(56) Given v+w+z=30; v+w-z=18; v−w+z=14.

(57) Given u+v=164; v+w=82; u+}w=136.

(59) Given 3(ax+by)=z; 5y=7(x+3a); 11x=}z+121.

Ans. x=2; y=4; z=3; u=3; t=1.

Elimination may be effected in a general form, and particular cases be resolved by substitution in this form.

We shall illustrate this with a system of three equations.

Eliminating among these three equations by any of the foregoing methods,

we find

x=

(b''c'b'c')k+(be" —b'c)k' + (b'c-bc')k"
(a'b"—a"b')c+(a"b—ab")c' +(ab′-a'b)c""

x=

(a'c" — a'c')k+(a'c—ac")k' +(ac' —a'c)k" y= The same denominator as in the value of x

2=

(a'b' —a'b'')k+(ab" —a’b)k' +(a’b—ab')k”
The same denominator as in the value of r

To apply this general form to a particular case, take (Example 53) above.
(1 × —3—4 × 6) (—15)+(—6×6—1 × 4) (—19)+[4 × 4—(—6 X—3)](—46)___1257_
(7X1—2×4)4+(2×—6—5×1)(−3)+(5 × 4—7—6)6
(42+6)(-15)+(8-30)(-19)+(-15-28)(-46)

y=

=

1676

419

= =3, 419

=4,

Changing the signs of k, k', k'', in order that they may be positive in the second member of the three proposed equations, and performing the multiplications indicated in the general values of x, y, and z, they may be written as follows:

x=

y=

z=

kb'c" —kc'b" +ck'b'' —bk'c' +bc'k" — cb'k"
ab'c"-ac'b" +ca'b" -ba'c" +bc'a" - cb'a"
ak'c"-ac'k" +ca'k" -ka'c"+ke'a" - ck'a"
The same denominator as that of x

ab'k" —ak'b'+ka'b” —ba'k" +bk'a" —kb'a"
The same denominator as before

By observing carefully the composition of the formulas for two and three equations, we may discover general rules by means of which we can calculate the formulas suitable for any number of equations.

FIRST RULE.-To find the common denominator in the values of all the unknown quantities. With the two letters a and b form the arrangements ab and ba, then interpose the sign between them, thus:

ab-ba.

If there are but two equations to resolve, place an accent on the 2o letter of each term, and the result, ab'-ba', will be the common denominator of the values of x and y.

If there are three equations, pass the letter c through all the places in each term of the expression ab—ba, taking care to alternate the signs; ab will thus give abc-acb+cab; also, —ba will give ―bac+bca—cba, and the whole

abc-acbcab-bac+bca-cba ;

then place one accent on the 2o letter of each term, and two on the 3o, and the resulting expression will be the common denominator of the values of x, y, and z.

If there are four equations, take the letter d, which is the coefficient of the fourth unknown u, and pass it through all the places in each term of the sexinomial above formed, taking care to alternate the signs of the terms furnished by each of them, beginning with + for those which result from a term preceded by the sign+, and with for those resulting from a term affected with the sign; finally, place one accent on the 2o letter, two on the 3o, and three on the 4°. The resulting polynomial is the common denominator of the four unknown quantities x, y, z, U.

ab'c"d""-ab'd"c""+ad'b"c""-da'b"c""

If there be a greater number of equations, proceed in the same manner. SECOND RULE.-The numerators may be derived from the common denominator. For this purpose, it is only necessary to replace, without touching the accents, the letter which serves for coefficient of the unknown quantity we wish to find, by the letter k, which represents the known term in the second member. Thus : change a into k, to have the numerator of x; b into k, to have that of y; and so on.

There remains still a method of elimination to be mentioned, which alone is applicable to equations of higher degrees, as well as to those of the first. It is called the method of the common divisor. It consists, where two equations are given, in dividing one by the other (after transferring all the terms to the first member in both), that divisor by the remainder, and so on till the letter of arrangement, which must be one of the unknown quantities, is exhausted from the remainders. The last remainder containing but the other unknown quantity, being put equal to zero, will present an equation from which the first unknown quantity is eliminated.

If there be three or more equations, eliminate one of the unknown quantities in this way between the first and second, then between the first and third, and so on.

The reason which may be given for this rule here, though a better one will be furnished hereafter, is, that the dividend being zero and the divisor zero, the quotient must be zero and the remainder zero.

Let us apply this method to Example (8) above. The two given equations are

Substituting this value in x+y-10=0, we obtain x=7.

(9y+10)x-(2y3+10y +98)x

(9y2+10) x- 273-10y-98

(9y+10)x+(36y3+40y)x—18y4—110y2-100|x+19y3+25y+49

(38y3+50y+98)x 18y-110y2—100÷2

(19y+25y+49)x- 9yt 55y?— 50

9y+ 10

(9y+10)(19y3+25y +49)x— 81y6-585y4-1000y—500

(972+10)(19y3+25y +49)x- 38y6—24044—1960 y3—25072 —ˆ040y--4802

433-345y+196073-750y+2940y+4802.