Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[ocr errors]

; or

Ex. 47. Given 3+x_. to find the values of x.

22 I-VX 4

Ans. x=4, or 1, or i+-7. Ex. 48. Given 4x4 +2=4.2°+33, to find the values of x.

= 2

3 1+(-43) Ans. x=2, or

2

4
§ II. SOLUTION OF ADFECTED QUADRATIC EQUATIONS,

INVOLVING TWO UNKNOWN QUANTITIES. 424. When there are two equations containing two unknown quantities, a single equation, involving only one of the unkoowo quantities, may sometimes be obtained, by the rules laid down for the solution of simple equations ; from which equation the values of the unknown quantity may be found, as in the preceding Section. Wheoce, by substitution, the values of the other may also be determined. lo many cases, however, it may be more convenient to solve one or both of the equations first ; that is, to find the values of one of the unknown quantities, in terms of the other and known quantities, as before ; when the rules for eliminating unkoown quantities, (§ I. Chap. IV), may be more easily applied.

The solution will sometimes be rendered more simple by particular artifices ; the proper application of which shall be illustrated in the following examples. Ex. 1. Given x+2y=7,1 to find the values of x and y. and x + 3xy-y=23, S

From the 1st equation x=7—2y;

..x=49-28y +4ya ; Substituting these values for x and zo in the 2d equation, then 49--26y+4y +217-6y2-y=23,

or 3y +-7y=49--23=26. (Art. 417), 36y +-84y+49=312+49=361; '. extracting the square root, 6y+7=19,

and 6y=19-7=12; y=2,

and x=7—2y=7-4=3. Ex. 2. Given 4xy=96—x'y, and x+y=6, to find the values of x and y.

From the first equation xʻya + 4xy+4=100, and extracting the root, ry+2=+10;

..xy=8, or -12. Now squaring the second equation,

22 + 2xy +y=36; but 4xy =32, or -48.

... by subtraction, -2xy+yo=4, or 84 ; and extracting the root, x-y=+2, or +784 ;

but x+y= 6; .. by addition, 2x=8, or 4, or 6+84 ;

whence, x=4, or 2, or 3+721 ; and by subtraction, 2y=4, or 8, or 6784 ;

y=2, or 4, or 3+V21. Es. 3. Given x+r+y=18-y, and xy=6, to find the values of x and y.

By transposition, x+y++y=18 ; and from the second equation, 2xy

=12

.. by addition, x2+2xy + y +x+y=30 ; and completing the square,

121 (x+y)'+(x+y)+1=30+1=... extracting the root, x+y+1=+,

and aty=5, or -6 ; whence, from the 1st equation, **+y=13, or 24 ;

but 2ry=12;

... by subtraction, x-2xy +y=1, or 12; is extracting the root, x-y=+1, or+2v3.

Now x+y=5, or-6; .. by addition, ex=6, or 4, or -6+2/3 ;

..x=3, or 2, or -3+V3; and by subtraction, 2y=4, or 6, or 67 3X3 i

.: y=2, or 3, or-37V2. Es. 4. Given x-vxy+y-vx+vy=0, and Vitvy = 5, to find the values of x and y. Completing the square in the first equation,

(Vxy) - (Vx-vy)+=; and extracting the root, x-vym=+j;

Virvy, =1 or 0, but from the second equation, vx+y=5;

.. by addition, 2/x=6, or 5,

5 and Vr=3, or Ž.'. 1=9, or 7

25 By subtraction 2vy=4, or 5 i. y=4, or

4 Es. 5. Given zys=2y2, and 8x3-y3=14, to find the values of x and y.

25

[ocr errors]

y.

=x ;

2

but x

From the 1st equation, 2}=2y; and :: frs=ył ; substituting this value in the second equation,

825-4x}=14 ; and :: 1675-x}=28 ; or, by changing the signs, rf_16x3=_-28 ; completing the square, 23 -16.25 +64=36 ;

and extracting the root, -8=+6;

xs=14, or 2, and x=2744, or 8. Ex. 6. Given zi+yš=3x, and 7++y} =r, to find the values of x and By squaring the second equation, «+2xy}+y

+y3 =3x ; :: by subtraction, 2-q8+2x+y)=x—-32 ;

but from the second equation, y;=r-x+; Let this value be substituted in the preceding equation ; then c-*+91-22=22—3c ;

.. by transposition, 2x=x-*;

and dividing by ~, 9=x^2}; completing the square, x-x1+1=2+1=;

and extracting the root x1-+=+1; :.

