4

Ans. x=4, or 1, or ±√−7.

Ex. 48. Given 4x+= =4x+33, to find the values of x.

2

§ 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 unknown 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. Whence, by substitution, the values of the other may also be determined. In 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 unknown quantities, (§. 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, 2 to find the values of x and y. and x2+3xy-y2=23,

From the 1st equation x=7—2y ; x2-49-28y+4y2; Substituting these values for x and x in the 2d equation, then 49-28y+4y2+21y—6y2—y2=23,

or 3y+7y=49-23=26.

(Art. 417), 36y2+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—x2y2, and x+y=6, to find the va

lues of x and y.

From the first equation x2y2+4xy+4=100, and extracting the root, xy+2=+10;

Now squaring the second equation,

..xy=8, or—12.

=32, or 48.

x2+2xy+y2=36; but 4xy

... by subtraction, x2-2xy+y=4, or 84; and extracting the root, x-y=+2, or ±√84; but x+y= 6;

.. by addition, 2x=8, or 4, or 6±84; whence, x=4, or 2, or 3±√21 ;

and by subtraction, 2y=4, or 8, or 684; ... y=2, or 4, or 3√21.

Ex. 3. Given x2+x+y=18-y3, and xy=6, to find the va

lues of x and y.

By transposition, x2+y2+x+y=18;

and from the second equation, 2xy

=12;

... extracting the root, x+y+1=±1, and x+y=5, or —6 ;

whence, from the 1st equation, x2+y=13, or 24;

Ex. 4. Given

but 2xy=12;

... by subtraction, xa—2xy+y3=1, or 12; ... extracting the root, x-y=+1, or+2/3. Now x+y=5, or-6;

.. by addition, 2x=6, or 4, or -6+2/3; ... x=3, or 2, or −3±√√3 ; and by subtraction, 2y=4, or 6, or 6x/3; . y=2, or 3, or 3√2. x-2/xy+y—√x+y=0, and √x+√y

5, to find the values of x and y.

Completing the square in the first equation,

(√x—√y)2 — (√x−√y)+1=1 ;

and extracting the root, x

√x-√y,=1 or 0,

but from the second equation, x+√y=5;

By subtraction 2y=4, or 5 ; .. y=4, or

25

4

25

4

Ex. 5. Given x3y=2y2, and 8x3—y3=14, to find the

values of x and y.

From the 1st equation, xa=2y3; and .. ‡x3=yś; substituting this value in the second equation,

8x-4x3=14; and .:. 16x3-x3=28; or, by changing the signs, x3 — 16x3——28 ; completing the square, x3—16x3+64—36 ; and extracting the root, x3—8=+6;

x3=14, or 2, and x=2744, or 8.

Ex. 6. Given x+y=3x, and x+y=x, to find the va

lues of x and y.

By squaring the second equation, x+2x3y+y3=x2 ;

... by subtraction, x-x3+2x3y3—x2—3x ; but from the second equation, y=x-x;

Let this value be substituted in the preceding equation; then x-x+2x-2x=x2-3x ;

.. by transposition, 2x=x2—x2;

and dividing by x, 2=x-x2

completing the square, x-x+1=2+1=1;

and extracting the root

.. x=2, or -1 ; and x=4, or 1.

By taking the former value of x, we have y3—x—x2 =4-2=2; ..y=8.

and by taking the latter value, y=x-x=1+1=2,

(since x ——1, —x3=+1); .. y=8.

1 I

Ex. 7. Given y—64=8x1y, and y-4=2yx to find the

values of

x and y.

I

From the first equation, y3--8x2y=64 ;

completing the square, y3-8x3y+16x=16x+64;

extracting the root, y-4x3=+4√(x+4) ;

and.. y=4x2+4√√/(x+4).

;

Also, from the second equation, y-2yx=4;

1

completing the square, &c. y=x3±√(x+4) ; multiplying by 4, 4y2=4x2+4√(x+4) ;

425. When the equations are homogeneous, that is, when x2, y2, or xy, is found in every term of the two equations, they assume the form of

ax2+bxy+cy2=d,

a2x2 + b'xy+c'y2=d; and their solution may be effected in the following manner :

Assume x=vy, then x2=vy; by substituting their values for x and x in both equations, we have

... (a'd—ad')v2+(b'd—bď′)v=cd—c'd; 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 2x2+3xy+y2=20, and 5a2+4y2=41, to find the values of x and y.

Let x=vy, then 2v3y2+3vy2+y=20;

... by division, completing the square, &c. v=3 or 3.

Consequently 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 -+:

y X

and x+y=12,

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

.. (6+v)3+(6—v)2=18(6+v) ×(6—v) ; or 432+3602=648-18v2;

and by transposition, 54v2=216;

..2=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 x+y=b, to find the values of x and y.

Let x-y-22; then xa+z, and y=a-z;

.. by substitution, (a+z)+(α-2)56, or, by involution and addition, 2a5+20a322+10az=b;

b+8a5.

10a

10α

−)], and y=a[¬a2±

Now, let x+y=6, and x+y=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 3-19. Or, by substituting the above values of a and b in the equation 10az1-20a z2+2a=b, it becomes 30*+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 x+4y-14, and y+4x-2y+11, to find the values of x and y..

Ans. x——46, or 2; and y=15, or 3.