with which the same process is to be repeated, and so on till there is only one equation, to be reduced by the rules for a single equation.

2. Or, as in the 2d rule of the same problem, find the value of one of the unknown quantities in one of the equations only; then substitute this value instead of it in the other equations; which gives a new set of equations to be resolved as before, by repeating the operation.

3. Or, as in the 3d rule, reduce the equations, by multiplying or dividing them, so as to make some of the terms to agree: then, by adding or subtracting them, as the signs may require, one of the letters may be exterminated, &c. as before.

EXAMPLES.

1. Given

x + y + z = 9
x + 2y +3x=16
x+3y+4z
= 31

; to find x, y, and z.

1. By the 1st method :

Transp. the terms containing y and z in each equa. gives

x= 9

x=16

x = 21

2y

Sy

3z,

4z;

Then putting the 1st and 2d values equal, and the 2d and 3d values equal, give

9

16

z = 16
21

3%= 2y

3y 4z;

=5

In the 1st trans. 9, z, and 2y, gives y = 7
In the 2d trans 16, 3z and 3y, gives y
Putting these two equal, gives 5
Trans. 5 and 2z, gives z = 2.

Hence y = 5 - z=3, and x = 9

z=7 2z;

y

y - z;

This value of x substit. in the 2d and 3d, gives

9+ y + 2z = 16,

9+ 2y + 3z = 21;

In the 1st trans. 9 and 2z, gives y

This substit. in the last, gives 23

=7

Trans. z and 21, gives 2 = z.

2z;

z = 21;

Hence again y = 7 — 2z = 3, and ≈ = 9 — y —— z = 4.

3dly. By

3dly. By the 3d method; subtracting the 1st equ. from the 2d, and the 2d from the 3d, gives

Subtr. the latter from the former, gives z = 2.
Hence y 5 –
-1= 3, and x = 9 — y − z = 4,

find x, y, and z.

x- y = 2, x — z = 3, and y z = 1; to Ans. x7; y = 5; z— 4,

2x + 3y + 4z = 34°

5. Given

3x+4y+5z = 46

; to find x, y, and z.

4x+5y+6z = 58

12.

Ans. x = 1, y ='20, z = 60.

A COLLECTION OF QUESTIONS PRODUCING SIMPLE

EQUATIONS.

QUEST. 1. To find two numbers, such, that their sum shall

be 10, and their difference 6.

Let x denote the greater number, and y the less*.

Then, by the 1st condition x + y = 10,
And by the 2d

Transp. y in each, gives

Put these two values equal,
Transpos 6 and — y, gives
Dividing by 2, gives
And hence

· X- y =

6,

y,

6+ y;

x = 10 and x = gives 6+ y = 10- y; 2y= 4;

* In all these solutions, as many unknown letters are always used as there are unknown numbers to be found, purposely the better to exercise the modes of reducing the equations: avoiding the short ways of notation, which though giving a shorter solution, are for that reason less useful to the pupil, as affording less exercise in practising the several rules in reducing equations.

QUEST. 2,

QUEST. 2. Divide 1001. among A, B, C, so that ▲ may have 201. more than B, and в 10. more than c.

In the 1st substit. y + 20 for x, gives 2y + z + 20 = 100; In this substituting z + 10 for y, gives 3z + 40

By transposing 40, gives

And dividing by 3. gives

3z = 60;

z= 20.

Hence y z + 10 = 30, and xy + 20 = 50.

100;

QUEST. 3. A prize of 5001. is to be divided between two persons, so as their shares may be in proportion as 7 to 8; required the share of each.

Put x and y for the two shares; then by the question,

7:8::x:y, or mult. the extremes

and the means, 7y = 8x,

and x+y=

Transposing y, gives x = 500

500;

y;

This substituted in the 1st, gives 7y
By transposing 8y, it is 15y= 4000;
By dividing by 15, it gives y= 2663;
And hence x = 500 — y =

2333.

QUEST. 4. What number is that whose 4th part exceeds

its 5th part by 10?

