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QUESTION 4, What is that number which being multiplied by 6, the product increased by 36, and that sum divided by 18, the quotient shall be 20 ? Solu. Let the number sought.

Then 6r = the number multiplied by 6.
Also Or + 36= the product increased by 38,

6.0 + 36
And

= that sum divided by 18 18

6.0 + 36 Hence, by the question,

= 20.

18
And 62+36 = 360, multiplying by 18.
or 6x = 360

36 = 324 ;
324

= 54, the number required.

6 QUESTION 5. A post is one-fifth in the earth, three-sevenths in water, and 13 feet out of the water; what is the length of the post ? Solu. Let x= the length of the post; Then = the part of it in the earth ;

=
3 x
Also

7

= the part of it in the water.

And 13 = the part of it out of the water ; But the part in earth + part in water + part out of water the whole post.

3.1 Therefore +

5+

+ 13 = 2. 7

15.0
Multiply by 5, then 3+ + 65 5x.

7
by 7, and 7 x + 150 + 455 = 35.2 ;
or

455 351-74 153 13%;
455
Hence x =

= 35 feet the length of the post.

13 QUESTION 6. Having paid away one-fourth and one-seventh of my money, I had eighty-five pounds left in my purse; how much money was in it previously to these disbursements ? Solu. Let s= the money id purse at first, then is the money

7 paid away Now, by the question,

+*+ +85

7

1

281877 +40+2380

17.r=2380

2380
Therefore, r= -=140The money at first in purse.

17

Eramples for practice. 7. The difference of two numbers is 10, and if to their sum 15 be added, the whole will be 43 ; it is required to find those two numbers ?

Ans. 9 and 19. 8. The difference of two numbers is 14, and if 9 times the less be subtracted from 6 times the greater, the remainder will be 33 ; what are those numbers ?

Ans. 17 and 31. 9. If to a certain number I add 20, and from two-thirds of this sum I take away 12, the remainder will be 10; what is that number?

Ans. 13. 10. What number is that, of which if I add one-third, one-fourth, and two-sevenths together, the sum shall be 73 ?

Ans. 84. 11. What number is that the third part of which exceeds its fifth by 72 ?

Ans. 540. 12. Two merchants, A and B, lay out equal sums of money in trade? A gains 1201., and B loses 801.; and now A.'s money is treble. of B.'s :- Pray what sum had each at first ?

Ans. 180.

(66.) The equations which we have hitherto considered contained but one unknown quantity; but, in the solutions of Algebraic Problems, equations frequently arise containing two or more unknown quantities ; as, ar + by =c, ax + ly + cx=d, &c. From the first of these it is evident, by the rules already laid down, that

by. or, X=C

- by

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ar =

[ocr errors]

is sa.

[ocr errors]
[ocr errors]

a

In this equation the value of x, which is found in terms of y; but y is an unknown quantity, consequently I still remains undetermined; whence it is manifest, that from the equation ar + by = 1, neither x nor y can be found. But if in addition to ax + by=C, we have another equation, as,

f - ey dr + ey=f, from which x may be found as before =

d c-by we should then have

each being equal to x.

d From this last equation, y may be found by the foregoing rules; and when y is found

becomes known.

fs ey

C- by

[ocr errors]

From what has been said, it aprears that in order to determine the values of two unknown quantities, two equations must be given. In

like manner, it may be shewn that three equations must be given to determine three unknown quantities ; and so on. It mig?t be further remarked, that these equations must be independent one upon the other. Thus, ar + ly = 0, and dr + ey s are independent, but ar + ly =, and ax =-- - by, are not independent, for the one may be derived from the other; and the two, though of a different form, amount, in effect, to no more than one equation. What we have now said will lead us to

PROBLEM I.

To exterminate two unknown quantities, or to reduce the two simple

equations containing them, to a single one. (67.) Rule 1.-1. Find the value of one of the unknown quantities in terms of the other, from each of the two given equations.

2. Make the two values thus found equal to each other, and there will arise a new equation containing only one unknown quantity, whose alue may be found as before.

2r +3y=23 Example 1. Given

5x – 2y=100

it is required to find x and y. 23-34

10+2y From the first equation x=

from the second x= 2

5 23-3y_10+2y Consequently

2 Or 115 — 15y=20+4y, Or 19y=115—20=95; 95

23—3y

23-15 And r=

34. 19

x+y=a 2. Given

it is required to find x and y.

