192. If there be three independent simple equations, and three unknown quantities, reduce two of the equations to one, containing only two of the unknown quantities, by the preceding rules; then reduce the third equation and either of the former to one, containing the same two unknown quantities; and from the two equations thus obtained the unknown quantities which they involve may be found. The third quantity may be found by substituting their values in any of the proposed equations.

The same method may be applied to any number of independent simple equations, in which the number of unknown quantities is the same as the number of equations. [For other methods see Note 7. Appendix 1.]

193.

That the unknown quantities may have definite values, there must be as many independent equations as unknown quantities. When there are more equations than unknown quantities, the value of any one of these quantities may be determined from different equations; and should the values thus found differ, the equations are incongruous; should they be the same, one or more of the equations are unnecessary. When there are fewer equations than unknown quantities, one of these quantities cannot be found, but in terms which involve some of the rest, whose values may be assumed at pleasure; and in such cases the number of answers is indefinite. Thus, if x + y = a, then xay; and assuming y at pleasure, we obtain a value of a such, that x + y

= a.

These equations must also be independent, that is, not deducible one from another.

Let x+y=a, and 2x + 2y = 2a; these are not independent equations, since the latter equation being deducible from the former, it involves no different supposition, nor requires any thing more for its truth, than that x + y = a should be a just equation.

[It is sometimes, however, not easy to discover at once whether proposed equations be independent or not. Thus in the equations

it is not obvious at first sight that the third equation is derived from the other two. But by multiplying the first equation by 4, and subtracting the second, the result is the third equation; and accordingly the usual process, being applied to find x, y, x, would certainly fail.]

PROBLEMS WHICH PRODUCE SIMPLE

EQUATIONS.

194. From certain quantities which are known to investigate others which have a given relation to them, is the business of Algebra.

When a question is proposed to be resolved, we must first consider fully its meaning and conditions. Then substituting one or more of the symbols x, y, z &c. for such unknown quantities as appear most convenient, we must proceed as if they were already determined, and we wished to try whether they answer all the proposed conditions or not, till as many independent equations arise as we have assumed unknown quantities, which will always be the case if the question be properly limited (Art. 193); and by the solution of these equations the quantities sought will be determined.

PROB. 1. A bankrupt owes A twice as much as he owes B, and C as much as he owes A and B together; out of £300, which is to be divided amongst them, what must each receive?

Let a represent what B must receive, in pounds;
then 2x = what A must receive,

and x+2x, or 3x = what C must receive;

amongst them they receive £300; therefore

x+2x+3x= 300

6 x = 300

2x 100£. what A must receive

3x= 150£. what C must receive.

PROB. 2. To divide a line of 15 inches into two such parts, that one may be three fourths of the other.

Let 4x= the number of inches in one part,

PROB. 3.

If A can perform a piece of work in 8 days, and B the same in 10 days, in what time will they finish it together?

Let a be the time required, in days; and w the work.

W

In one day 4 performs th part of the work, or

; there

8

PROB. 4. A workman was employed for 60 days, on condition that for every day he worked he should receive 15 pence, and for every day he played he should forfeit 5 pence; at the end of the time he had 20 shillings to receive; required the number of days he worked.

Let a be the number of days he worked,

then 60 is the number he played,

20x240 + 300 = 540

.. x = 27, the number of days he worked,

60 − x = 33, the number of days he played.

PROB. 5. How much rye, at four shillings and sixpence a bushel, must be mixed with 50 bushels of wheat, at six shillings a bushel, that the mixture may be worth five shillings a bushel ?

Let x be the number of bushels required;

then 9 the price of the

=

rye

in sixpences

600= the price of the wheat......

=

(50+) 10 the price of the mixture............ .. 9x+600

= 500 + 10x

100 = x, the number of bushels required.

PROB. 6. A and B engage together in play; in the first game A wins as much as he had and four shillings more, and finds he has twice as much as B; in the second game B wins half as much as he had at first and one shilling more, and then it appears that he has three times as much as A; what sum had each at first?

Let a be what A had, in shillings,

y what B had.