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EXAMPLES.

1. Given og at 18}

2x+3y=23

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and y.

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23 – 3y Here, from the first equation, f=

2

10+2y And from the second, x=

5

23-3y 10+2y® Whence we have

2

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

28. 15
That is, y=
=5, and a

-4.
19

x+y=a? 2. Given

to figd the values of x and y
C-y=bS
Here, from the first equation, x=d- - Y,

And from the second,
Whence a-y=b+y; or 2y=a-b,

}

.

$b+y,

amb

And therefore y=

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2

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3. Given to find the values of. x .and y.

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Here, from the first equation, x =14

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Therefore, by equality, 14

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- ,

gy And consequently 42– 2y=72

2
Or, by multiplication 84- 4y=144-9y;
And, therefore, also by=144 - 84560,

24
Or, by division, y= =12, and x =14=
5

3

60

6.

RULE II.

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Find the value of either of the unknown quantities in that equation in wbich it is the least involved; then substitute this value in the place of its equal in the other equation, and there will arise a new equation with only one unknown quantity in it; the value of which may be found as before.

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and y.

From the first equation, x=17-2y ; which value, being substituted for xc, in the second,

gives 3(17-2y)-y=2, Or 51 - 6y-y=2, or 7y=51 — 249,

49 Whence y=

=7, and x=17-2y=3.

7 2. Given x+y=137

to find the values of x and y. x-y= 3 From the first equation, x=13-y; which value being substituted for x, in the second, Gives 13-y-y=3, or 2y=13—3=10,

10 Whence y=

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

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Here the analogy in the first, turned into an equa.

tion,

ay
gires bx=ay, or x=

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

apy?
=

tya=C, 62

ay

gives (**)*+ya =c, or

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Whence we have a2 ye+boy=bac, or ya =

al+-62

C

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And, consequently, y=bw 22 462, and x =av ant to

a+

RULE III.

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.

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EXAMPLES

1. Given {

3x+5y=40
x+2y=14

}

to find the values of

and y.

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

Then, subtract the first equation from this, and it will give by - 5y=42 - 40, or y=2.

Whence, also, x=14-2y=14-4=10.

2. Given

{ =

5x-3y=9
2x+5y=16

}

to find the values of 3

and y.

Multiply the first equation by 2, and the second by 5 ; then 10x – 6y=18, and 10x+25y=80. And if the former of these be subtracted from the lat.

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

2.

31 Whence, by the first equation, x=

9+3y_15

=3. 5 5

EXAMPLES FOR PRACTICE.

1. Given 4x +y=34, and 4y+x=16, to find the values of x and y.

Ans. x=8, y=2 2. Given 2x+3y=16, and 3x-2y=11, to find the values of x and y.

Ans. x=5, y=2 3x 2y

61 3. Given

and
+

to find the 5 4 20' 4 5 120 values of x and y.

1 Ans. x=

2.C

9

+

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y=14

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3

4

4. Given * +7y=99, and? +7x=51, to find the val

7 ues of x and y.

Ans. x=7,

2y-5. Given

=
– 12=+8, and put in the

-85 2

+27, to find the values of x and y.

Ans. x=60, y=40 6. Given x+y=s, and x2 - y2=d, to find the values of x and y.

92 +0 82Ans. x=

2s 7. Given xty : a :: -y:b, and 32 --Y2=, to find the values of x and y.

a+b
Ans. x= V
ab

ab

C

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23

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2

2

8. Given ax+by=c, and dx +ey=f, to find the values of x and y.

af-dc Ans. x=

ae

ce-bf
-bd2
y=

ae-bd 9. Given x+y=d, and x2 - ya=b, to find the values of

a2+6 a2 - 6

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x and y.

Aps. x= - y
2a

2a 10. Given x2 +xy=a, and ya +xy=b, to find the val

6 ues of x and y.

Ans. x=

vatb

a

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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 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. 40known quantities bé exterminated from the same num

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