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The rules of Algebra, therefore, make not only known certain contradictions, which may be found in the enunciations of problems of the first degree ; but they still indicate their rectification, in rendering sub

tractive certain quantities which we had regarded as # additive, or additive certain quantities which we had

regarded as subtractive, or in giving for the unknown quantities affected with the sign

Hence, it follows, that we may regard as forming, properly speaking, but one question, those whose enunciations are connected to one another in such a manner, that the solution which satisfies one of the enunciations, can, by a simple change of the sign, satisfy the other.

We must nevertheless observe that we can make upon the signs and values of the terms of an equation, hypotheses which do not agree with the enunciation of a concrete question, whereas the change which we will make in this enunciation might be always represented by the equation.

These principles, which will be illustrated by examples, are applicable to equations of all degrees, and to determinate equations containing many unknown quantities. The question which conducts to the equation

ar+b=cx+d, is not well enunciated for a>c, and 6>d, since the first member is greater than the second. Thus the formulæ,

d-b

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gives for x a negative value ; but by rendering the unknown x negative, the equation is changed into the following

b--ax=d-ex,

a

which is possible under the above relations between a and c, b and d, and which gives then for x an absolute value.

If we have b>d and c>d, the two subtractions become impossible in the formula

d-b

; but in order to resolve the equation, let us subtract cm +-6 from both members, which would be impossible, because that co+b is greater than each of the two members :- we must therefore, on the contrary, take away axtd from both sides, and it becomes

b--d=C --UX; from whence we deduce

b-d

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Ca

This formulæ, compared to thc preceding, differs from it in this, that the signs of both terms of the fraction, are changed.

We may therefore conclude, that we can operate on negative isolated quantities, as we would do if they had been positive.

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199. These principles will be clearly elucidated, when we come to treat of the solutions of Prob. lems producing simple Equations: we shall now proceed to illustrate the Roles in the preceding Section, by a variety of practical examples.

3x -11 5x -5 Ex. 1.. Given

21+
16

8 97-7X

to find the value of x. 2

3

9

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Multiplying both sides of the equation by 16, the least common multiple of 16, 8, and 2, we shall have 336+ 3x – 11=10x -10+776-56x;. .. by transpostion, 3x - 10x + 56x=11-10+776-336,

or 49x=441;

441 by division x= x=9.

49 3x 5

2x- -4 Ex. 2. Given +

-12

to find the 2

3 value of x.

Multiplying both sides of the equation by 6, the product of 2 and 3, which is the least common multiple, we have

6x +92—15=72-4x+8; by transposition, 6x +92 +4x=72+3+15,

or 19x=95 ;

95
by division, x= .. I=

=5. 19

2x - 4 In this example, when the fraction

is

3

12xC 24 multiplied by 5, the result is

(40

3 8)=-4x+8, or which is the same thing, when the sign stands before a fraction, it may be transformed, so that the sign + may stand before it, by changing the sign of every term in the numerator; therefore, we make the above step - 4x+8, and not 4x -- 8.

-1

2x2 Ex. 3. Given 43

=x+ +24, to find 2

5 the value of x.

Multiplying by 10, the least common multiple, and we have,

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40x — 5x+5=10x+42–4+240, by transposition, 40x45x -- 10x - 4x=240–4-5,

or, 40. — 13x=231 ;
and 21x=231,

231
by division, x=

x=11

21

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Ex. 4. Given 2x+1=5x.—2, to find the value .

=–

2

of x.

or

Multiplying by 2; we have,

4x-.+2=102-4, .. by transposition, 4x-x-10x=-4_2,

–7x=-6, by changing the signs, 7x=6, '. by division, x=

7 Ex, 5. Given 3ax -- 2bx=36-a, to find the value

6

of c.

Here, 3ax 2bx=(32—26)x, by collecting the coefficients of x. Therefore,

(3a-2b)x=3b--a,

3b-a by division, x

3a-26 Ex. 6. Given bx tox=2x+3a, to find the value

of x.

в За

by transposition betor-2c=-3a,

or (6-1)x=3a, .. by division,

b-1

200 Ex. 7; 'Given

-4x+

to find the va

d lue of x. Multiplying by abd; we have,

3bdr---abcd+adx=4abdr+-2abx,

ct

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d

by transposition, 3bdx tadx-4abdx-2abx=abcd, or (36d+ad-4abd - 2ab) x=abcd,

abcd .. by division,

3bd tad - 4abd - 2abo Ex. S. Given +=b+c, to find the value

5 6 6

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of x.

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Multiplying by 30, the product of 5 and 6, the product becomes

60—5% +50=306 +300; by transposition, 6x — 5x=306+300—50,

and ..x=306 +300-59. 12

14+ x Ex. 3. Given .

: 5x

::1:8, to find 9

3 the value of x. Multiplying extremes and means, we have

96-8x 14 ta

-50-
9

3
Multiplying by 9, the least common multiple,

96 - 8x=45x – 42-3x, by transposition, -45%-8x +3r=-96-42, by changing the signs, 450+8x— 3x=96+42,

or 50x=138,

138 19 .. by division, a -2

50

25 -b 620 bx-a Ex: 10. Given +

to find the 4 3 2 3 value of x.

Multiplying by 12, the least common multiple of the denominators, and the equation will become,

3a2---35+4=666-4bx+4a, by taking away 4a from each member, we shall have

3ax-36=6bx46x=2bx,

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