From this equation determine the value of the remaining unknown quantity, and substitute it in either of the proposed equations; the two results are the values of the two unknown quantities in these equations.

Find the values of x and y in each of the following examples;

1st..

x+y=15)

x-y=7

} x=11, y=4.

OF EQUATIONS OF THE FIRST DEGREE INVOLVING THREE OR MORE

UNKNOWN QUANTITIES.

55. Let there be three unknown quantities, x, y, z, and three equations,

ax+by+cz=d
a'x+by+c'z=ď
a'x+by+c"z=d'

1

2

3

And let it be required to assign the values of x, y, z, in terms of the given quantities a, a, a"; b, b' b"; c, c', c''; d, d', d'.

z may be eliminated between equations 1, 2 and equations 1, 3 by one of the methods (Art. 54). Adopting the method by reduction, the equation in x, y, obtained from equations 1, 2, is,

(ac'-ca')x+(bc'-cb')y=de'-cd'

and from equations 1, 3,

(ac"-ca")x+(bc"-cb")y=dc"-cd"

The values of x, y ought to satisfy equations 4 and 5.

4

5

To eliminate y from equations 4, 5, it is necessary to multiply the terms of equation 4 by be"-cb", and the terms of equation 5 by be-cb', and to subtract the second product from the first.

The expression of the remainder (the multiplication being merely indicated) is,

{(ac'—ca') (be"—cb′)—(ac”—ca′′) (be′—cb')} x=(dc′—cd'′) (bc′′—cb′′)

-(de"-cd'') (bc'-cb').

Performing the multiplications, reducing the like terms, and dividing every term by the common factor c, (ab'c"-acb"+ca'b"-ba'c'"+bc'a"-cb'a") x=db'c"-dc'b"+cd'b"-bdc"

+bed"-cb'd".

db'c"-de'b'+cd'b"-bd'c"+bc'd"—cb'd" •.• x=ab'c" —ac'b" +ca'b"—ba'c' +bc'a" —cb'a"

6

Eliminating x, z; x, y, in the same manner,

ad'c''-ac'd''+ca'd' —da'c''+dc'a"-cd'a" y= ab'c''—ac'b" +ca'b"-ba'c' +bc'a" —cb'a" ab'd'-ad'b'+da'b" —ba'd'+bd′a"-db'a"

ab'c"-ac'b'+ca'b''—ba'c'"+bc'a'—cb'a"

• 7

8

From the investigation of Article 52 it follows that these values of x, y, z, must satisfy equations 1, 4, 5.

Now equation 2 is a consequence of equations 1, 4, and equation 3 is a consequence of equations 1, 5.

Therefore the values of x, y, z, which satisfy equations 1, 4, 5, must also satisfy equations 1, 2, 3.

By generalising the process which has been employed in the case of three equations, a rule for the resolution of a number of equations m, involving m unknown quantities, may be found.

The rule is: Eliminate one unknown quantity between the first and each of the remaining m-1 equations; m-1 equations of the first degree, involving m-1 unknown quantities, are thus obtained.

Proceed, with the m2-1 equations, to eliminate one unknown quantity between the first of the m-1 and each of the remaining m-2 equations; m-2 equations of the first degree, involving m-2 unknown quantities, are thus obtained.

Continue this series of operations with the m-2, the m-3... the m-(m-1) equations; an equation of the first degree involving only one unknown quantity is the last result.

From the last equation deduce the value of the unknown quantity

which it contains.

Substitute this value in the last equation but one; the value of a second unknown quantity is determined.

Substitute these values in the last equation but two; the value of a third unknown quantity is obtained; and by returning thus through the equations upwards to the (m-1)th, and lastly to the mth, the values of all the unknown quantities are determined.

Example 1. Given 10x-20y+302=60,

to find the values of x, y, z.

8x+12y-16z=80,

27x-18y+45z=234,

In the first of these equations the factor 10 is common to all the terms; also, in the second the factor 4, and in the third the factor 9, is common to all the terms.

