Thus an equation of the first degree, which ought to be satisfied by the quantity y, is obtained. The value of y can consequently be deduced from equation 5, by Article 49.

For, removing the denominators a'c-a'by-ac'-ab'y, and transposing ab'y-a'by-ac'-a'c, or (ab'—a'b)y=ac'-a'c,

ac'-a'c ..yab'-a'b

Substituting this value of y in either of the equations 3, 4,

c-by_c b

x=

=

a

ay

becomes

bac'-a'c
X
a a ab'-a'b'

ab'c-a'bc-abc'+a'bc

aab'-aa'b

a(b'c—bc') ___ b'c—be'
a(ab'-a'b)—ab'—a'b'

It is indifferent which of the equations 3, 4 is employed to determine the value of x; for the value of y, which is deduced from equation 5, ought to render the two members of equation 5 identical; and the two expressions of the value of x from equations 3 and 4 form the members of equation 5. Consequently the values of the unknown quantities x, y are

x=

b'c-be' ab'-a'b'

ac'-a'c y=ab'—a'b

From the manner in which these values have been found, it is certain that they must satisfy the two proposed equations, for they satisfy equations 3 and 4; and as equations 3 and 4, by the removal of their denominators and the transposition of the terms in y, are made to reproduce the proposed equations 1, 2, it follows that these values of x, y must also satisfy equations 1, 2.

Hence the assumption made in commencing the investigation of the values of x and y, namely, that there are values of x, y which satisfy the two equations, is true.

The same conclusion is obtained by replacing x, y by their values in either of the equations 1, 2.

For since ar+by=c,

b'c--be ac-a'c

by substitution a. ́ab'—a'b+b·ab'—ab=c,

therefore ab'c-abc'+abc'-a'bc-ab'c-a'be,

or ab'c-a'bc=ab'c-a'bc.

The result, which is an identical equation, shows that these values of x, y satisfy the equation ax+by=c.

An equation involving two unknown quantities, x, y, can always be reduced to the form ax+by=c, in which a expresses the algebraic sum of the known quantities by which x is multiplied, b that of the known quantities by which y is multiplied, and c that of the known quantities which are connected with x, y by the signs + and

If it is, besides, made a condition that a, b, c, d, b, c, may express subtractive as well as additive quantities, the equations ax+by=c, a'x+b′y=c' are rendered perfectly general.

This method of elimination, which is named the method by Comparison, has the disadvantage of rendering both the members of equation 5 fractional, unless a, a' are equal to 1 or to each other,

If b=b' or b=b'=1, it will be more convenient to eliminate y from the proposed equations.

51. Resuming the general equations,

ax+by=c
a'x+b'y=c′

1

2

than x

and considering x, y, as before, to represent two numbers which satisfy these equations, if the value of x,

c-by

x=

a

3

which is obtained from the first equation, is substituted for x in the second equation, the result is,

c-by

+b'y=c'

a

4

This equation contains only the unknown quantity y, consequently the value of y can be determined from it.

For by Article 49, a'c-a'by+ab'y=ac',

ab'y-a'by-ac-d'c,

or (ab'―a'b)y=ac-a'c,

ac-a'c •'•y=ab'-a'b

Substituting this value instead of y in equation 3, and reducing

5

- 6

in

These results correspond with the values of x, y, obtained by the method of comparison; but, notwithstanding this coincidence, it is necessary to prove that this method must give values which satisfy the equations proposed. b'c-bc' ac-d'e Now the value a= being obtained by making y=ab'—ab ab'-a'b equation 3, it follows that these values of x, y must satisfy equation 3. Multiplying by a, and transposing by to the first member, equation 3 becomes equation 1; therefore the values which have been determined for r, y must satisfy equation 1.

ac-a'c

Again, the value y=ab-ab must satisfy equation 4, from which it is

deduced.

But in equation 4 the fraction which is multiplied by a (viz. b) is the

a

second member of equation 3. This fraction therefore gives the value of x ac-a'c

Consequently equation 2 is also satisfied by these values,.