[ocr errors]
[ocr errors]

5

That is y=

59

2

2

{

a-6

's and

From the first equation x=a-y, and from the second x=b+y. Therefore a-y=+y, or 2y=a-V, consequently y=

2 and x = a- y

a+b Orr=a

[ocr errors]

2

2

Lys } it is required to find x and y.

ser+ fy=7 3. Given

zr+ || 18
у

2y From the first equation x = 14

3

Зу And from the second x=24 i

2 2y

3y There ore 14 = 24 , Aud 42 – 2y = 72

3

2

[merged small][ocr errors][subsumed][ocr errors]
[blocks in formation]
[ocr errors]

= 6.

+

20

2y

24 And I=14 S 14 = 6. 3

3 4. Given 45 ty 34, and 4y + x 16, it is required to find x and

Ans. x=8, and dy.

ly

= 2. 21 Зу 9

3.1 2y

61 5. Given + and

, it is required 5 4

4 5 120

] to find x and y.

Ans. x = and y =

y=

3 6. Given x+y=s, and x-yo=d, it is required to find x and y.

fa+d sa-d Ans. I= and

y= 2s

2s 7. Given 1-y = d, and so y

n : m, it is required to find s and y.

Rule 2.1. Consider which of the unknown quantities must be first terminated, and find its value in that equation where it is least involved.

2. Substitute the value, thus found, for its equal in the other equation, and a new equation will arise with only one unknown quantity, whose value may be found as before

x+2y=171 Ex. 1. Given 38-y= 2 it is required to find r and y. .

31= From the first equation r=17--2y. And this value substituted for s in the second, gives 3(17-2y)-y=2, Or 51-6y - y=2, or 51—7y=2; that is 7y = 51—2=49;

49

= 7, and x = 17-2y=17-14 = 3.

{

'

Whence y=3

{x+y=1} }it is required to find 8 and 3

.

2. Given
-y 3

y From the first equation r=13-y.

And this value, substituted for x in the 2d, gives 13-y-y=3, O 13-2y=3, That is, 2y=13-3=10,

10 Whence y=

=5, and x=13-y=13-3=8, 2

Sa:b:: X: Y 3. Given Pro+y=c it is required to find x and y.

a'y The firs: analogy turned into an equation is bx=ay, or x=

b ay

= C, or

And this value of x substituted in the 2d, gives( *%)*+3* =

[ocr errors]
[ocr errors]

+y'=c,

[ocr errors]

Whence y=

[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

5

3

[ocr errors]

Ans. (da

062 Or aoyo +boy' =.cb, or ye=

ch2
102772)t, and == 6

a? +62 4. Given 2x+3y=16, and 3x-2y=11, it is required to find r and y.

Ans. x=5 and y=2. 5. Given + 7y=99, and +70=51, it is required to find and y +

Ans. x=7 and y=14. 6. Given 3—12=X+8, and**+ -8=24*+ 27, it is re

++y y

2y=

,

4 quired to find u and y.

Ans. x = 60 and y=40. 7. Given a : 6:: 1 : y, and way=d, it is required to find x and y.

d13 F=x, and

)=y. .

-63 Rule 3.-1. Multiply or divide the given equations by such numbers or quantities as will make the term which contains one of the unknown. quantities the same in both equations.

2. Then, by adding or subtracting the equations, according as may oe required, a new equation will arise with only one unknown quantity as before.

$3.x+5y=402 Ex. 1. Given Bir 2x+2y=143 it is required to find x and

y. First, multiply the 2d equation by 3, and it will give 3x+6y = 42.

Then from the last equation subtract the first, and it will give oy-5y=42-40, or y=2, .: x=14–2y=1444=10.

2. Given {2r+ 5y=16 } it is required to find x and y.

1 Let the first equation be multiplied by 2, and the 2d by 5, and we shall have 10x - 6y=18, 10x +25y=80 ; And if the former of these be subtracted from the latter, it will give

62 Sly=62, or y= 2,

31

3y And consequently r =

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9+34, by the first equation,

5

Orx

–9+6 15

3.

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Another method.
Let the first equation be multiplied by 5, and the second by 3,

=

and we shall have { 25.4 I 157=45

Now, let these two equations be added together,

and the sun will be 31r=93,

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