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9. A cistern was filled by two pipes in 12 hours and by one in 20. Required the time in which it would be filled by the other alone.

Put x for the time required.

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would be the quantity supplied by the first in 12

12

and the quantity supplied by the second in 12 hours.

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But in 12 hours the two running together fill the pipe.

12 12
+ =1;

XC 20

and 240+12x=20x.

..8x=240,

and x=30, the number of hours required.

10. Find two numbers in the proportion of 2 to 1, and such that if 4 be added to each, the two sums shall be as 3: 2.

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By transposition, x=4.

..8 and 4 are the numbers required.

CHAP. XXII.

OF SIMPLE EQUATIONS INVOLVING TWO OR MORE
UNKNOWN QUANTITIES.

202. It often happens in the solution of Algebraical Problems, that we are obliged to introduce into our calculations two or more unknown quantities, usually represented by the last letters of the Alphabet, as x, y, z: and if the question be determinate, we must have as many separate equations as there are unknown quantities; and, as we are as yet considering those equations, which contain only the first power of the unknown quantities, the form of these equations will evidently be,

ax+by+cz=d.

203. We will begin first by considering those equations which involve two unknown quantities, as x and y; and will propose to ourselves two general equations, from which the values of each may be separately determined, viz.

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

where a, b, a, b, c and d are supposed to signify known quantities. :

204. Now there are three different methods by which the values of x and y may be determined from these two equations. 1y The first and most natural method will readily present itself to the mind; which is to find a value of one of the unknown quantities, asr, in each equation, and then to consider the equality of those two values; for we shall then arrive at an equation in which the other unknown quantity, y, will alone be contained, and of which we shall then be able to deter

mine the value, by the rules already given in the preceding chapter. The value of y being thus found, we have then only to substitute that value in either one of the equations, in order to determine the value of x.

Proceeding therefore by this rule with the equations given, the operation will be as follows; viz.

From the first equation, x=c—by

a

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Now since things that are equal to the same are equal to one another, these two values of x are equal; and we have therefore this new equation;

c_by_d-b'y;

a

involving y only, and which, being cleared of fractions, becomes,

a'c-aby-ad-aby;

by transposition, ab'y-a'by-ad-a'c,

or (ab'a'b). y=ad-a'c.

ad-a'c :.y=ab-ab

Substituting this value of y in the first equation, x=

we shall have, first,

c-by

α

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The second method proceeds upon the principle, that if each side of an equation be multiplied by the same quantity, the equation will remain unaltered. Now, in the two equations above given, it is evident that if the first be multiplied by the coefficient of x in the second, and the second by the coefficient of x in the first, two new equations will arise, in each of which, x and its coefficients are precisely similar, while the values of x and y are still the same, as in the equations originally given; viz.

aa'x+aby=a'c

aa'x+aby=ad.

Now, if the first equation be subtracted from the second, and its coefficients will disappear, and a new equation will arise, as before, involving only the unknown quantity y; viz.

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the same value as before, and which being substituted in the first equation in like manner, will give the same value of x as was found by the first method.

The third method differs but little from the first, except, that instead of finding a value for x in both equations, a value for it is found first in one, which is then substituted for x itself in the other, creating thereby, as in each of the preceding methods, a new equation containing y only. For instance;

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Clearing this equation of fractions,

ac-aby+aby=ad

or aby-abyad-a'c

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Either of these three different methods may therefore be resorted to in the solution of equations of this nature, with equal success. Which of them it may be most convenient to adopt will depend in a great measure upon the questions proposed for solution, and will easily be seen by practice.

It will be seen for instance immediately, that if in the question proposed the coefficients of either of the unknown quantities be the same in both equations, the second will be the readiest method to adopt; for we may then begin directly, by subtracting or adding the equations, according as the sign of such unknown quantities is, in the two cases, the same, or different; and we shall obtain at once, as we have already seen, a new equation, involving only one unknown quantity. If, for example, the question proposed were

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from which value of y, x may be determined in the manner already shewn.

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from which value of x, y may also be determined.

In general, if the equations proposed be of the form

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