-1 and x=4, or 1. By taking the former value of x, we have y}={-} =4--2=2; ..y=8. and by taking the latter value, yš:

= (since x-=-1,-~*=+1); :: y=8. Ex. 7. Given ga—6=8xły, and y—4=2yic; to find the values of x and y.

From the first equation, yo–8xảy=64 ; completing the square, y: -8x2y+16x=16x+64; extracting the root, y–4x3=+41(x+4);

and :: y=4x2 + 4x(x+4).

x =2, or

[ocr errors]

-*=1+1=2,

[ocr errors]
[ocr errors]

2

Also, from the second equation, y—2y3x3 =4 ;

, .. completing the square, &c. yo=x2 + (x+4);

yvx) multiplying by 4, 4y ==4x2+4/(x+4) ;

and y=16. 12 y4

3 But, from the second equation, až

8 2y

9

by involution, x: 425. When the equations are homogeneous, that is, when a>, , or xy, is found in every term of the two equations, they assume the form of

ax?+bxy+cyard,

a'r?+b'xy+c'ya d'; and their solution may be effected in the following manner :

Assume x=vy, then x=v‘yo ; by substituting their values for xa and x in both equations, we have

d avoy2 +bvy? +cy?=d; .. y'=

(1), ava+botc

d' a'voya+b'vyo+c'y =d'; .. ?P=a'z2+b'vyté

(2). d

ď Hence

; av +bute av +10+ c' .. (a'd-ad')u2+(b'dbd)v=cd-cd; which is a quadratic equation, from whence the value of v may be determined. Having the value of v, the value of y may be found from either of the equations (1) or (2); and then the value of x, from the equation x=vy.

Ex. 8. Given 2x+3.xy+y=20, and 522 +4y=41, to find the values of x and y. Let x=vy, then 20-ya +3vyo+y=20;

20 .. yo=

and 5v4y +4y=41;
2v2 +30+1'
41

41
y=
Hence

603-410
502 +4
2v2 +30+1

' -13. ... by division, completing the square, &c. v=° or .

41 369 Let v=}, then y=

=9;..Y=3, or --3, 5v2 +-4

and x=vy=1, or -1. 13

164

13 164 Again, let v= then

and xt861'

20

41

;

y=IV

2

[ocr errors]

2

Also, since

[ocr errors]

Coộsequently there are four values, both of x and y, which satisfy the proposed equations.

426. When the unknown quantities in each equation are similarly involved, the operation may sometimes be shortened, by substituting for the unknown quantities the sum and difference of two others. Ex. 9. Given +

+39=18,

to find the values of x and y. Y

and x+y=12, Assume xsztv, and y=2-0;+y=2z=12 : or 2=6; .. x=6+v, and y=6-v.

g2 y2

+2=18, 23+y+18xy ;

Y ::(6+y)+(6-1)=18(6+0) X(640);

or 432+360=648-18v2; and by transposition, 54vé=216 ; ..v=4 ; and v=+2; ..x=6+2=8, or 4 ;

and y=6+2=4, or 8. 427. In all quadratics of this kind, in which x may be changed for y, and y for x, in the original equations, without altering their form, the two values of one of the quantities may be taken for the values of the two quantities sought.

Ex. 10. Given x+y=2a, and 26 tyø=b, to find the values of x and y.

Let 2-y=22 ; then x=atz, and y=a42; i by substitution, (a+z)+(a-2)5=b, or, by involution and addition, 2a5+20aʼza +16az=b; b2a5

16-7-8a ...24+2az's , and r=#V[~*+ve -)]. 10a

100 :x=a+V1–ao+v++8a%)], and y=aFV[-a?

100 6+80%

-)]. 100 Now, let aty=6, and ** +yo=1056 ; then by substituting 3 for a, and 1056 for b, in the formulæ of roots, the values of x and y will be found ; that is, x=3+1, or 3+/-19; and y=3+1, or 377–19. Or, by substituting the above values of a and 6 in the equation 10az_20a 2 + 2as=b, it becomes 3024 +540z +486=1056 ; from which the values of z may be found ; whence, by substitution, the values of x and y will be determined, as before.

Ex. 11. Given 3+4y=14, and ya +-4x=2y+-11, to find the values of x and

y

Aos. =-46, or 2 ; and y=15, or 3.

z

[ocr errors]
« ΠροηγούμενηΣυνέχεια »