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or circumstance which has no necessary connection or relation to the result required.

The next step is to assume letters to denote the data and quæsita of the problem, taking care to avoid ambiguity or contradiction in the assumptions.

After this designation of the known and unknown quantities, the problem must, with the aid of symbols of operation, be translated into algebraical language ; that is, the conditions of the problem must be denoted by equations, making no distinction between the known and unknown quantities ; and when this is effected it will then appear whether the problem is determinate or indeterminate.

Having formed the equations, the last process consists in their solution. It will be found that some methods are more operose and tedious than others. The most direct and simple methods of solution are to be preferred when such are possible; but these are not always subject to any general rule, but rather depend on the skill which is acquired by experience.

In the solution of equations involving more than one unknown quantity, it will sometimes be found that they are so involved that they cannot easily be separated by any of the ordinary methods. In such cases some artifice is generally possible, by means of which the equations may become simplified and their solution rendered practicable. It is impossible to lay down any rules for the employment of artifices in the solution of equations. The manner in which the unknown quantities are involved only can suggest what artifices are possible for rendering the equations more easy of solution. Instead of assuming single letters to denote unknown quantities, it will be found convenient in some cases to assume for them the sum and difference of the two unknown quantities, in others the product and the quotient, &c.

All homogeneous equations of two unknown quantities may be simplified by assuming for one of them some unknown multiple of the other, and making the substitution in the two equations. By this means both of the unknown quantities can be eliminated, and the resulting equation will involve only the quantity which denotes the unknown multiple.

The same method can be extended to equations involving three or more than three unknown quantities.

It has been pointed out that equations proposed for solution may be deducible from, or contradictory to, one another. In the former case, in the attempted solution, some quantity may be found equal to itself, and in the latter some absurdity will follow, as a greater quantity equal to a less. And this dependency or inconsistency of equations is not always obvious until the identity or absurdity appears in the final result of the operation.

SIMPLE EQUATIONS. Art. 1. Simple equations, or equations of the first degree, may involve one, two, three, or more than three unknown quantities.

If a simple equation involve only one unknown quantity, the equation admits of only one value of that quantity.

The solution of all equations in general will be found to depend on the axioms stated on the eighth page of Section IV.

In the application of the second and third axioms in the solution of equations, it will be seen that if an equal quantity be added to or subtracted from each side of an equation, the process is the same as simply transposing or removing the quantity from one side to the

of the equation with its algebraical sign changed; as the addition of ta to each side of the equation mx—a=b, is mx—ata =b+a; and is equivalent to the removal of -a from one side to ta on the other, as mx=b+a. And the subtraction of ta from each side of the equation mx +a=b, is muta-b=b-a, and is equivalent to the removal of ta from one side to - a on the other, as mx =

= 6-a. Hence the second and third axioms are united in the general rule, that any quantity may be transposed from one side to the other side of an equation by changing its sign. By the application of this rule the terms involving the unknown and known quantities can be separated and placed together, the former on one side and the latter on the other side of the equation; and when the equation is reduced to the form MX = C, by the seventh axiom the unknown quantity is determined to

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And by this axiom an equation may be simplified if every term of it contain the same common factor which does not involve any unknown quantity.

Also by the sixth axiom an equation may be cleared from fractions by multiplying each side of the equation by the denominators of the fractions, or by their least common multiple, as the equation +=o becomes bx+ax = abo by multiplying each side by a and 1, the denominators of the fractions.

And further, the signs of all the terms of an equation may be changed, as the process is equivalent to multiplying the terms on each side by -1.

It is also obvious that if the same quantity with the same sign be found on both sides of an equation, it may be removed from each side.

If a term involving the unknown quantity in an equation be affected with a fractional index, by means of the eleventh axiom the fractional index can be reduced to an integral index, and the separation of the known and unknown quantities become possible, as if (mx+b)*=c, by raising each side of the equation to the second power, it becomes mx+b=, and the known quantity b can be separated from x the unknown quantity.

2. To solve the general equations a x+by=C, aqx+by = C,, involving two unknown quantities, or to find the values of x and y which satisfy both equations.

When there are given two independent equations of the first degree which simultaneously involve two unknown quantities x and y, it is obvious that, in order to determine the value of one of them, there must first be deduced from the two given equations an equation which shall contain only one of the unknown quantities.

There are three methods by which the values of x and y may be determined from these equations, aqx+b1y = C, Qqx+b.y = Cz.

First method: By finding the value of one of the unknown quantities in terms of the other and known quantities, from one equation, and substituting it in the other equation. Let the value of x be found from the first equation ax+by=0.

6,-by By transposing by, a,=61-by, and x = then substituting

ay this value of x in the second equation, and+by=cy, it becomes ag(91-by)

+by=ca, clearing the equation from fractions, ay Ag(ab,y+abay = @ica, and by transposition (a,b,—azbe)y=a,c—0,01,

0,09--0901 and the value of x is found by substituting this value of y in either of the two given equations.

