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.

{

1. Given x2+y2=65) xy=28

}

to find the values of x and y.

Here, from the second equation, we have y

Whence, by the common rule before given, we have

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

extracting the root x=±±

65 33

7, or-4, and con

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.) x2+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, x+2xy+y3= 121, and 2-2xy+y=9. Whence by extracting the square roots of each of these last equations, there will arise, x+y=11, and x-y=+3, and consequently by adding and subtracting these we shall have 2x⇒±14, or x=7, or -7, and y=4, or 4. It will also sometimes facilitate the operation 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.

x2+xy=56

2. Given {xy+2y=60

}

to find the values of x and y.

Here, agreeably to the above observation, let x=vy, then vay2+vya≈56, and vy2+2y2=60, whence, from the

first of these equations, y2=

56

2

and from the second

Therefore, by equating the right hand mem

ber of these two expressions, we shall have

= ́v+2 v2+v"

or 60v2+60v=56v+112. And, by transposing 56v, and

1 38

dividing the result by 60, v2+:

19 15

Hence by the

common rule, for quadratics, we have v=

1

And, consequently, by the form

er part of the process, y2=;

4

60

=18, or y=√

60
v+2 11+2

(18)=3/2, and x=vy=× 8 √√2=42. The work

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 quantities, in each equation, are similarly involved

x2 y2
+=18

3. Given y x

x+y=12

}

to find the values of x and y.

Here, according to the above observation, let there be assumed x=z+v, and y=z-v. Then by adding these two equations together, we shall have x+y=2z=12, or z=6, also, since x=6+v, y=6-v, and by the first equation, x+y=18xy, we shall obtain, by substitution, (6+ v)+(60)3=18(6+v)(6—v), or, by involving the two parts of the first member, and multiplying those of the second, 432+362-648-18v3, whence, by transposition

216

54 216; and by division, v2=? -4; or v2.

54

And therefore, by the first assumption, and the first part of the process, we have x=z+v=6±2=8, or 4, and y=z-v=6±2=4, or 8.

QUESTIONS PRODUCING QUADRATIC
EQUATIONS.

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

4+✔(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.

Let x= the greater part,

Then will 60-x= the less,

And x(60-x)=60x-x2=864, by the question, Or by changing the signs on both sides of the equation x2-60x=-864,

Whence x=30(900-864)

the rule,

30/36=3016, by

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 the greater of the two numbers, Then will a-x= the less,

And x2+(a-x)2=2x2-2ax=a2b, by the question, Or 2x2-2axb-a2, by transposition,

b-a2

And x2 -ax

by division.

2

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

x=

10 1

2

+2✓(116—100)=7, the greater number,

10 1

And 10-x=

2

(116-100)=3, the less.

4. Having sold a piece of cloth for 241., I gained as much per cent. as it cost me; what was the price of the cloth ?

Let x pounds the cloth cost,

Then will 24-x= the whole gain,
But 100 x:x : 24-x, by the question,
Or x2100(24-x)=2400-100x,

That is, x+100x=2400,

Whence x-50+(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 4 more for the same money, he would have paid 17. less for each; how many did he buy?

Let a represent the number of sheep,

Then will

80

be the price of each,

x+4

And 90x+320=80x+x+4x, by the same, Or, by leaving out 80% on each side. x2+4x=320, Whence x=2+√(4+320)=−2+18, by the rule, And consequently x16, 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 x= the greater number and y= the less.

Then {+3} by the question.

Hence 1:

x+y

xy, or xy+1, by 2d equation.

And (y+1)+y=y(y+1), by 1st equation,
That is, 2y+1=y2+y; or y2+y--1,

1

Whence y=+ √ / (+1)

2

1

+1)=2+2✓5, by the rule,

3 1

And

2 2

xy+1+5=2.6180...

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

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