Let x denote the number sought.
Then by the question ‡x − x = 10;

By mult. by 4, it becomes x

x = 40;

By mult. by 5 it gives x 200, the number sought.

QUEST. 5. What fraction is that to the numerator of which if I be added, the value will be; but if one be added to the denominator, its value will be ?

y+1;

The 1st mult. by 2 and y, gives 2x + 2
The 2d mult. by 3 and y + 1 is 3x =
The upper taken from the under leaves -
By transpos. 2, it gives x = 3.
And hence y

2x + 2 = 8; and the fraction is

= y;

2 = 1;

QUEST. 6:

0

QUEST. 6. A labourer engaged to serve for 30 days on these conditions: that for every day he worked, he was to receive 20d. but for every day he played, or was absent, he was to forfeit 10d. Now at the end of the time he had to receive just 20 shillings, or 240 pence. It is required to find how many days he worked, and how many he was idle?

Let x be the days worked, and y the days idle.
Then 20x is the pence earned, and 10y the forfeits;
Hence, by the question

= 30,

x + y =

and 20x

10y = 240;

The 1st mult, by 10, gives 10x + 10y = 300;
These two added give 30x=540;
This div. by 30, gives

Hence

y=30

x= 18, the days worked; -x= 12, the days idled.

QUEST. 7. Out of a cask of wine, which had leaked away 1, 30 gallons were drawn; and then, being gaged, it appeared to be half full; how much did it hold?

Let it be supposed to have held x gallons,
Then it would have leaked 4 gallons,

Conseq. there had been taken away + 30 gallons,
Hence r =4x+30 by the question.

Then mult, by 4, gives 2x = x + 120 ;

And transposing x, gives x = 120 the contents.

QUEST. 8. To divide 20 into two such parts, that 3 times the one part added to 5 times the other may make 76.

QUEST 9. A market woman bought in a certain number of eggs at 2 a penny, and as many more at 3 a penny, and sold them all out again at the rate of 5 for two-pence, and by so doing, contrary to expectation, found she lost 3d.; what number of eggs had she?

=

Let x number of eggs of each sort.
Then will x = cost of the first sort,

And

cost of the second sort;

But

But 5: 2 : 2x (the whole number of eggs) : fx ;
Honce fr == price of both sorts, at 5 for 2 pence;
Then by the question x +

Mult. by 2, gives

2

x + 3x

And mult by 3. gives 5x

r = 3;

3
- x= 6;

5

= 18;

Also mult. by 5, gives x 90, the number of eggs of each sort.

QUEST. 10. Two persons, A and B, engage at play. Before they begin, A has 8 guineas, and B has 60. After a certain number of games won and lost between them, a rises with three times as many guineas as B. Query, how many guineas

did a win of B?

Let x denote the number of guineas a won.
Then A rises with 80 + x,

Theref. by the quest. 80 + x = 180

Transp. 80 and 3x, gives 4x = 100;

And dividing by 4, gives x = 25, the guineas won.

QUESTIONS FOR PRACTICE.

1 To determine two numbers such, that their difference may be 4, and the difference of their squares 64.

Ans. 6 and 10.

2. To find two numbers with these conditions, viz. that half the first with a 3d part of the second may make 9, and that a 4th part of the first with a 5th part of the second may make 5. Ans. 8 and 15.

3. To divide the number 20 into two such parts, that a 3d of the one part added to a fifth of the other, may make 6.

4. To find three numbers such, that the and 2d shall be 7, the sum of the 1st and 3d 8, the 2d and 3d 9.

Ans. 15 and 5.

sum of the Ist and the sum of Ans. 3, 4, 5.

5. A father, dying, bequeathed his fortune, which was 28001. to his son and daughter, in this manner; that for every half crown the son might have, the daughter was to have a shilling. What then were their two shares?

Ans. The son 20001, and the daughter 8001.

6. Three persons, A, B, C, make a joint contribution, which in the whole amounts to 4001. of which sum в con

tributes