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When there are more equations and unknown quantities than one, a single equation, involving only one of the unknown quantities, may sometimes be obtained by the rules before laid down for the solution of simple equations; and, in this case, one of the unknown quantities being determined, the others may be found by substituting its value in the remaining equations.

EXAMPLES.

28

; and

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y2 1. Given

23 to find the values of x and y.

XY
Here, from the second equation, we have y

784
by substituting this in the first, 3+ 65, or 24

6502

ac 2 784. Whence, by the common rule before given, we have 65

4225 x=&v

784 2

4 Or, by reducing the parts under the last radical, and ex

33 the

I 7, or -4,

4, and con2

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H

or

28

28 sequently, y=7

:4, or -7.

4 Or the solution, in cases of this kind, may often be more readily obtained by some of the artifices frequently made use of upon these occasions; which can only be learned from experience: thus, taking, as before, (1.) 22 + y2 = 65, (2.) xy

28, we shall have, as in the former method, by multiplying by 2, 2xy == 56, and, by adding this equation to the first, and subtracting it from the same, 22 + 2xy + ya 121, and 22

2xy + y2 == 9. Whence, by extracting the square roots of each of these last equations, there will arise x+y=+ll, and x — y=+3, and consequently, by adding and subtracting these, we shall have 2x + 14, or x =

-7, and y = 4, or - 4. It will also sometimes facilitate the operations, by substituting for one of the unknown quantities the product of the other, and a third unknown quantity; which method may be applied with advantage whenever the sum of the dimensions of the unknown quantities is the same in every term of the equation.

56

to find the values of x and y. xy+ 2y=60S

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or

2. Given I my

}

vy,

Here, agreeably to the above observation, let = then raya + vye = 56, and vy? + 2y = 60, whence, from the

56 first of these equations, yo = and from the second, ya

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60

Therefore, by equating the righthand member

v+2

or

9

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(206

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60

56 of these two expressions, we shall have

v +0 60v2 + 600 = 560 + 112. And, by transposing 56v, and

1 28 dividing the result by 60, 12 +

Hence, by the 15 15

1

1 common rule for quadratics, we have v

IV 30

900 28

1 41 4 十

+ 15/

And, consequently, by the former 30 30 3 60

60 part of the process, ya

18, or y=V ot-2

15 +2

4. (18)= 3 V 2, and x =

VY x 32=472. The work

3 may also be sometimes shortened, by substituting for the unknown quantities, the sum and difference of two other quantities; which method may be used when the unknown quanti.ies, in each equation, are similarly involved.

22 yo
+

18 3. Given

to find the values of x and y. Y

x + y = 12 Here, according to the above observation, let there be assumed x = z tv, and y =>— v. Then, by adding these two equations together, we shall have a toy 22 312, or 3=6, also, since x= 6 +0, y =6

6 + v, y=6 -- v, and by the first equation 23 + 1

18xy, we shall obtain, by substitution, (6 + 10)3 + (6 - 0) = 18 (6 + 0) (6 – u), or, by involving the two parts of the first member, and multiplying those of the second, 432 + 36v2 -= 648 - 18vo, whence, by transposition,

216 5402 = 216; and by division, za

4; or v = +2.

54 And therefore, by the first assumption, and the first part of the process, we have x=2 +0= 6+2=8, or 4, and Y=2 V=6+2 -- 4, or 8.

QUESTIONS PRODUCING QUADRATIC

EQUATIONS.

Let X =

The methods of expressing the conditions of questions of this kind, and the consequent reduction of them, till they are brought to a quadratic equation, involving only one unknown quantity and its square, are the same as those already given for simple equations.

1. To find two numbers such that their difference shall be 8, and their product 240.

Let x equal the least number.

Then will x + 8 = the greater,
And x (x + 8) = x2 + 8x = 240, by the question,
Whence x = .4+v (16 + 240) 4 + ✓ 256 by the

common rule, before given,
Therefore x = 16 - 4 = 12, the less number,

and x +8= 12 +8= 20, the greater. 2. It is required to divide the number 60 into two such parts, that their product shall be 864.

the greater part, Then will 60

ihe less, And x (60 —x) = 60x 22 = 864, by the question, Or, by changing the signs on both sides of the equation,

22 – 60x =

864, Whence x = 30 + V (900

30 + v (900 – 864) == 30 + V 36 = 30 + 6,

by the rule, And consequently, x = 30 + 6 = 36, or 30 - 6=24, the two

parts sought. 3. It is required to find two numbers such, that their sum shall be 10 (a), and the sum of their squares 58 (b).

