Multiplying the first equation by c, and the second by a,

and 43719x = 4y-2x;
and by transposition, 437 17x+4y.

and 17x297+3y=297 +60=357;

Multiplying the first equation by 5+2y,

80+300x+32y + 120 x y
3y-1

40x+16xy

.. by trans", 40x + 107 =

values of x and y.

=

80 + 300x + 32y + 120xy;

and multiplying by 3 y − 1,

3y-1

120xy 40x+321 y107 = 80+300x +32y+120xy;

And from the second equation,

27x2

12y2+15x+2y + 2 = 27 x2 — 12y2 + 38 ;

.. by transposition, 15x+2y=36;

whence, the coefficients of x having aliquot parts, multiplying the first equation by 3, and the second by 68,

and

y

+ 2x : y − 2x :: 12x+6y −3 : 6y – 12x — 1..

to find the values of x and y.

(18. Cor. 1.) multiplying the first equation by 420, 252x+168y-700x+105y−140=420x+42y−84x − 240x+60y,

and by transposition, 171y-544x=140.
From the second equation, (Wood's Alg. 182.)
2y 4x 12y-4: 24x-2,

and (Alg. 184.) y 2x: 6y-2: 12x-1;
.. (21) 12xy-y=12xy - 4x;

(17. Cor. 3.) y=4x.

which value of y being substituted in the first equation,

684x-544x=140,

or 140x140;

.. x = 1,

and y=4x4.

F

to find the values of x and y.

(18. Cor. 1.) multiplying the first equation by 15y,

.. 45y-21y-6x=75 y −25 x − 45 ;
and by transposition, 51y- 19x=45.

Multiplying the second equation by 2x+5,

and 321x10732x + 340 y +80;

and by transposition, -187340y-289x.

The coefficients of y in this case having aliquot parts; mul

tiplying the first by 20, and the last by 3,

1020y - 380 x=

900,

and 1020y 867 x =

— 561;

.. by subtraction,

487 x

= 1461,

and x = 3;

consequently, 51 y=45+19 x = 45 +57 = 102;

.. y = 2.

(24.) If there be three unknown quantities, their values may be found from three independent equations.

For from two of the equations, a third, which involves only two of the unknown quantities, may be deduced by the preceding rules; and from the remaining equation, and one of the others, another which contains the same two unknown quantities. 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. In some particular equations, two unknown quantities may be exterminated

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, 2y = 16;

(18. Cor. 1.) multiplying the third equation by 12, the least

common multiple of 2, 3, and 4,