Page images
PDF
EPUB
[graphic]

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. before.

EXAMPLES.

Z,

42 ;

[ocr errors]

x + y + z = 9 1. Given x + 2y + 3 = 16 ; to find x, y, and z,

x + 3y + 4z = 31 1. By the 1st method : Transp. the terms containing y and z in each equa. gives

X = 9

y x = 16

2y 3z, * = 21

Зу Then putting the 1st and 2d values equal, and the 2d and 3d values equal, give 9 y Z= 16

2y 3z,
16
2y

3z = 21 · 3y – 42;
In the 1st trans. 9, Z, and 2y, gives y = 7
In the 2d trans 16. 3z and 3y, gives y = 5
Putting these two equal, gives 5 - Z=7

2=;
Trans. 5 and 2z, gives z = = 2,
Hence y = 5 – Z = 3, and x = 9

y - Z=4

[ocr errors]
[ocr errors]

22;

21 ;

2dly. By the 2d method :

From the 1st equa. x = 9 - Y -
This value of x substit. in the 2d and 3d, gives

9 + y + 2z= 16,

9 + 2y + 3z
In the 1st trans. 9 and 2z, gives y = y 2z;
This substit. in the last, gives 23 -- Z=21;

Trans, z and 21, gives 2 – Z.
Hence again y=7–2z = 3, and x = 9 - Y - Z 5 s.

3dly. By

y + 2z

y = 5

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

7,

y + z = 5;
Subtr. the latter from the former, gives z = 2.
Hence y =

- I = 3, and x = 9 - y - %= 4,

x + y + z = 18
2. Given

x + 3y + 2z 38 ; to find x, y, and z.
x + jy + fz = 10

Ans. 2 * 4, y = 6, z = 8.
x + 3y +
3. Given

x + y + zz 20 ; to find x, y, and z.
x + y + z

Ans. x = 1, y = 20, z = 60. 4.

y = 2, X — z = 3, and y Z-l; to find x, y, and z.

Ans. x = 7;y

5 ; 2= 4,

{
{

[ocr errors]

= 27

}

= 16

1.7 Given x

5. Given

2x + 3y + 4z = 34
3.x + 4y + 5z = 46
4x + 5y + 62 = 58,

}

; to find x, y, and z.

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 ist condi:ion x + y 10,
And by the ad

6,
Transp. y in each, gives = 10

Y, and x =

6 + y;
Put these two values equal, gives 6 + y = 10 Y;
Transpos 6 and

y, gives 2y = 4;
Dividing by 2, gives
And hence

6+ y = 8,

[ocr errors]

Y = 2.

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.

z + 10

= 60;

[ocr errors]

QUEST. 2. Divide 1001. among A, B, C, so that A may have 201. more than B, and B 101, more than c.

Let x = A's share, y = B's, and z = c's.
Then x + y + z = 100,

x = y + 20,

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

100; By transposing 40, gives

32 And dividing by 3. gives

20. Hence y = 2 + 10 = 30, and x = y + 20 = 50.

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 8.x,

and x + y = 500;
Transposing y, gives x = 500 - Y;
This substituted in the Ist, gives 7y = 4000 – 8y;
By transposing 8y, it is 15y = 4000;
By dividing by 15, it gives y = 2663;
And hence r = 500 - y

233;. 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 fx – fx= 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 h; but if one be added to the denominator, its value will be ?

y + 1

Let denoté the fraction.

y

36 +1
Then by the quest.
-= j, and

f
y
The 1st mult. by 2 and y, gives 2x + 2 = y;
The 2d mult. by 3 and y + 1 is 3x = y +1;
The upper taken from the under leaves x 2 1;
By transpos. 2, it gives x = 3.
And hence y = 236 +2 = 8; and the fraction is

QUEST. 6:

1

$

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

x + y

= 30,

and 20x 10y 240 ;
The Ist mult. by 10, gives 10x + 10y = 300;
These two added give 300 = 540;
This div. by 30, gives x= 18, the days worked ;
Hence

= 12, the days idled.

y = 30

QUEST. 7. Out of a cask of wine, which had leaked away 4,
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 fx gallons,
Conseq. there had been taken away 4x + 30 gallons,
Hence for = x + 30 by the question.
Then mult, by 4, gives 2x =x + 120;
And transposing x, gives x = 120 the contents.

CL

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

= 76.

Let x and y denote the two parts.
Then by the question

2 + y 20,

and 3x + 5y Mult. the 1st by 3, gives

3x + 3y Subtr. the latter from the former, gives 24

16; And dividing by 2, gives

y : Hence, from the 1st,

x = 20

[ocr errors]

= 8. y = 12.

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 3= cost of the first sort,
- cost of the second sort ;

But

[ocr errors]

3 ;

But 5 : 2 :: 2x (the whole number of eggs): **;
Hunce jr

price of both soris. ai 5 for 2 pence ;
Then by the question fx + fx
Mult. by 2, gives x + x
And mult by 3. gives 5 r

18;
Also mult. by 5, gives x > 90, the number of eggs

each sort.

- 6;

of

Quest. 10. Two persons, A and B, engage at play. Before they begin, A has 8 guineas, and B has 60. Afie, 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,
And B rises with 60
Theref. by the quest. 80 + x = 180
Transp. 80 and 3r, gives 4x =
And dividing by 4, gives x = 25, the guineas won.

32 ;

100;

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.

Ans. 15 and 5.

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

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 twn 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 B. con

tributes

« PreviousContinue »