This method of elimination is termed the method by Substitution. Like the method by comparison it has the disadvantage of introducting fractional expressions into the first equation, which contains only one unknown quantity, unless one of the coefficients a, a, b, b'=1. When this happens the method is very convenient.

ax + by=c a'x+b'y=c'

1

2

a is equal to a', the unknown quantity z can be eliminated by subtracting the members of one equation from the other, or adding the members of one equation to the other, according as the terms in r are affected with the same or contrary signs.

Under this hypothesis the result obtained is

by by=c+c, or (bby-ce',

c+c' and.'. y=b+b'

If a is not equal to a', it is evident that the coefficients of x in equations 1, 2 can be rendered equal by multiplying the terms of equation 1 by a', the coefficient of x in equation 2, and the terms of equation 2 by a, the coefficient of r in equation 1.

When the coefficients of the terms in r have been thus rendered equal, the elimination of x is effected in the same manner as in the particular case of a=a'.

If from (ax+by=c)×a', or aa'x+a'by=a'c,
is taken (ax+by=c')xa, or aa'x+aby=ac',
the remainder is (a'b-ab')y=a'c-ac';

a'c-ac'

・・・ y=a'b-ab"

The value of a may be found by substituting this value of y in equation 1 or equation 2, or by eliminating y in the same manner as x has been eliminated; thus,

From equation 1 multiplied by b', or ab'x+bb'y=b'c,
take equation 2 multiplied by b, or a'bx+bb'y=bc';
therefore (ab'-a'b)x=b'c-bc',

b'c-bc'

and x= ̄ab'—a'b

This method of elimination is sometimes called the method by Reduction, because the coefficients of the same unknown quantity in both equations are reduced to equality; and sometimes the method by Addition and Subtraction, because, when the coefficients of one unknown quantity have been rendered equal in both equations, that unknown quantity is eliminated by taking the sum or the difference of the two equations according as the terms rendered equal are affected with the same or with contrary signs.

As in the preceding methods, it is necessary to prove that the results obtained by the method of reduction satisfy the proposed equations in x, y.

To render the proof general, let equation 1 be multiplied by m, and equation 2 by n:

Subtracting

(ax+by=c) xm=max+mby=mc
(ax+b'y=c)xn=na'x+nb'y=nc'
(ma-na)x+(mb―nb')y=mc-nc′

3

4

It is to be shown that equation 5, and either of the proposed equations 1, 2, have the same solutions as the equations 1, 2.

For this purpose let the terms of equation 1 be multiplied by m, and let the product be subtracted from equation 5:

(ma-na')x+(mb―nb')y=mc—nc′
max+mby=mc
-na'x-nb'y――nc

and dividing all the terms by -n,

which is equation 2 reproduced.

a'x+b'y=c',

Therefore, in the same manner as the system of equations 1 and 5 is a consequence of the system 1 and 2, so, reciprocally, the system 1, 2 is a consequence of the system 1, 5; consequently values of x, y which satisfy equations 1, 5, must also satisfy equations 1, 2.

Now, if the multipliers m, n are chosen in such a manner that ma―na'=0, and consequently that equation 5 may not contain x; if from this equation (5) the value of y is deduced, and substituted in equation 1, for the purpose of deducing the value of x, it is evident that values of x, y, which satisfy equations 1, 5, must be obtained; therefore these values must satisfy equations 1, 2 also.

When the value of x is determined in the same manner as that of y; to prove that both values satisfy equations 1, 2, suppose that the terms of equation 1 are multiplied by m', the terms of equation 2 by n', and that the second product is subtracted from the first, the result is,

(m'a-n'a')x+(m'b-n'b')y=m'c-n'c'

6

Multiplying now the terms of equation 5 by m', the terms of equation 6 by m, and subtracting the second product from the first,

(m'ma-m'na)x+(m'mb―m'nb')y=m'mc-m'nc'
(mm'a—mn'a')x+(mm'b—mn'b')y=mm'c-mn'c'
(mn'a'—m'na)x+(mn'b'—m'nb)y=mn'c'm'nc'

or, (mn'—m'n)a'x+(mn' —m'n)b′y=(mn' —m'n)c'

or a'x+b'y=c.

