10;

2

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

174—6y-62 +4y+32=120,

and by transposition, 2y+32=54 L; in which, substituting the value of y found above,

2(33-22)+3z=54;

or 66 -42+3=54; ... by transposition, z=12; whence y=33---2z=33—2459,

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 2, 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 y and z may be deduced by any of the rules in the preceding section, and hence, the value of x, can be readily ascertained.

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 a+y+z=31

to find the values x+y-z=25

of x-y-=9

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

.'. X=20. Subtracting the second from the first, 2z=6;

..z=3; and subtracting the third from the second,

2y=16; ..y=8.

x-y=2,1

-
Ex. 3. Given x-2=3,

to find x, y, and z.

y-z=1, Here, subtracting the first equation from the second, we have y-z=1; which is identically the third.

Therefore, the third equation furnishes no new condition; but what is already contained in the other two; and, consequently, the proposed equations are indeterminate ; or, what is the same, we may obtain an infinite number of values which will satisfy the conditions proposed.

This can be easily verified, by comparing the proposed equations with those of (Art. 207), and substituting in the formulæ of roots, (Art. 215); for,

0 0

0 then we shall find x=

and
0
0

09

Z

254. It is proper to remark, that in particular cases, Analysts make use of various other methods, besides those pointed out in the practical rules; in the resolution of equations, which greatly facilitate the calculation, and by means of which, some equations of a degree superior to the first, may be easily resolved, after the same manner as simple equations.

We shall illustrate a few of those artifices, by the following examples.

1

1 1 Ex. 4. Given

C

8'
1
9'
1

Z

-十-十

z

By adding the three equations, we shall have

2 2 2 1 1 1121
十一十 +-+
y

8 9 10 360 Or, dividing by 2,

1 1 1 121
y

720 From this subtracting each of the three first equations, and we shall have 1 720

7 -23

;
720"

31
1 :41
720

23

;
. y
720'
41

41
1 49
720

34

x=14-
720
49

49

31

or z

31

or y=

y=17

or X

4z=x+y+u, of x, y, z, and u.

Ex. 5. Given 2x=y+z+u,

3y=x+ztu, to find the values +

, , , and u=*-14, By adding x to each member of the first equation, y to the second, and a to the third, we shall get

x+y+ztu=3x=4y=52 ;
30

3x and from thence, z= and

y=-; 5

4 which values being substituted in the first equation, we have

3x

13x + tu;.. u=

i 4 5

20 but, by the fourth equation, u=x-14;

13x 5.X-14= or 20x— 280=132

20 whence x=40: consequently y=*=30, z=24

=, 2

4 and u=2-14=26.

33

and z.

Es. 6. Given 4x-44-4z=24,7 to find the values

} 6y-22-2z=24,

of

X, Y, and 7z- y- x=24, By putting x+y+z=S, the proposed equations become 8x-4$=24, 8y-25=24, 3z-S=24;

..x=3+ S, y=3+1S, z=3+ S. By adding these three equations, we have

x+y+z=9+ZS; whence S=72. Substituting this value for S, in x, y, and z, we shall find

x=39, y=21, and z=12.

lues of x, y,

Ex. 7. Given x+y+z=90, to find the va

2x + 40=3y+20, and 2x 42+40=10,

and z. Ans. x=35, y=30, and z=25. Ex. 8. Given xta= y +z,

to find the vayta=2+2z, lues of x,y, and and z+a=3x+3y,) 2.

z
5а

na Ans.

and 2=
11'
y=

11
Ex. 9. It is required to find the values of x, y,
and z, in the following equations :
x+y=13, x+z=14, and y+z=15.

Ans. x=6, y=7, and z=8.

Z

Ex. 10. In the following it is required to find the values of x, y, and 2.

+9+ +=124, 23 4

x=48, y tot =94,

Aps.

y=120,

z=240. + 4 5

Z

,

4

+ +

to

276,

= 4,

lues of x, y,

2.

Z

Ex. 11. Given x+y+z=26, to find the va-
XY

and and x-2

= 6,

Ans. x=12, y=8, and z=6.
Ex. 12. Given x+y+ z= 9,) to find the va-

xt
x+2y+3z=16, lues of x, y,

,
and at y-2z= 3, and z.
xt

Ans. x=4, y=3, and z=2. Ex. 13. Given + y + z=12, to find the 1+2y+3z=20,

values of x, and fr+ y + 2= 6,) y, and 2. z

z Ans. x=6, y=4, and z=2. Ex. 14. Given x+y=z=3, «+z-y=9, and y+z-x=10; to find ihe values of x, y, and 2.

Ans. x=0}, y=9, and 2=9. Ex. 15. Given x+y=100, y+}z=100, and z+c=100; to find the values of x, y, and z.

Ans. =64, y=72, and z=84. Ex. 16. Given 4x+3y+z_2y+2z -x+1

=5+ 10

15 1-2-5

5 9x + 5y--22 2x+y - 3z_76+2+3, 1

7h 12

4

11 6' 5y+3z 2x + 3y -and

2+22=y-1+

3x +2y +7 12 12

6 to find the values of x, y, and z.

Ans. x=9, y=7, and z=3. Ex. 17. Given x+iy=357, y=}z=476, ztu 3595, and utx=714; to find the values of X,

+

4+?

Y, 2, and u.

Ans. x=190, y=334, 2=426, and u=676: