1

3y-x

5

7x+6, 4y-9
+

and 3x+42y-3:: 5: 3,

to find the values of x and

Ans. x=9,

and y=? 13-x

=4.

=3x

11

3

2

y.

5x+13

Ans. x=7, and y=9. 8y-3x-5

Ex. 18. Given

=9+

Ans. x=5, and y=4.

Ans. x 45, and y=60.

Ex. 21. Given x+150:y-50:3: 2, ? and x-50:100:: 5 : 9, §

to find

the values of x and y.

Ans. x300, and y=350.

Ex. 22. Given (x+5).(y+7)=(x+1)(y−9)+ 112,

Involving three or more unknown Quantities.

252. When there are three independent simple equations involving three unknown quantities.

RULE.

From two of the equations, find a third, which involves only two of the unknown quantities, by any of the rules in the preceding Section; and in like manner from the preceding equation, and one of the others another equation which contains the same two unknown quantities may be deduced.

;

Having therefore two equations, which involve only two unknown quantities, these may be determined and, by substituting their values in any of the original equations, that of the third quantity will be obtained.

253. If there be four unknown quantities, their values may be found from four independent equations. For from the four given equations, by the rules in the last Section, three may be deduced which involve only three unknown quantities, the values of which may be found by the last Article ; and hence the fourth may be found by substituting in any of the four given equations, the values of the three quantities determined.

n

If there be n unknown quantities, and n independent equations, the values of those quantities may be found in a similar manner. For from the n given equations, n-1 may be deduced, involving only ni unknown quantities; and from these n-1, n-2 may be obtained, involving only n-2 unknown quantities; and so on, till only one equation remains, involving one unknown quantity; which being found, the values of all the rest may be determined by substitution.

y+2z=33.

(A).

Multiplying the third equation by 12, the least

common multiple of 2, 3, and 4,

6x+4y+3z=120

multiplying the 1st equ" by 6, 6x+6y+6z=174;

... by subtraction, 2y+3x=54;

but, multiplying equation (A) by 2,

2y+4z=66;

by subtraction, z=12.

From equation (A), by transposition, y=33-2z; ... by substitution, y=33-24, or y=9. From the first equation, by transposition, x=29-y-z;

.. by substitution, x=29-9-12,

and x=29-21, ... x=8.

In like manner, had the first equation been multiplied by 2, and subtracted from the second, an equation would have resulted, involving only x and z; and had it been multiplied by 4, and subtracted from the third when cleared of fractions, another equation would have been obtained, involving also x and z; whence by the preceding rules, the values of x and 2 could be found, and consequently the value of y also, by substitution.

Or if the first equation be multiplied by 3, and the second subtracted from it, an equation would arise, involving only x and y ; and if the first when multiplied by 3, be subtracted from the third when cleared of fractions, another would arise involving only x and y; whence the values of x and y might be determined. And hence the third, that of z might be found.

SECOND METHOD.

From the first equation, x=29-y-z; substituting this value of x in the second equation, 29-y-z+2y+3z=62;

... by transposition, y=33-2z. Also substituting, in the third equation, the value

of x found from the first,

équati

multiplying this equation by 12, the least common multiple of 2, 3, and 4,

174-6y-6x+4y+3z=120,

and by transposition, 2y+32=54;

in which, substituting the value of y found above,

2(33—2z)+3x=54 ;

or 66-4z+32=54;

by transposition, z=12;

whence y=33-22-33-24-9, and x=29-y-z-29-9-12=8. It may be observed, that there will be the same variety of solution, as in the last case, according as x, y, or z, is exterminated.

THIRD METHOD.

The values of x, found in each of the equations, being compared, will furnish two equations each involving only y and z; from which the values of and z may be deduced by any of the rules in the preceding section, and hence, the value of x, can be readily ascertained..

y

The same observation applies to this method of solution, as did to the last.

In some particular equations, two unknown quantities may be eliminated at once. Ex. 2. Given x+y+z=31

x+y-z=25

x-y-z=9 S

to find the values

of x, y, and z.

Adding the first and third equations, 2x=40;

... x=20.

Subtracting the second from the first, 2z=6;

and subtracting the third from the second,

z=3;

2y=16; .. y=8.