7

9

In like manner if equation 5 is multiplied by n', equation 6 by n, and the second result is subtracted from the first,

(n'm-nm')ax+(n'm—nm')by=(n'm-nm')c
or ax+by=c.

10

Wherefore the pairs of equations 1, 2, and 5, 6, are mutually consequences of each other.

Now the multipliers m, n, m', n', may be so chosen that equation 5 shall not contain a, and equation 6 shall not contain y, it is therefore evident that the values of x, y, obtained from equations 5, 6, must satisfy equations 1, 2.

It has been assumed that the terms in x have the same signs in both of the equations 1, 2, and also the terms in y, and that the elimination must in both instances be made by subtraction. If the signs are different they can be made the same by affecting the factors m, n with contrary signs.

It has also been assumed that mn'—m'n is not equal to zero.
If mn'—m'n=0, equations 9, 10 are reduced to 0=0.

Whence, in replacing equations 1, 2 by equations 5, 6, it is necessary to choose m, n, m, n', in such a manner that mn'-m'n may not be equal to zero, or, which is the same thing, that mn' may not be equal to m'n or m: m'::n: n'.

U

The operation by which the coefficients of the unknown quantity to be eliminated are made equal is analogous to that for the reduction of fractions to the same denominator, and it admits of the same simplifications.

the coefficients of y can be decomposed into the products 7×3 and 7× 11. The coefficients of y in both equations are consequently rendered equal by multiplying the terms of the first equation by 11, and the terms of the second equation by 3.

In like manner the coefficients of x are rendered equal by multiplying the terms of the first equation by 3, and the terms of the second by 2.

In eliminating by the method of reduction the equation in a alone, or in y alone, is not incumbered with fractional expressions, as it commonly is when the method of comparison or of substitution is employed.

Whichever method is followed, after the elimination of one unknown quantity, the resulting equation evidently ought always to give the same value of the remaining unknown quantity. Hence, if by means of a process of elimination one unknown quantity is got rid of, and the resulting equation is brought to the form ar=b (a and b being known quantities), it is certain that every other process must give either the equation ar=b, or an equation differing from it by only a factor common to both its members; otherwise the values of a would not be the same.

x

53. It is shown in Article 55 that the preceding methods of elimination can be applied to the determination of the values of three unknown quantities from three independent equations of the first degree, four unknown quantities from four equations, &c. The principle of the methods is the same, namely, that each leads to one equation involving one unknown quantity, and, by consequence, admitting of only one value. By the substitution of this value in the equation which involves the unknown symbol whose value has been determined and another unknown quantity, this equation is converted into an equation with one unknown quantity, which also can have only one value. By the substitution of these values in the equation which involves the two quantities whose values have been determined and a third unknown quantity, the single value of this is obtained, and so on. Consequently it follows that there exists only a single system of values which can satisfy two or more independent equations of the first degree involving the same number of unknown quantities, and that this system of values can be determined by any of the three preceding methods.

54. To find the values of two unknown quantities from two independent simple equations involving these quantities,

Rule. Determine, from each of the proposed equations, the value of the same unknown quantity in terms of the data and the other unknown quantity, and put the one value equal to the other;

Or, find, from one of the proposed equations, the value of one unknown quantity in terms of the data and the other unknown quantity, and substitute this value for its symbol in the other equation;

Or, multiply the terms of the first of the proposed equations by the coefficient of either of the unknown quantities in the second equation, the terms of the second equation by the coefficient of the same unknown quantity in the first equation, and take the difference or the sum of the results according as the coefficients of the unknown quantity which is to be eliminated have the same or different signs.

An equation of the first degree involving only one unknown quantity is obtained by each of the methods.