с —b,y , ,у ъ ас. — 1,0
ay an an az az abz-azbi'

bac, -6,62

a, b, a,b, And these values of x and y, when substituted in the given equations, .satisfy both equations by reducing each to an identity.

a,(6,61–6,ca), b(,c—a,0)
Thus ajx+by=

+
ab,-azbi a,bz-azba
a,b,c,—a,b,cz +a,b,c2-a,b,cı

a,b,a,b,
(a,b,a,b,) ei

abe-Agby

az(bzę, —6,62), bz(a,02—Agci) And ac+by =

a, b, - a,b,

+

a,be-a,b, a,b,c,-a,b,c,+a,bacı-a,b261

cybe-azbi

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or =

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ac
(3,62—azb;)og

a, b,- ab Second method: By equating the values of the same unknown quantity from each equation.

From the first equation, ajz+b,y=C1,
From the second, aqx+bay=C2, X=

17bay Next equating these values of x,

4-by_c9-by
a

A2
Clearing the equation from fractions, azcı - a,b y = 0,62 -- ,by,

and (a,b– azb.)y=a,0,– Azę, by transposition,

Az

..

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a, b, -a,bi'
and the value of x can be found as before,

620, – 6,62

a,b,- azbi Third method: By equalising the coefficients of one of the unknown quantities in each equation, and then eliminating it by addition or subtraction of the equations.

In the given equations a,x+by=C1, azx+bzy=if an, ay, the coefficients of x, be prime to each other, multiply the first equation by 22, and the second equation by an,

then ayant +a,b y = 22€, and a7@gx tab.y =@. Next add or subtract these equations according as the signs of the coefficients of x are either one positive and the other negative, or both positive or both negative, and the resulting equation is

(a,b, a;bx)y = @qc; - Q762, ас, - а.с.

as before. a,b, - ab

abı-ab It may be remarked that the signs of the terms in the values of x and y appear with signs contrary to the terms in the values found by the other methods. This circumstance arises from subtracting the second equation from the first. If the first had been subtracted from the second, the signs would not have been changed.

The values of x and y are of the same value in both instances, as a fraction remains unchanged in value when the signs of every term of the numerator and denominator are changed.

Also, if the known quantities involved in the equations be positive or negative integers or fractions, the values of x and y in general will

Y be either integers or fractions positive or negative.*

..y=

and x=

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* The values of w and y in two simple equations when reduced to the form

m

or

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m

m

:

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Let the value of « assume the form o

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It is possible, however, that the relation of the constants in the equations to each other may be such that the values of u and y may

0

0 assume the forms of

o' If the values of one of the unknown quantities assume the form of and that of the other

0

0
then from x=

6,0,-8,62

a, b, - a,b, b261 - 6.09 = 0, and a,b- azb, = 0,

6. .. b,c, = b;cy, and _b

';

Comba
also a,b2 = azbı, and
b9_1 Hence

Ci Qy

Cg dz Next let the value of y assume the form

then from y=

a,C2 – AqC1

a, b, - a,b, a, b, a,b, = 0, and a Cy—2,0 = m,

ay .. a,ba=azbi, and ; also 2,0g=m+a,cı, and

1.

+

Az A2C, C2 But from the former a, _", C , which is impossible,

а. , C2 C2 аус, с. if m have a definite value, therefore the given equations are incompatible, and do not admit of any definite values of x and y.

If, however, the values of x and y both assume the form of 0

then a/cz-2,0 = 0, a,b,-a,= 0, and 6,4-6,4, = 0, o' a, b, a, a b G1

&,
then

ne,
Q2
6
az c, b,
Cg а.

02 and a, = na,, b, = nby, 4 = ne,, which shews that the co-efficients of one equation are respectively equimultiples of the corresponding coefficients of the other. Hence it follows that one of the equations may be deduced from the other, and the two equations do not express two independent conditions.*

m

a, b, a, b,

m

=

+

if Q

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a,x+b,y=, a,x+b2y=ca, can be found by substitating the particular for the general values of the constants in the values of x and y, as, for example, To find the values of x and y in the equations 5x + 8y = 31, 2x – 3y =0. Generally x=

6,0, -6,67

a, b, -a,b a,b,-a,b,
LIere a, =5, 6, 8, c, = 31; 0,=2, 6, = -3, C, = 0.
-93-0 -93

+3,
-15-16 - 31

0 - 62 - 62 y=

+2.

- 15 - 16 -31 .: 3 and 2 are the values of x and y in the equations 5x+8y=31, 2x – 3y=0.

* The subject of indeterminate equations arises from the fact of some problems involving more unknown quantities than independent conditions to determine them.

x =

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