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4

x =

ax

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.'. Ar

a

с

a

20x

20x Multiplying by 20, -200=

-60,

5

5x — 200=4x-60, by transposition, 5x - 4x=20a-60,

= 200-60. bx 2a Ex. 5. Given

to find the value of x. 5 5 5

5ax 5bx 5 x 2a Multiplying by 5,

5 5 5

-bx=2a, by collecting the coefficients, (a-b)x=2a,

2a ... by division, x=

-6° 2ax 36x 5x Ex. 6. Given +

+3, to find the value

2 of x.

Here aac, the product of 2, a, and c, being the least common multiple, Multiplying by aac, 4a2x +3abcx=10cx+6ac, by transposition, and collecting the coefficients, we shall have (4a2 + 3abc-10c)x=6ac,

бас .. by division, x=

4a +3abc--10c

4 5x +14 Ex. 7. Given 3x

-4 4

3 the value of x. Multiplying by 12, the least common multiple,

we have 36x -3x+12-48=20x+56-1, by transposition, 36x-3x-20x=56-1+48--12,

or 13=91,

133 91 by division,

x=7. 1313

56 63 Ex. 8. Given

to find the value 5x+3 1400 of x.

Ans. X=1.

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19, to find

.

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-5°

Ex. 9. Given 571

+

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x+ x +2 x +3

16

to find 3

4 the value of x.

Ans. x=

=13. X-3

19 Ex. 10. Given + -20

to find the 2 3

2 value of x.

Ans. x=231.

11 19Ex. 11.. Given x +

to find the 3

2 value of x.

Ans. x=5. C -5

284 Ex. 12. Given

+6x

to find the 4

5 value of x.

Ans. x=9.

2x+6 11x37 Ex. 13. Given 3x + =5+

, to find 5

2 the value of x.

Ans. x=7. 6x 4 18 - 4x Ex. 14. Given

tx, to find 3

3 the value of x.

Ans. x=4. ar -3 bx + 2 9 Ex. 15. Given

5
3
2

3

87 find the value of x.

Ans. -=

106 +30-60

-1 x+3 2x + 1 3 Ex. 16. Given

to 7 2

4 find the value of x.

94.

2x

X-1

,to

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14

Ans.

RULE IV.

198. If the unknown quantity be involved in a proportion, the proportion must be converted into an equation (Art. 190); and then proceed to resolve this equation according to the foregoing Rules.

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143 •x=24

Ex. 1. Given 3x-2:4:: 5x -9:: 2, to find the value of x. Multiplying extremes and means, we have

2(3x - 2)=4(5x-9),

or 6x — 4=20x - 36, by transposition, 3.0 — 20x=-36+4,

or - 14x=-32, by changing the signs, 14x=32,

14x 32 by division,

14 Ex. 2. Given 3a : x : :b+5 : x 9, to find the value of x. Multiplying extremes and means, we have

3a.(x--9)=r.(6+5),

or Sax - 27a=bx+5., by transposition, 3ax-bx-5x=27a, collecting the coeff's., (3a-6-5)x=27a,

27a .. by division, x=

3a-b-5°

2 3 Ex. 3. Given

5 :

to find the 4

3° 4'
value of x.
Multiplying extremes and means, we have

2
ž•(x— 5),

3
3x - 15 2ic — 10
16

3
by clearing of fractions, 9x - 45=32x – 160,

by transposition, 9x — 32x=45 – 160, collecting and changing signs, 23x=115,

23x

115 by division,

; 'x=5,

23 23 Ex. 4. Given 2x — 3: -1 :: 4x : 2x + 2, to find the value of x.

x - 5

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Multiplying extremes and means, we shall have

(2x-3).(2x+2)=40(x-1),

or 4x2 – 2x—6=4x2 - 4x, by transposition, &c., 2x=6,

.. by division, x= =3. Ex. 5. Given atx:b::c-x:d, to find the value of x in terms of a, b, c, and d.

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Multiplying extremes and means, ad+dx=bc-bx,

by transposition, +d=bc-ad, or (6+d)x=bc-ad,

bc-ad .. by division, x=

b+do -1

3 Ex. 6. Given

: * +2 :1, to find the 3

4 4 value of x.

X - 1 3x +6
Multiplying extremes, &c.,

3

4 clearing of fractions, 4x --4=9x+18,

by transposition, 4x --- 9x=18+4, changing the signs, &c., 5x=-22,

22 .. by division, x=- =-4.

5

3x Ex. 7. Given 23 -1: x+1::

to find the 3 :

2 4' value of x.

Aps. x=-115

1 Ex. 8. Given x+3:a::b to find the value

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Involving only one unknown Quantity. 198. It is necessary to observe that an equation expressing but a relation between abstract numbers or quantities, may agree with many questions whose enunciation would differ from that of the one proposed : but the principles of the resolution of equations being independent of any hypothesis upon the nature and magnitude of quantities; it follows, therefore, that the value of the unknown quantity substituted in the equation, will always reduce it to 0=0, although it may not agree with the particular question. This is what will happen when the value of the unknown quantity shall be negative ; for it is evident that when a concrete question is the subject of inquiry, it is not a negative quantity which ought to be the value of the unknown, or which could satisfy the question in the direct sense of the enunciation.

The negative root can only verify the primitive equation of a problem, by changing in it the sign of the unknown; this equation will therefore agree then with a question in which the relation of the unknown to the known quantities shall be different from that which we had supposed in the first enunciation. We see therefore that the negative roots indicate not an absolute impossibility, but only relative to the actual enunciation of the question.

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