Let x = the greater of the two numbers,

Then will a And 22 + (a − x) = 2x2 2ax + a = b, by the question, Or 2202 2ax = 6 - a2, by transposition,

b - a
And 23

by division.
2
b ca

1 Whence x =

£ ✓ (2b - am). 2

by the rule,
And if 10 be put for a, and 58 for b, we shall have

10 1
2
töv (116 – 100)= 7, the greater number,
2

10 1 And 10 - 20

V (116 -- 100) == 3, the less. 2

the less,

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a

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ll

Let x =

X

4 Having sold a piece of cloth for 341., I gained as much per cent. as it cost me; what was the price of the cloth ?

pounds the cloth cost,
Then will 24 the whole gain,
But 100: 0 :: X : 24 - X, by the question,
Or, x2

100 (24 — x) = 2400 100x,

That is 22 to 100x --- 2400, Whence a 50 + 0 (2500 + 2400) = 50 +- 70 == 20,

by the rule, And consequently, 201. - price of the cloth. 5. A person bought a number of sheep for 801., and if he had bought four more for the same money, he would have paid ll. less for each; how many did he buy? Let s represent the number of sheep,

80
Then will be the price of each,

320,

80 And

price of each, if x + 4 cost 801. ic +4

80 80 But

+ 1, by the question, 8 +-4

80x Or, 80

+ x, by multiplication.

XC + 4 And 80x + 320 = 803c + 202 to 4x, by the same, Or, by leaving out 80x on each side, 22 + 4x Whence x = 2 + 1 (4 + 320) = 2 t. 18, by the rule,

And consequently, = 16, the number of sheep. 6. It is required to find two numbers, such that their sum, product, and difference of their squares, shall.be all equal to each other. Let 2 = the greater number and

y =

= the less.
Şu to y = XY
Then

wಣಿ - 9*.

x2 - y2 Hence 1

y, or x = y + 1, by 2d equation. And (y + 1) +y=y(y + 1), by 1st equation, That is, 24 to 1 = y + y; y + y

=1. 1

1 1 +

tv5, by the rule, 2

2 2

1 Therefore, y=

V5 = 1.6180 ...

} by the question.

i ty

C

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+

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And x =

3
y+1= +15= 2.6180 ...

2
Where ... denotes that the decimal does not end.

7. It is required to find four numbers in arithmetical progression, such that the product of the two extremes shall be 45, and the product of the means 77.

Let x = least extreme, and y = common difference, Then X, x + y, x + 2y, and x + 3y, will be the four numbers,

$*(x + 3y) = x + 3xy Hence

{(x+) 6+ 2y) = x + 3xy + 2y = 77}, by the question,

And 2y2 = 77 45 = 32, by subtraction,

32
Or, ya
16 by division, and

y= V16 = 4. 2 Therefore, 22 + 3xy = x + 12x = 45, by the 1st equation, And consequently, i = 6 + (36 + 45) 6+9= 3,

by the rule. Whence the numbers are 3, 7, 11, and 15. 8. It is required to find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84.

and be the three numbers,
Then, uz=yo, by the nature of proportion,
And {

272727284} by the question,
Hence, x +2= 14 y, by the second equation,
And 2a + 2zx + 7 = 196 -- 28y + y, by squaring both sides,
Or x2 + 2 + 2y = 196 – 28y + y by putting 2yfor its

equal 2xz,
That is, a + y2 + 7 = 196 - 28y, by subtraction,

Or, 196 - 28y = 84, by equality,
196 84

= 4, by transposition and division. 28

16. Again, xz y = 16, or x = by the 1st equation,

16. And x + y +-% = +4+2= 14, by the 2d equation,

Or, 16 + 42+ 22 14z, or 22 - 102 Whence z=5&v (25 - 16)=5+3=8, or 2 by the rule,

Therefore, the three numbers are 2, 4, and 8.

Let x, y,

2,

Hence y

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-16,

9. It is required to find two numbers, such that their suni shall be 13 (a), and the sum of their fourth powers 4721 (6). Let x = the difference of the two numbers sought,

1 1 a + x Then will, a + x, or = the greater number,

2

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