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84. To avoid fractions, and simplify the calculation, we may vary the process of elimination in the following manner.

Let there be the equations

a x + by = c,

a' x + b' y = c',

it is evident, that if one of the unknown quantities, x, for example, has the same coefficient in the two equations, we have only to subtract one of these equations from the other, in order to make this unknown quantity disappear. This may be seen at once in the equations

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It is evident, that the coefficients of x may be immediately made equal in the equations

a x + by = c,

a' x + b'y = c',

by multiplying the two members of the first by a', the coefficient of x in the second, and the two members of the second by a, the coefficient of x in the first; we thus obtain,

a a x + a' by = a' c,

a a' x + a b' y = ac'.

Then subtracting the first of these from the second, the unknown quantity disappears; and we have

(a b'a' b) y = ac' — a'c,

an equation, which contains only the unknown quantity y; from this we may deduce,

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The method, we have just employed, may always be applied to equations of the first degree, to exterminate any one of the unknown quantities.

By exterminating, in the same manner, the unknown quantity find the value of x.

y, we may

If we apply this process to three equations, containing x, y, and z, we may first exterminate x from the first and second, then from the first and third; we thus obtain two equations, which contain only y and z, from which we may exterminate y.

When this calculation is performed, the equation containing z,

to which we arrive, will have a factor common to all its terms, and consequently will not be the most simple, which may be obtained.

85. Bézout has given a very simple method for exterminating at once all the unknown quantities except one, and for reducing the question immediately to equations, which contain one unknown quantity less, than the equations proposed. Although this process is necessary, only when equations with three unknown quantities are employed, we shall, in order to give a complete view of the subject, begin by applying it to those, which contain only two.

Let there be the equations

a x + by = c,

a' x + b' y = c;

multiplying the first by any indeterminate quantity m, we have

a mx + b my = mc ;

subtracting from this result the equation

a' x + b'y = c',

there remains

am x ax + b my· b'

y = cm — c',

or

(am- a') x + (bm — b') y = cm — c.

Since m is an indeterminate quantity, we may suppose it to be such, that bm = b'. In this case, the term multiplied by y disappears, and we have

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b', we make a m = a',

the term,

If, instead of supposing bm =

which contains x, will vanish, and we shall have

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the value of m will not be the same as before; for we shall have

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and by substituting this in the expression for y, we find

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If we change the signs of the numerator and denominator of this value of y, the denominator will become the same, as that in the expression for x, since we shall have

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we shall be led, by an obvious analogy, to multiply the first of these equations by m, and the second by n, m and ʼn being indeterminate quantities, to add together the results, and from the sum to subtract the third; by this means, all the equations will be employed at the same time, and the two new quantities m and n, which we may dispose of, as we please, will admit of any determinate value, which may be necessary to make both the unknown quantities disappear in the result. Having proceeded in this manner, and united the terms by which the same unknown quantity is multiplied, we shall have

(am + a'n — a'') x + (b m + b' n — b') y + (cm + c' n — c'') z = dm + d'n —

d".

If we would make the unknown quantities x and y disappear, we must take the equations

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From the two equations, in which m and n are the unknown quantities, it is easy to deduce the value of these quantities, by means of the results obtained in the preceding article; for it is only necessary to change in these results x into m, y into n, and to write instead of the letters

a, b, c, a', b', c',

the letters

Sa, a', a",
b, b', b",

which gives

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Substituting these values in the expression for z, and reducing all

the terms to the same denominator, we have,f

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If we had made the terms containing x and z to disappear, we should have had y; the letters m and n would have depended upon the equations

am + a' n = a", cm + c' n = and proceeding as before, we should have found

y

=

-

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d(c' a" a' c') + d' (a c" — ca") — d" (a c'—ca') b (c' a'' — a' c'') + b' (a c" — c a") — b'' (a c' — ca')' Lastly, by assuming the equations

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we make the terms multiplied by y and z to disappear; and we have

— c b') — d' (b c' — c b')

c

— c b") — a" (b c' — c b')'

d(c' b'b' c'') + d' (b c" x = a (c' b'' —b'c'') + a' (b c" These values being developed in such a manner, as to make the terms alternately positive and negative, if we change, at the same time, the signs of the numerator and denominator, in the first and third, we shall give them the following forms;

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if we multiply the first by m, the second by n, the third by p, and from the sum of their products subtract the fourth, we shall have

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The preceding equations, which must give m, n, and p, may be resolved by means of the formulas found for the case of three unknown quantities. This method will appear very simple and convenient; but the nature of the results obtained above will furnish us with a rule for finding them without any calculation. 88. To begin with the most simple case, we take an equation with one unknown quantity, ax = b; from this we find

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in which the numerator is the whole known term b, and the denominator the coefficient a, of the unknown quantity.

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The denominator in this case also is composed of the letters a, a', b, b', by which the unknown quantities are multiplied. We first write a by the side of b, which gives a b; we then change the order of a and b, and obtain ba; prefixing to this the signwe have ab ba; lastly, we place an accent over the last letter in each term, and the expression becomes a b' — ba' for the denominator.

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From this expression we may find the numerator. To obtain that for x, we have only to change each a into c, and each b into c for that of y, putting an accent over the last letter as before; in this way we find c bbc for the one, and a c-ca' for the

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