SECT. VII. T1, --.EDECTION-Op rQT.IATIONS INVOLVING MORE As the unknown quantities may be combined together in very different ways, so as to constitute an equation, the methods most proper for their elimination must therefore be various. The three following, however, are of general application, and the last of them may be used with advantage, not only when the unknown quantity to be eliminated rises to the same power in all the equations, but also when the equations contain different powers of that quantity.

Method 1. Observe which of the unknown quantities is the least involved, and let its value be found from each equation, by the rules of last section.

Let the values thus found be put equal to each other, and hence new equations will arise, from which that quantity is wholly excluded. Let this operation be now repeated with these equations, thus eliminating the unknown quantities one by one, till at last an equation be found which contains only one unknown quantity.

Ex. Let it be required to determine x and y from these Ex. Let the given equations, as in last method, be Method 3. Let the given equations be multiplied or divided by such numbers or quantities, whether known or unknown, that the term which involves the highest power of the unknown quantity may be the same in each equation.

Then, by adding or subtracting the equations, as occasion may require, that term will vanish, and a new equation emerge, wherein the number of dimensions of the unknown quantity in some cases, and in others the number of unknown quantities will be diminished ; and by a repetition of the same or similar operations, a final equation may be at last obtained, involving only one unknown quantity.

Ex. Let the same example be taken, as in the illustration of the former methods, namely, To eliminate x, let the first equation be multiplied by 5, and the second by 2 ; thus we have Here the term involving x is the same in both equations ; and it is obvious, that by subtracting the one from the other, the resulting equation will contain only y, and known numbers ; for by such subtraction we find 19y = 95, and therefore y=5.

Haying got the value of y, it is easy to see how x may be found from either of the given equations ; but it may also be found in the same manner as we found y. For let the first of the given equations be multiplied by 2, and the second by 3, we have Let the second equation be now subtracted from the first, and the third from the second, and we have Next, to eliminate y, let the first of these equations be multiplied by 3, and the second by 5 ; hence, Subtracting now the latter equation from the form:, 10x=240 and x=24,

By subtraction. we have But x2 + y2 + z2 - xy - xz - yz is the sum of the three given expressions, and equal to a2 + b2 + c2.

In the same manner may y be determined, by multiplying the first of the given equations by d, and the second by a; for we find adx +bdy = cd , adx + afy = ay .

And taking the difference as before, we get bdy, - afy = cd - ay , And therefore Y = cd - ag bd - af • This example may be considered as a general solution of the following problem. Two equations expressing the relation between the first powers of two unknown quantities being given, to determine those quantities ; for whatever be the number of terms in each equation, it will readily appear, as in Art. 55, that by proper reduction they may be brought to the same form as those given in the above example.

Let us next consider such equations as involve three unknown quantities.

Multiply the first by ,/x, .the second by Vy, and the third by ,Jz, and add two and two. There results Multiplying any two of these we get one of the unknown quantities : x= (a +e - b) (a +b - c), &c.