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OF QUADRATIC EQUATIONS,

WITH TWO UNKNOWN QUANTITIES.

176.) For the solution of Quadratic Equations with two unknown quantities, the three following methods, which are generally used, will be found applicable to a vast variety of questions.

The FIRST METHOD is performed by this general

(77.) Rule.- Find the value of one unknown quantity in terms of the other, in one of the given equations, and substitute this value for it in the other equation ; from which there will arise a quadratic equation soluble by the methods already given.

Eramples. 2.r+y '1. Given

92 3

to find the values of x and y. and 3ry = 210 Solu.-Multiply by 3, and from the first equation 2x +y=27.

27 .. 2r=27-y; and rz

= 9

y 2

Hence, 3.ry=3(2=)y=210;

Or, 3 (27-y) xy=420, multiplying by 2.
That is, 8ly-3y* =420.

Or, .. 27y-yo=140, dividing by 3
. ye-27y=-140, changing the signs.

729_ 729 729-560 Comp. the sq. ye_274 + 4

-140 =

4

169
4

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Or,

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= 20 or 7;

=

+13

27 +13
;
2

2
Consequently, =
27-y_27 ---20 or 7

34 or 10. 2

2 2. A certain number consists of two digits ; the left hand digit is equal to three times the right hand digit; and if we subtract 12 from the number itself, the remainder will be equal to the square of the left hand digit; it is required to find the number. Solu.-Put I =left hand digit then will 10x +y be

and y=right band one the number.

y+

Hence, by the question, r=3y,

and 10.r + y-12=r; Therefore, 30y +y-12=9ys by substitution; Or, gyo-31y=-12;

31

12 Whence yo -y=

j 9

9
31
And y' -
961

12
961 961 – 432

529
324
324 9 324

324'
31_+23 31 +23 54

8

4

-= 3 or
18 18
18 18 18

9
And r = 3y = 3 X 3 = 9; consequently, the number is 93.

3. Let there be two numbers such that, if the less be taken from 3 times the greater, the remainder shall be 35 ; and if 4 tines the greater be divided by 3 times the less plus 1, the quotient will be equal to the less number; to find these two numbers. Ans. 13 and 4.

4. There is a certain number, and the sum of its digits is 15; now if 31 be added to their product, the digits will be inverted ;. it is required to find that number.

Ans. 78.

Hence, y

and y =

or

5. Given ty= 8
y

20
and +1=
y

y

to find the values of I and y.

:4;

Ans. x = 16,
or x = 15, y

5.

(78.) The SECOND METHOD :

When those parts of the given equations that contain the two unknown quantities bear a particular relation to one another; as, for instance, their sum and difference; their sum and product; their sum and the sum of their squares; their product and the sum of their squares; their sum and the sum of their cubes, &c. &c. The equations may then be managed by some one of the following solutions.

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Then, the square of the first equation, =r? + 2.ry+y'=a' (A) And the second equation x by 4

4xy

=46 (B). By subtracting (B) from (A) .—2xy +ye=a2-40. Extracting now the square root 1-y=+va-41,

but

x+y=a. Hence, by addition, 2x=a+va-4b, or rs

atva'-46

2

2

a F var-46 And, by subtraction, 2y=a F va-46, or y= (81.) 3. Let x +y =a

2e +y'=1

Then the square of the first equation is ze + 2xy +yo=a* (A); From which take

2 +y==; Therefore

2xy=a'-6 But multiply by 2, and then we have 4xy=2a'—26 (B). Subtract (B) from (A), and x-2xy + y'=a-2a-26=26-a.

Hence, 1-y=+_26a. Having now x+y and s-y, we should proceed by the first Example fArt. 79) to find the values of x and y. (82.) 4. Let **+y=a

and xy=b

Then

2xy=26.
Therefore 3* + ary+y=a+21, or x+y=+ya+2b.
Subtract now 4xy =46;
Then will -2xy +ye=a+26–46=2-26,

Orx-y+va-26.
Hence, & and y may now be found as in the preceding exa
(83.) 5. Suppose that I +y =a

and x+y=1

Then will x+3xy +3.xy? +y*=a , cubing the first equa
Subtract deco

+y=1, the second equation

Then the difference will be 3.roy + 3xy'=a'-6,
Or
3.xy x (x+y)=a'-;

a
But x+y=a; :: 32 x xy=a'-l, or xy=

За

أو

-)

Having now found x+y and xy, the values of x and y may he found as in the second Example.

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Raise rty to the 4th power, and x* +4x*y+6roy? + 4xyo+y=a;
Subtract from this the 2d equation, m*

+y=b.
and the remainder will be 4x®y + 6xưys + 4rys = a - b.

Let now atub=0

= (A)

).

Divide the remainder by xy, and we have 4x? + Oxy +4y' =

xy Now the first equation squared is .... ...+ 2xy + y=a; Therefore,

4.x2 +8xy +4yo=4a* (B) From (B) the greater take (A) the less, then 2xy=4a?

ту
Clear the remainder of the fraction, and we get 2xły?=4aoxy-C.
By transposition and division, xạy2– 2aRxy=-c.

Hence we have ry—a®= Va-c,
And we get my

=a? +vat-c.
Wherefore it and y may now be found as in Example second.
(85.) 7. Let x+y=a, and x + y'=1.
Then the first equation raised to the 5th power,

X+5x*y +10.x*y* +10x®y + 5zy'+y=a';
The second ditto is
}

+y^=^.
by the question,
Therefore the difference, 5x*y + 10x®y? + 10x®yo +5xy*=a'-l=c,

5xy x 2 + 2x®y + 2xy! + ys=c;
Therefore, I'+2xoy+2xy? +y'=

5xy
Now 28+3x®y+3xy2+y8=a;
Hence, by subtraction, .. x+y+ ryo=a'— 584

Or,

C

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с

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с

(

Or,
xy x (x+y)=a-

5ry
And
axy=a' -

;

5xy
From which equation the value of xy may be found

And then x and y may be found as in Example 2. (86.) 8. Given x+y=s, and xy=p; to find the value of x +yo, 1+y'; **+y* ; &c. in terms of (s) and (p).

First, the square of the 1st equation, x + 2xy +yo=s* ;
Therefore, by transposition, x+y=se-2xy=s-2p.
Secondly.....

(ro+y') x (x+y) = (5—2p),
Or, which is the same, x'+xy x (x+y)+y = 59-2ps,
That is,

x' +ps+yo

59-2ps; Therefore,

x+y = 5-3ps. And, thirdly,

(r+yo) x (x+y) (5-3ps), Or, which is the same, **+ry x (+y)+y = 5*—3pse,

That is, x+px (so—2p)+y* St-3ps,
Therefore, ..

r* +y = $-4ps'+2ps.

Examples for Practice. 9. There are two numbers, the sum of which is 19, and the surr of their squares is 193 ; required those numbers.

Ans. 12 and 7 10. The sum of the squares of two numbers is 261, and their pro'duct is 90; required those two numbers.

Ans, 15 and 6 11. The sum of two numbers is 9, and the sum of their cubes is 351 ; required those two numbers.

Ans. 7 and 2 12. The sum of two numbers is 7, and the sum of their 4th powers is 641; what are those numbers ?

Ans, 5 and 2 13. The sum of two numbers is 6, and the sum of their 5tt powers is 1056; what are those numbers ?

Ans. 2 and 4. 14. The sum of two numbers is 13, and their product is 30; it is required to find the sum of their 4th powers.

Ans. 10081. à (87.) The THIRD METHOD.- Equations of the form

Odry + eys in which the sum of the indices of the unknown quantities is the same in every term of both equations, are managed by

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