32. Given values of x.

3

xn

= 4, to find the value

3

Ans. (32) 20

+10x3+54x2+145x=602, to find the

1. Given 2x-y=15, 2

and 2x-y=9, to find the values of x and y.

From the second equation, x=

9+y2

2

;

substituting this value of x in the first equation, 9+ y2―y=15;

by transposition, y2—y=6;

completing the square, yo—y+—=6+1 = 20;

From the first equation, y=2x-15;

substituting this value of y in the second equa

tion, 4x2-62x=-234;

2

62

4

completing the square, x2-6x+31 =

2. Given 4x+4y=56,

961

16

to find the values of a

and 6x+3y=xy+12, and y.

Dividing the first equation by 4,

x+y=14;. y=14-x;

substituting this value of y in the second equa

tion, 6x+42-3x=14x-x2+12;

by transposition, x2-11x=-30;

completing the square, x2-11x+·

121 121

4

3. Given x2+ y2=50—x—? -y, to find the values of

and y=10,

Transposing the first equation,

x2+y2+x+y=50;

but, 2xy

x and y.

=40; (4 times 2nd. eq.)

by addition, x2+2xy+y2+x+y=90;

1

or (x+y)2+(x+y)=90;

completing the square, (x+y)2+(x+y)+

1

19

extracting the root, x+y+2=± 22;

by transposition, x+y=9, or —10;
squaring both sides of this equation,
x2+2xy+y2=81, or 100;

but 4xy =80,

by subtraction, x2-2xy + y2=1, or 20; extracting the square root,

x-y=1, or ±√20;

but x+y= 9, or -10;

by addition, 210, or 8, or -10±2/5;

.. x = 5, or 4, or ➡5±√5;

but by subtraction, 2y=8, or 10, or —10—2/5; .. y=4, or 5, or 5/5.

4. Given (x+y)2+2x=195—2y, to find the values and a2+y=97, of x and y.

Transposing the first equation, (x+y)2+2.(x+y)

=195;

completing the square, (x+y)2+2.(x+y)+1=196; extracting the root, x+y+1=±14;

by transposition, x+y=+14-1=13, or -15; squaring each side, x2+2xy+y2=169, or 225; but x2 + y2 = 97;

by subtraction,

2xy

72, or 128;

by subtraction, x2−2xy+y2=25, or —31;

.. x—y=±5, or ±√√—31; but x+y= 13, or —15;

by addition, 2x=18, or 8, or -15±√√—31;

but by subtraction, 2y= 8, or 18, or ‡√—31—15;

Dividing the first equation by 2, and completing the square, x1y2-2x2y+1=2208+1=2209; extracting the root, xy-1-47;

by transposition, x2y=48, or

-46;

multiplying the second equation by y,

x2y+y2=60—y;

substituting the affirmative value of xy, into this equation, 48+y2=60−y ;

by transposition, y2+y=12;

completing the square, y2+y+1=12+ 1 = 4;

But if the negative value of x2y be substituted, -46+y260-y;

by transposition, y2+y=106;

completing the square, y2+y+1=106+

EXAMPLES FOR PRACTICE

In Adfected Quadratics of two unknown quantities.

1. Given x2-xy=60, to find the values of x and and x+y=19,

5

y.

2

y= 7, or 214.

2. Given xy+18x=1728, to find the values of x

and 8y=9x,

3. Given 5x+4y=35, and 9y―xy=10x,

4. Given 2x+y=7, and x2+3y=13,

}

Ans.

and y.
Sx= =32, or
y=36, or

to find the values of r

and y.