Suppressing the common factors, the proposed equations are reduced to

x-2y+3z=6

2x+3y-4z=20

3x-2y+5z=26

Replacing, in the general formula,

a, b, c, d, by 1,-2, 3, 6,
a, b, c, d, by 2, 3,-4, 20,
d', b, c, d, by 3,-2, 5, 26,

and performing the operations indicated,

x=

90-48+(-120)—(−200)+208-234_96
15-8+(-12)-(-20)+24-27 =12=8
100-(-104)+156-60+(-72)—180__48_

y= 15-8+(-12)-(-20)+24-27-12
12=4

2=

=

78-(-40)+(-24)-(—104)+(-120)—54_24
15-8+(-12)-(-20)+24-27

12-2

Or the values of the unknown quantities may be found by the general rule, as follows:

Multiply 1st equation by 4, and 4x-8y+12z=24

Multiply 1st equation by 5, and 5x—10y+15z=30

3d

Therefore, by addition

Therefore, by subtraction

or

From equation 4

subtract equation 5

9x =72, and x==8.

Substituting this value of x in equation 5,

8+y=12, and therefore y=12-8=4.

Substituting these values of x, y in the 1st equation, 8-8+32=6; therefore 32=6, and z=g=2.

Consequently x=8, y=4, z=2 are the values of x, y, z in the proposed equations.

56. If all the unknown quantities do not enter into each equation, the process of elimination, though conducted by the same principles, can be abridged.

Example. Let 7u-13z=87

3u+14x=57

10y-3x=11
2x-11z 50

1

2

3

4

If u is eliminated between the first and second equations, the resulting equation will contain only x, z and known quantities; consequently from this and the fourth equation, which also contains only r, z and known quantities, the values of x, z can be determined.

Substituting this value of z in equation 4,

2x-11z 50 becomes 2x-11x(-4)=50,

..2x+44=50, 2x=50—44=6, and x=g=3.

Substituting 3 for x in equation 2,

3u+14x=57 becomes 3u+14x3=57, or 3u+42=57,
...3u=57-42=15, and u==5.

Lastly, substituting 3 for r in equation 3,

10y-3x=11 becomes 10y-3x3=11, or 10y-9=11

.*.10y=11+9=20, and y=20=2.

Whence u=5, y=2, x=3, z=-4, are the values of u, y, z, z, which satisfy the proposed equations.

57. In resolving equations of the first degree, if the number of equations is equal to the number of unknown quantities, it follows, from Article 53, that there is one system of values of the unknown quantities, and only one, which satisfies the proposed equations.

This general conclusion is to be taken with some exceptions, which are noticed in Article 80.

If the number of equations is greater than the number of unknown quantities, (if, for example, there are five equations between three unknown quantities, x, y, z,) taking three of these equations, a single system of values can be determined for x, y, z. These values being found, it will be requisite to try whether they verify the two remaining equations also. Now, unless these equations have been chosen in a particular manner, the verification cannot be made, and consequently there do not exist values of the unknown quantities which can at once verify all the five equations.

If, on the other hand, there are more unknown quantities than equations, (for example, five unknown quantities, u, v, x, y, z, and three equations,) arbitrary values may be given to two of the unknown quantities, u, v; then, by means of the three equations, the corresponding values of x, y, z may be determined. Changing the values of u, v, another system of values may be found for x, y, z; and by thus changing the arbitrary values of u, v, an indefinite number of values may be obtained for x, y, z;

Or, without assigning particular values to u, v, the three equations may be resolved as if u, v were known quantities, and x, y, z unknown.

In this manner values of x, y, z, involving the quantities u, v, are obtained; and from these can be deduced those values of x, y, z which correspond to particular values of u, v.

Find the values of the unknown quantities in each of the following examples:

1st.

2x+3y+4z=16]

3x+2y-52-8

x=3, y=2, z=1.