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5

Cor. 2. Hence also, if every term on both sides have a common divisor, that common divisor may

3x, a+6 2x +7 be taken away; thus, if +

then,

5 5 multiplying every term by 5, we shall have 3x ta +6=2x+7, or x=1-a.

b 3 7-2

--= then multiplying by c, we shall have ax-6+3=7-X, or ax+x=b+4.

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Also, if

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188. If every term on each side of an equation be divided by the same quantity, the results will be equal : Because, by dividing every term on each side by any quantity, the value of the whole side is divided by that quantity; and, (Art. 51), if equals be divided by the same quantity, the products will be equal.

Thus, if 6a2 + 3x=9; then, dividing by 3, 2aa + 0 =3. Also, if axa tbx=acx; then, dividing every term by

a x2 bx the common multiplier x, we shall have +

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аст

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or ax+b=ac. Cor. 1. Hence, if every term on both sides have a common multiplier, that common multiplier may be taken away.

Thus, if axtad=ab, then, dividing every term by the common multiplier a, we shall have x+d=b.

ab
Also, if
+

;
then dividing by the com-

ax

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с

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a

c

mon multiplier or (which is the same thing) multiplying by, we shall have a+b=4ax.

Cor. 2. Also, if each member of the equation have a common divisor, the equation may be reduced by dividing both sides, by that common divisor.

Thus, if ava?r=ab--ab, or (ax-a2)x= (ax-a%); then, it is evident that each side is divisible by axa”, whence x=b.

Again, if x2 -ao=rta; then, because xa --a =(+a).(x-a), it is evident that each side is divi.

x? - a sible by ta; and hence we have

xta

cta x ta' or x-a=1, and x=a+1.

2

189. The unknown quantity may be disengaged from a divisor or a coefficient, by multiplying or dividing the remaining terms of the equation by that di. visor or coefficient.

b

b Thus, if 2x+4=b, then x+2

and a

2.

2 Also, let +9=17; then, multiplying by 2, we shall have X2+18=17X 2,

, 2

or x +18=34, ... X=34-18. Again, let ax+bx=c-d, or, which is the same, let (a+b)x=c-d; then, dividing both sides by atb, the coefficient of x, and we shall have

cod atb

.

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a

Finally, let =c+d; then, the equation may

6 be put under this form,

x=c+d:

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a

ba

; therefore

1 1 and dividing each side by we shall have <=

b?
1
;

may

be still farther re1 1 duced, because

b ab

b- -a x=(c+d)

ab

ab or x=(c+d) x

b abctabd

a

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a

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190. Any proportion may be converted inio an equation ; for the product of the extremes is equal to the product of the means.

Because, if a :6::x:d; then = (Art. 24), and .. (Art. 187), ad=bx, by clearing of fractions, Let 3x : 5x : : 2x : 7; then 7 X 3x=2x X 5%,

or 21x=10x2 ; and .. 21=10x.
Again, let 5x+20: 4x+4 ::5:x+o1; then,

(5x+20) x(x+1)=5 X(4.0+4);
or, 5x +25x+20=20x+20;
and (Art. 186), 5x2 +25x=20x ;

.. (Art. 188), 5x+25=20. 191. When an unknown quantity enters into, or forms a part of an equation ; and if the equation can be so ordered, that the unknown quantity may stand by itself on one side, with its simple or first power, and only known quantities on the other, the quantity that was before unknown, will then become known.

Thus, suppose 3x+18=5x—2; then, by transposing 3.c and -2, we shall have 18+1=50-3x, or 20=2x;

20
there, x= =10.

2 Here, in the above equation, the value of the unknown quantity x, becomes known, and 10 is the value of a that fulfils the conditions required, which we can readily see verified, by subtituting this value of x in the given equation; thus,

3x=3x10=30, and 5x=5 X 10=50; hence, 3.0 +-18=30+18=48, and 5x-2=50-2 =48; therefore 10 is the true value of x, which answers the condition required, and this value of x is called the root of the equation.

192. Hence the root of an equation is such a number or quantity, as being substituted for the unknown quantity will make both sides of the equation vanish or equal to each other : Thus, in the Simple equation

3.-9+6=0; the value of x must be such, that if substituted for it, both sides must vanish, because the right-hand side is 0; but this value is found to be 1, for by transposition

3x=9-6=3, and dividing by 3, we shall have

3

or x=1;

3 3' therefore 1 is the root of the given equation, which can be easily verified by substituting it for x; thus,

3x −9+6=3X1-9+6=3–9+6=9-9=0.

Hence, the value of the unknown quantity being substituted in the equation, will always reduce it to 0=0.

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§ II. RESOLUTION OF SIMPLE EQUATIONS,

Involving only one unknown Quantity.

193. The resolution of simple equations is the disengaging of the unknown quantity, in all such , expressions, from the other quantities with which it is connected ; and making it stand alone, on one side of the equation, so as to be equal to such as are known on the other side, or, (which is the same thing,) the value of the unknown quantity cannot be ascertained till we transform the given equation, by the addition, subtraction, multiplication, or division of equal quantities, so that we may finally arrive at the conclusion,

x=n, n being a number, or a tormula, which indicates the operations to be performed upon known numbers. This number n being substituted for æ in the primitive equation, has the property of rendering the first member equal to the second. And this value of the unknown quantity, as has been already observed, is called the root of the equation, this word has not here the same acceptation as in (Art. 15).

194. In the resolution of simple equations, involving only one unknown quantity, the following rules, which are deduced from the Articles in the preceding Section, are to be observed.

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