=79, and 2y = 9+ to find x and

y. 2 Ans. * = 10, and y

= 3. Given 30x + 40y= 270, and 50x + 30y 340, to find

Ans. 2 = 5, y.

Y :

3. 4. Given 3x 3y = 2x + 2y, and a +y:xy :: 3:5, to find x and y.

Ans. x = 10, and 2.

Y 5. Given x'y + xy' = 30, and 2 + y = 35, to find x and y.

Ans. x = 3, and

Y

2.

5, and

and <=

ac ca to the last rule, when the results, so determined, will be y

ab ba'' cb'bc'

ab'basi which solution may be applied to any particular case of this kind, by substituting the numeral values of a, b, a', b', in th place of the letters, and observing, when either of them is negative, to change the signs accordingly.

Where the numerator is the difference of the products of the opposite coefficients in the order in which y is not found, and the denominator is the difference of the products of the opposite coefficients taken from the orders that involve the two unknown quantities. Coefficients are of the same order which either affect no unknown quantity, as cand c'; or the same unknown quantity in the different equations, as a and a'. Coefficients are opposite when they affect the different unknown quan. tities in the different equations, as a and b', á' and b.--Ed.

2x + y

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atb

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a

с

2

of x and y.

се

af de

de

ae

Ans. 2 =

.

2a

2a

a

Ans. *

5y

2y Y 6. Given

3, and 8

-2 5

4 2 3 to find x and y.

Ans. X = 12,
='12, and

Y

= 6. 7. Given x + y:a:: *

y:b, and x2 - y2 = c, to find the values of x and y.

- b
Ans. x =


, у


ab

2 ab. 8. Given ax + by = c, and dx + ey=f, to find the values

bf
Ans. x =

bd
Y

bd 9. Given a +y = a, and x2 - ya=b, to find the values of w and y.

a2 tb

aa_5

, Y 10. Given x2 + xy = Q, and ya + xy

= Q, and ya + xy=b, to find the values of x and y.

b

(a + b) ✓(a + b) Of the resolution of simple equations, containing three or more

unknown quantities. When there are three unknown quantities, and three independent simple equations containing them, they may be reduced to one by the following method. *

Rule.--Find the values of one of the unknown quantities in each of the three given equations, as if all the rest were known; then put the first of these values equal to the second, and either the first or second equal to the third, and there will arise two new equations with only two unknown quantities in them, the values of which may be found as in the former case; and thence the value of the third

Or, multiply each of the equations by such numbers, or quantities, as will make one of their terms the same in them all; then having subtracted any two of these resulting equations from the third, or added them together, as the case may

+ b)?Y

* The necessity for observing that the given equations in this and other similar cases are so proposed as to be.independent of each other, will be obvious from the following example :-

2—24+z=5; 2x+y --2=7; 2+31-23=2; where, if it were required to determine the value of x, y, and z, it will be found, by eliminating c from each of them, and then equating the results, that

51 32 -3, and 54 -- 3z 3; which equations, being identical, or both the same, furnish no determi.. nate answer. And, in effect, if the three equations be properly. examined, it will be found, that the third is merely the difference of the first and second, and consequently involves no condition but what is contained in the other two,

require, there will remain only two equations, which may be resolved by the former rules.

And in nearly the same way may four, five, &c., unknown quantities be exterminated from the same number of independent simple equations; but, in cases of this kind, there are frequently shorter and more commodious methods of operation, which can only be learnt from practice.*

EXAMPLES.

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2y - 32.

22,

2

x + y + 2 1. Given

8C + 2y + 32 = 62 to find x, y, and z.

2x + y +12 10
Here, from the first equation, x = 29 -- Y - 2.
From the second, a = 62

2 1
And from the third, x

20

39-22 Whence 29 - Y z=62 – 2y 32,

2 1 And, also, 29-y-=20

3Y~2. From the first of which, y= 33

3 And from the second, y= 27

22,

3 Therefore 33 - 22 = 27

or 2 =
22,

--- 12,
Whence, also y 33 - 22= 9,
And x = 29

y - 2

8. 2:0 + 4y

3z 22. 2. Given 410

2y + 5z = 18 to find x, y, and z.

6x + 7y * The values of the unknown quantities in the three literal equations,

a.+by+cz=d; a'xtb'y + c'z=d;a"x+bʼy +=d"; may be exhibited in general terms, like those before mentioned, as follows:

db'c'' - dcb+cd'b' - bd'c' tbc'a'' - cb'd '

-ac'b'' +ca'b'' ba'ć'' + bc'a" - coʻa"
ad'c' ac'd" toca'il da'c"+dc'a' cd'a'
Y

ac'b' fcab" ta'ć'' + bc'a
abi adb" da'b' - ba'd' #bd'a''-db'a"
ab'c'' ac'6" +-ca'b'_ba'c" +bc’a

cb'a"" ! which formula, by substitution, may be employed for the resolution of any numeral case of this kind, as in the instance of two equations before given. The numerator of any of these equations, such as z, consists of all the different products, which can be made of three opposite coeflicients taken from the orders in which z is not found; and the denominator consists of all the products that can be made of the three opposite coefficients taken from the orders which involve the three unknown quantities.

= 63

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abic

ab'c

cha

X, Y,

and z.

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15, and

Here multiplying the first equation by 6, the second by 3, and the third by 2, we shall have

12x +- 24y

18% = 132, 123 6y + 15z = 54, 12x + 147

22 126. And, subtracting the second of these equations successively from the first and third, there will arise

30y 332 = 78,

20y 172 = 72. Or, by dividing the first of these two equations by 3, and then multiplying the result by 2,

204

22% = 52,

20y - 172 = 72, Whence, by subtracting the former of these from the latter, we have 5z = 20, or 2 = 4. And, consequently, by substitution and reduction,

y=7, and x = 3. 3. Given x+y+z=53, x + 2y +32= 105, and x + 3y+ 42 = 134, to find the values of

Ans. X = 24, y =

24, y = 6, and 2 = 23. 1 1

1 1 1 4. Given x + 24 + 3

3a + y = 52 1 1 1 4 = 12, to find the values of x, y,

Ans. x = 12, y = 20,2= 30. 5. Given 78 +- 5y + 2z = 79, 8x + 7y + 9% = 122,

and x + 4y + 5z = 55, to find the values of x, y,

Ans. x =

4, y = 9,2= 3. 6. Given x + y = a, x +z=b, and y +2=

C, to find the values of x, y, and z.

batc

and 2 ==
2
2

2
2
Y

Y kry. Given

Y + 62,7

+ , 2 3 4

4 5

4 38, to find x, y, and z.

60, and 2 = 120. 8. Given z ty = x + 100, y -- 2x = 22 – 100, and % + 100 = 3x + 3y, to find 2, y, and 2.

Ans. x =
x = 0, 3

45.67, and 2 5347 9. Given x +y+z= 7, 200 3=y + 3z, and 5x + 5z - 3y + 19, to find x, y, and z.

Ans. x

4, y = 2 and 2 = 1. 10. Given 3x + 5y 42

25, 5x

2y + 3z = 46, and 3y + 5z

62, to find x, y, and 2.

Ans. x=7, y

7, y = 8, and >= 9

x + 7

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and z.

and %.

a

btc

a + b

C

Ans. Y

Ans. 30

24, y

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